2.1: Functions and Limits

In this section we study limits of real-valued functions of a real variable. You studied limits in calculus. However, we will look more carefully at the definition of limit and prove theorems usually not proved in calculus.

A rule \(f\) that assigns to each member of a nonempty set \(D\) a unique member of a set \(Y\) is a {}. We write the relationship between a member \(x\) of \(D\) and the member \(y\) of \(Y\) that \(f\) assigns to \(x\) as

\[ y=f(x). \nonumber \]

The set \(D\) is the {} of \(f\), denoted by \(D_f\). The members of \(Y\) are the possible {} of \(f\). If \(y_0\in Y\) and there is an \(x_0\) in \(D\) such that \(f(x_0)=y_0\) then we say that \(f\) {} or {} the value \(y_0\). The set of values attained by \(f\) is the {} of \(f\). A {} is a function whose domain and range are both subsets of the reals. Although we are concerned only with real-valued functions of a real variable in this section, our definitions are not restricted to this situation. In later sections we will consider situations where the range or domain, or both, are subsets of vector spaces.

Example \(\PageIndex{1}\)

Example 2.1.1 The functions f , g, and h defined on .1; 1/ by f .x/ D x 2 ; g.x/ D sin x; and h.x/ D e x have ranges Œ0; 1/, Œ1; 1, and .0; 1/, respectively. Solution

Add text here.

Example \(\PageIndex{2}\)

Example 2.1.2 The equation Œf .x/2 D x (2.1.1) does not define a function except on the singleton set f0g. If x < 0, no real number satisfies (2.1.1), while if x > 0, two real numbers satisfy (2.1.1). However, the conditions Œf .x/2 D x and f .x/ 0 define a function f on Df D Œ0; 1/ with values f .x/ D p x. Similarly, the conditions Œg.x/2 D x and g.x/ 0 define a function g on Dg D Œ0; 1/ with values g.x/ D p x. The ranges of f and g are Œ0; 1/ and .1; 0, respectively. Solution

Add text here.

It is important to understand that the definition of a function includes the specification of its domain and that there is a difference between f , the name of the function, and f .x/, the value of f at x. However, strict observance of these points leads to annoying verbosity, such as “the function f with domain .1; 1/ and values f .x/ D x.” We will avoid this in two ways: (1) by agreeing that if a function f is introduced without explicitly defining Df , then Df will be understood to consist of all points x for which the rule defining f .x/ makes sense, and (2) by bearing in mind the distinction between f and f .x/, but not emphasizing it when it would be a nuisance to do so. For example, we will write “consider the function f .x/ D p 1 x 2,” rather than “consider the function f defined on Œ1; 1 by f .x/ D p 1 x 2,” or “consider the function g.x/ D 1= sin x,” rather than “consider the function g defined for x ¤ k (k D integer) by g.x/ D 1= sin x.” We will also write f D c (constant) to denote the function f defined by f .x/ D c for all x. Our definition of function is somewhat intuitive, but adequate for our purposes. Moreover, it is the working form of the definition, even if the idea is introduced more rigorously to begin with. For a more precise definition, we first define the Cartesian product X Y of two nonempty sets X and Y to be the set of all ordered pairs .x; y/ such that x 2 X and y 2 Y ; thus, X Y D ˚ .x; y/ ˇ ˇ x 2 X; y

A nonempty subset \(f\) of \(X\times Y\) is a function if no \(x\) in \(X\) occurs more than once as a first member among the elements of \(f\). Put another way, if \((x,y)\) and \((x,y_1)\) are in \(f\), then \(y=y_1\). The set of \(x\)’s that occur as first members of \(f\) is the of \(f\). If \(x\) is in the domain of \(f\), then the unique \(y\) in \(Y\) such that \((x,y)\in f\) is the {}, and we write \(y=f(x)\). The set of all such values, a subset of \(Y\), is the range of \(f\).

Arithmetic Operations on Functions

The essence of the concept of limit for real-valued functions of a real variable is this: If \(L\) is a real number, then \(\lim_{x\to x_0}f(x)=L\) means that the value \(f(x)\) can be made as close to \(L\) as we wish by taking \(x\) sufficiently close to \(x_0\). This is made precise in the following definition.

We emphasize that Definition~ does not involve \(f(x_0)\), or even require that it be defined, since excludes the case where \(x=x_0\).

The next theorem says that a function cannot have more than one limit at a point.

Suppose that holds and let \(\epsilon>0\). From Definition~, there are positive numbers \(\delta_1\) and \(\delta_2\) such that \[ |f(x)-L_i|<\epsilon\mbox{\quad if \quad} 0<|x-x_0|<\delta_i, \quad i=1,2. \nonumber \] If \(\delta=\min(\delta_1,\delta_2)\), then \[\begin{eqnarray*} |L_1-L_2|\ar= |L_1-f(x)+f(x)-L_2|\\ \ar \le|L_1-f(x)|+|f(x)-L_2|<2\epsilon \mbox{\quad if \quad} 0<|x-x_0|<\delta. \end{eqnarray*} \nonumber \] We have now established an inequality that does not depend on \(x\); that is, \[ |L_1-L_2|<2\epsilon. \nonumber \] Since this holds for any positive \(\epsilon\), \(L_1=L_2\).

Definition~ is not changed by replacing with \[\begin{equation}\label{eq:2.1.8} |f(x)-L|<K\epsilon, \end{equation} \] where \(K\) is a positive constant, because if either of or can be made to hold for any \(\epsilon>0\) by making \(|x-x_0|\) sufficiently small and positive, then so can the other (Exercise~). This may seem to be a minor point, but it is often convenient to work with rather than , as we will see in the proof of the following theorem.

From and Definition~, if \(\epsilon>0\), there is a \(\delta_1>0\) such that \[\begin{equation}\label{eq:2.1.14} |f(x)-L_1|<\epsilon \end{equation} \] if \(0<|x-x_0|<\delta_1\), and a \(\delta_2>0\) such that \[\begin{equation}\label{eq:2.1.15} |g(x)-L_2|<\epsilon \end{equation} \] if \(0<|x-x_0|<\delta_2\). Suppose that \[\begin{equation}\label{eq:2.1.16} 0<|x-x_0|<\delta=\min (\delta_1,\delta_2), \end{equation} \] so that and both hold. Then \[\begin{eqnarray*} |(f\pm g)(x)-(L_1\pm L_2)|\ar= |(f(x)-L_1)\pm (g(x)-L_2)|\\ \ar \le|f(x)-L_1|+|g(x)-L_2|<2\epsilon, \end{eqnarray*} \nonumber \] which proves and .

To prove , we assume and write \[\begin{eqnarray*} |(fg)(x)-L_1L_2|\ar= |f(x)g(x)-L_1L_2|\\[.5\jot] \ar= |f(x)(g(x)-L_2)+L_2(f(x)-L_1)|\\[.5\jot] \ar \le|f(x)||g(x)-L_2|+|L_2||f(x)-L_1|\\[.5\jot] \ar \le(|f(x)|+|L_2|)\epsilon\mbox{\quad (from \eqref{eq:2.1.14} and \eqref{eq:2.1.15})}\\[.5\jot] \ar \le(|f(x)-L_1|+|L_1|+|L_2|)\epsilon\\[.5\jot] \ar \le(\epsilon+|L_1|+|L_2|)\epsilon\mbox{\quad from \eqref{eq:2.1.14}}\\[.5\jot] \ar \le (1+|L_1|+|L_2|)\epsilon \end{eqnarray*} \nonumber \] if \(\epsilon<1\) and \(x\) satisfies . This proves .

To prove , we first observe that if \(L_2\ne0\), there is a \(\delta_3>0\) such that \[ |g(x)-L_2|<\frac{|L_2|}{2}, \nonumber \] so \[\begin{equation} \label{eq:2.1.17} |g(x)|>\frac{|L_2|}{2} \end{equation} \] if \[ 0<|x-x_0|<\delta_3. \nonumber \] To see this, let \(L=L_2\) and \(\epsilon=|L_2|/2\) in . Now suppose that \[ 0<|x-x_0|<\min (\delta_1,\delta_2,\delta_3), \nonumber \] so that , , and all hold. Then

This proves .

Successive applications of the various parts of Theorem~ permit us to find limits without the \(\epsilon\)–\(\delta\) arguments required by Definition~.

-.2em The function \[ f(x)=2x\,\sin\sqrt{x} \nonumber \]

satisfies the inequality \[ |f(x)|<\epsilon \nonumber \] if \(0<x<\delta=\epsilon/2\). However, this does not mean that \(\lim_{x \to 0} f(x)=0\), since \(f\) is not defined for negative \(x\), as it must be to satisfy the conditions of Definition~ with \(x_0=0\) and \(L=0\). The function \[ g(x)=x+\frac{|x|}{ x},\quad x\ne0, \nonumber \] can be rewritten as \[ g(x)=\left\{\casespace\begin{array}{ll} x+1,&x>0,\\ x-1,&x<0;\end{array}\right. \nonumber \] hence, every open interval containing \(x_0=0\) also contains points \(x_1\) and \(x_2\) such that \(|g(x_1)-g(x_2)|\) is as close to \(2\) as we please. Therefore, \(\lim_{x\to x_0} g(x)\) does not exist (Exercise~).

Although \(f(x)\) and \(g(x)\) do not approach limits as \(x\) approaches zero, they each exhibit a definite sort of limiting behavior for small positive values of \(x\), as does \(g(x)\) for small negative values of \(x\). The kind of behavior we have in mind is defined precisely as follows.

6pt

12pt

Figure~ shows the graph of a function that has distinct left- and right-hand limits at a point \(x_0\).

Left- and right-hand limits are also called {}. We will often simplify the notation by writing \[ \lim_{x\to x_0-} f(x)=f(x_0-)\mbox{\quad and \quad}\lim_{x\to x_0+} f(x)=f(x_0+). \nonumber \]

The following theorem states the connection between limits and one-sided limits. We leave the proof to you (Exercise~).

With only minor modifications of their proofs (replacing the inequality \(0<|x-x_0|<\delta\) by \(x_0-\delta<x<x_0\) or \(x_0<x<x_0+ \delta\)), it can be shown that the assertions of Theorems~ and remain valid if $\lim_{x\to x_0}$'' is replaced by\(\lim_{x\to x_0-}\)’‘or ``\(\lim_{x\to x_0+}\)’’ throughout (Exercise~).

Limits and one-sided limits have to do with the behavior of a function \(f\) near a limit point of \(D_f\). It is equally reasonable to study \(f\) for large positive values of \(x\) if \(D_f\) is unbounded above or for large negative values of \(x\) if \(D_f\) is unbounded below.

Figure~ provides an illustration of the situation described in Definition~.

6pt

12pt

We leave it to you to define the statement ``\(\lim_{x\to-\infty} f(x)=L\)’’ (Exercise~) and to show that Theorems~ and remain valid if \(x_0\) is replaced throughout by \(\infty\) or \(-\infty\) (Exercise~).

We will sometimes denote \(\lim_{x\to\infty}f(x)\) and \(\lim_{x \to-\infty}f(x)\) by \(f(\infty)\) and \(f(-\infty)\), respectively.

-.3em The functions \[ f(x)=\frac{1}{ x},\quad g(x)=\frac{1}{ x^2},\quad p(x)=\sin\frac{1}{ x}, \nonumber \] and \[ q(x)=\frac{1}{ x^2}\sin \frac{1}{ x} \nonumber \] do not have limits, or even one-sided limits, at \(x_0=0\). They fail to have limits in different ways:

The kind of behavior exhibited by \(f\) and \(g\) near \(x_0=0\) is sufficiently common and simple to lead us to define {}.

Throughout this book, “\(\lim_{x\to x_0} f(x)\) exists” will mean that

\[\lim_{x\to x_0} f(x)=L,\quad where L is {\it finite\/}.} \nonumber \]

To leave open the possibility that \(L=\pm \infty\), we will say that \[ \lim_{x\to x_0} f(x)\quad\mbox{\it exists in the extended reals.\/} \nonumber \] This convention also applies to one-sided limits and limits as \(x\) approaches \(\pm\infty\).

We mentioned earlier that Theorems~ and remain valid if $\lim_{x\to x_0}$'' is replaced by\(\lim_{x\to x_0-}\)’‘or ``\(\lim_{x\to x_0+}\).’’ They are also valid with \(x_0\) replaced by \(\pm\infty\). Moreover, the counterparts of , , and in all these versions of Theorem~ remain valid if either or both of \(L_1\) and \(L_2\) are infinite, provided that their right sides are not indeterminate (Exercises~ and ). Equation and its counterparts remain valid if \(L_1/L_2\) is not indeterminate and \(L_2\ne0\) (Exercise~).

-.3em A function \(f\) is {} on an interval \(I\) if \[\begin{equation}\label{eq:2.1.18} f(x_1)\le f(x_2)\quad\mbox{whenever $x_1$ and $x_2$ are in $I$ and $x_1 <x_2$}, \end{equation} \] or {} on \(I\) if \[\begin{equation}\label{eq:2.1.19} f(x_1)\ge f(x_2)\quad\mbox{whenever $x_1$ and $x_2$ are in $I$ and $x_1 <x_2$}. \end{equation} \] In either case, \(f\) is on \(I\). If \(\le\) can be replaced by \(<\) in , \(f\) is {} on \(I\). If \(\ge\) can be replaced by \(>\) in , \(f\) is {} on \(I\). In either of these two cases, \(f\) is {} on \(I\).

6pt

12pt

In the proof of the following theorem, we assume that you have formulated the definitions called for in Exercise~.

We first show that \(f(a+)=\alpha\). If

\(M>\alpha\), there is an \(x_0\) in \((a,b)\) such that \(f(x_0)<M\). Since \(f\) is nondecreasing, \(f(x)<M\) if \(a<x<x_0\). Therefore, if \(\alpha=-\infty\), then \(f(a+)=-\infty\). If \(\alpha>-\infty\), let \(M=\alpha+\epsilon\), where \(\epsilon>0\). Then \(\alpha\le f(x)<\alpha+\epsilon\), so \[\begin{equation} \label{eq:2.1.20} |f(x)-\alpha|<\epsilon\mbox{\quad if \quad} a<x<x_0. \end{equation} \] If \(a=-\infty\), this implies that \(f(-\infty)=\alpha\). If \(a>-\infty\), let \(\delta=x_0-a\). Then is equivalent to \[ |f(x)-\alpha|<\epsilon\mbox{\quad if \quad} a<x<a+\delta, \nonumber \] which implies that \(f(a+)=\alpha\).

We now show that \(f(b-)=\beta\). If \(M<\beta\), there is an \(x_0\) in \((a,b)\) such that \(f(x_0)>M\). Since \(f\) is nondecreasing, \(f(x)>M\) if \(x_0<x<b\). Therefore, if \(\beta=\infty\), then \(f(b-)=\infty\). If \(\beta<\infty\), let \(M=\beta-\epsilon\), where \(\epsilon>0\). Then \(\beta-\epsilon< f(x)\le\beta\), so \[\begin{equation} \label{eq:2.1.21} |f(x)-\beta|<\epsilon\mbox{\quad if \quad} x_0<x<b. \end{equation} \] If \(b=\infty\), this implies that \(f(\infty)=\beta\). If \(b<\infty\), let \(\delta=b-x_0\). Then is equivalent to \[ |f(x)-\beta|<\epsilon\mbox{\quad if \quad} b-\delta<x<b, \nonumber \] which implies that \(f(b-)=\beta\).

The proof is similar to the proof of

(Exercise~).

Suppose that \(f\) is nondecreasing. Applying

to \(f\) on \((a,x_0)\) and \((x_0,b)\) separately shows that \[ f(x_0-)=\sup_{a<x<x_0}f(x)\mbox{\quad and \quad} f(x_0+)=\inf_{x_0<x<b}f(x). \nonumber \]

However, if \(x_1<x_0<x_2\), then \[ f(x_1)\le f(x_0)\le f(x_2); \nonumber \] hence, \[ f(x_0-)\le f(x_0)\le f(x_0+). \nonumber \]

We leave the case where \(f\) is nonincreasing to you (Exercise~).

We now introduce some concepts related to limits. We leave the study of these concepts mainly to the exercises.

We say that \(f\) is {} on a set \(S\) if there is a constant \(M<\infty\) such that \(|f(x)|\le M\) for all \(x\) in \(S\).

Since \(S_f(x_1;x_0)\) is the supremum of \(\set{f(t)}{x_1<t<x_0}\), there is an \(\overline x\) in \([x_1,x_0)\) such that \[ f(\overline x)>S_f(x_1;x_0)-\epsilon/2. \nonumber \] This and imply that \(f(\overline x)>\beta-\epsilon\). Since \(\overline x\) is in \([a_1,x_0)\), this proves

.

Now we show that there cannot be more than one real number with properties and . Suppose that \(\beta_1<\beta_2\) and \(\beta_2\) has property ; thus, if \(\epsilon>0\) and \(a_1\) is in \([a,x_0)\), there is an \(\overline x\) in \([a_1,x_0)\) such that \(f(\overline x)>\beta_2-\epsilon\). Letting \(\epsilon=\beta_2-\beta_1\), we see that there is an \(\overline x\) in \([a_1,b)\) such that \[ f(\overline x)>\beta_2-(\beta_2-\beta_1)=\beta_1, \nonumber \] so \(\beta_1\) cannot have property . Therefore, there cannot be more than one real number that satisfies both and

.

The proof of the following theorem is similar to this (Exercise~).

In this section we study continuous functions of a real variable. We will prove some important theorems about continuous functions that, although intuitively plausible, are beyond the scope of the elementary calculus course. They are accessible now because of our better understanding of the real number system, especially of those properties that stem from the completeness axiom.

The definitions of \[ f(x_0-)=\lim_{x\to x_0-}f(x),\quad f(x_0+)=\lim_{x\to x_0+}f(x),\mbox{\quad and \quad} \lim_{x\to x_0} f(x) \nonumber \] do not involve \(f(x_0)\) or even require that it be defined. However, the case where \(f(x_0)\) is defined and equal to one or more of these quantities is important.

The following theorem provides a method for determining whether these definitions are satisfied. The proof, which we leave to you (Exercise~), rests on Definitions~, , and .

From Definition~ and Theorem~, \(f\) is

continuous at \(x_0\) if and only if \[ f(x_0-)=f(x_0+)=f(x_0) \nonumber \] or, equivalently, if and only if it is continuous from the right and left at \(x_0\) (Exercise~).

6pt

12pt

The function \(f\) defined in Example~ (see also Figure~) is continuous on \([0,1)\) and \([1,2]\), but not on any open interval containing 1. The discontinuity of \(f\) there is of the simplest kind, described in the following definition.

6pt

12pt

The reason for the adjective ``jump’’ can be seen in Figures~ and , where the graphs exhibit a definite jump at each point of discontinuity. The next example shows that not all discontinuities are of this kind.

Theorems~ and imply the next theorem (Exercise~).

Let \(f\) be defined on a deleted neighborhood of \(x_0\) and discontinuous (perhaps even undefined) at \(x_0\). We say that \(f\) has a at \(x_0\) if \(\lim_{x\to x_0}f(x)\) exists. In this case, the function \[ g(x)=\left\{\casespace\begin{array}{ll} f(x)&\mbox{ if } x\in D_f\mbox{ and } x\ne x_0,\\[2\jot] \dst{\lim_{x\to x_0}} f(x)&\mbox{ if } x=x_0,\end{array}\right. \nonumber \] is continuous at \(x_0\).

We have seen that the investigation of limits and continuity can be simplified by regarding a given function as the result of addition, subtraction, multiplication, and division of simpler functions. Another operation useful in this connection is {} of functions; that is, substitution of one function into another.

The next theorem says that the composition of continuous functions is continuous.

Suppose that \(\epsilon>0\). Since \(g(x_0)\) is an interior point of \(D_f\) and \(f\) is continuous at \(g(x_0)\), there is a \(\delta_1>0\) such that \(f(t)\) is defined and \[\begin{equation}\label{eq:2.2.4} |f(t)-f(g(x_0))|<\epsilon\mbox{\quad if \quad} |t-g(x_0)|< \delta_1. \end{equation} \] Since \(g\) is continuous at \(x_0\), there is a \(\delta>0\) such that \(g(x)\) is defined and \[\begin{equation}\label{eq:2.2.5} |g(x)-g(x_0)|<\delta_1\mbox{\quad if \quad}|x-x_0|<\delta. \end{equation} \] Now and imply that \[ |f(g(x))-f(g(x_0))|<\epsilon\mbox{\quad if \quad}|x-x_0|<\delta. \nonumber \] Therefore, \(f\circ g\) is continuous at \(x_0\).

See Exercise~ for a related result concerning limits.

A function \(f\) is {} on a set \(S\) if there is a real number \(m\) such that \[ f(x)\ge m \mbox{ for all } x\in S. \nonumber \] In this case, the set \[ V=\set{f(x)}{x\in S} \nonumber \] has an infimum \(\alpha\), and we write \[ \alpha=\inf_{x\in S}f(x). \nonumber \] If there is a point \(x_1\) in \(S\) such that \(f(x_1)=\alpha\), we say that \(\alpha\) is the {}, and write \[ \alpha=\min_{x\in S}f(x). \nonumber \] Similarly, \(f\) is {} if there is a real number \(M\) such that \(f(x)\le M\) for all \(x\) in \(S\). In this case, \(V\) has a supremum \(\beta\), and we write \[ \beta=\sup_{x\in S}f(x). \nonumber \] If there is a point \(x_2\) in \(S\) such that \(f(x_2)=\beta\), we say that \(\beta\) is the {}, and write \[ \beta=\max_{x\in S}f(x). \nonumber \] If \(f\) is bounded above and below on a set \(S\), we say that \(f\) is {} on \(S\).

Figure~ illustrates the geometric meaning of these definitions for a function \(f\) bounded on an interval \(S=[a,b]\). The graph of \(f\) lies in the strip bounded by the lines \(y=M\) and \(y=m\), where \(M\) is any upper bound and \(m\) is any lower bound for \(f\) on \([a,b]\). The narrowest strip containing the graph is the one bounded above by \(y=\beta= \sup_{a\le x\le b}f(x)\) and below by \(y=\alpha=\inf_{a\le x\le b}f(x)\).

6pt

12pt

6pt

12pt

Suppose that \(t\in [a,b]\). Since \(f\) is continuous at \(t\), there is an open interval \(I_t\) containing \(t\) such that \[\begin{equation}\label{eq:2.2.7} |f(x)-f(t)|<1 \mbox{\quad if \quad}\ x\in I_t\cap [a,b]. \end{equation} \] (To see this, set \(\epsilon=1\) in , Theorem~.) The collection \({\mathcal H}=\set{I_t}{a\le t\le b}\) is an open covering of \([a,b]\). Since \([a,b]\) is compact, the Heine–Borel theorem implies that there are finitely many points \(t_1\), \(t_2\), , \(t_n\) such that the intervals \(I_{t_1}\), \(I_{t_2}\), , \(I_{t_n}\) cover \([a,b]\). According to with \(t=t_i\), \[ |f(x)-f(t_i)|<1\mbox{\quad if \quad}\ x\in I_{t_i}\cap [a,b]. \nonumber \] Therefore, \[\begin{equation}\label{eq:2.2.8} \begin{array}{rcl} |f(x)|\ar =|(f(x)-f(t_i))+f(t_i)|\le|f(x)-f(t_i)|+|f(t_i)|\\[2\jot] \ar\le 1+|f(t_i)|\mbox{\quad if \quad}\ x\in I_{t_i}\cap[a,b]. \end{array} \end{equation} \] Let \[ M=1+\max_{1\le i\le n}|f(t_i)|. \nonumber \] Since \([a,b]\subset\bigcup^n_{i=1}\left(I_{t_i}\cap [a,b]\right)\), implies that \(|f(x)|\le M\) if \(x\in [a,b]\).

This proof illustrates the utility of the Heine–Borel theorem, which allows us to choose \(M\) as the largest of a {} set of numbers.

Theorem~ and the completeness of the reals imply that

if \(f\) is continuous on a finite closed interval \([a,b]\), then \(f\) has an infimum and a supremum on \([a,b]\). The next theorem shows that \(f\) actually assumes these values at some points in \([a,b]\).

We show that \(x_1\) exists and leave it to you to show that \(x_2\) exists (Exercise~).

Suppose that there is no \(x_1\) in \([a,b]\) such that \(f(x_1)=\alpha\). Then \(f(x)>\alpha\) for all \(x\in[a,b]\). We will show that this leads to a contradiction.

Suppose that \(t\in[a,b]\). Then \(f(t)>\alpha\), so \[ f(t)>\frac{f(t)+\alpha}{2}>\alpha. \nonumber \]

Since \(f\) is continuous at \(t\), there is an open interval \(I_t\) about \(t\) such that \[\begin{equation}\label{eq:2.2.9} f(x)>\frac{f(t)+\alpha}{2}\mbox{\quad if \quad} x\in I_t\cap [a,b] \end{equation} \] (Exercise~). The collection \({\mathcal H}=\set{I_t}{a\le t\le b}\) is an open covering of \([a,b]\). Since \([a,b]\) is compact, the Heine–Borel theorem implies that there are finitely many points \(t_1\), \(t_2\), , \(t_n\) such that the intervals \(I_{t_1}\), \(I_{t_2}\), , \(I_{t_n}\) cover \([a,b]\). Define \[ \alpha_1=\min_{1\le i\le n}\frac{f(t_i)+\alpha}{2}. \nonumber \] Then, since \([a,b]\subset\bigcup^n_{i=1} (I_{t_i}\cap [a,b])\), implies that \[ f(t)>\alpha_1,\quad a\le t\le b. \nonumber \] But \(\alpha_1>\alpha\), so this contradicts the definition of \(\alpha\). Therefore, \(f(x_1)=\alpha\) for some \(x_1\) in \([a,b]\).

The next theorem shows that if \(f\) is continuous on a finite closed interval \([a,b]\), then \(f\) assumes every value between \(f(a)\) and \(f(b)\) as \(x\) varies from \(a\) to \(b\) (Figure_{, page}).

6pt

12pt

Suppose that \(f(a)<\mu<f(b)\). The set \[ S=\set{x}{a\le x\le b \mbox{\quad and }\ f(x)\le\mu} \nonumber \] is bounded and nonempty. Let \(c=\sup S\). We will show that \(f(c)=\mu\). If \(f(c)>\mu\), then \(c>a\) and, since \(f\) is continuous at \(c\), there is an \(\epsilon>0\) such that \(f(x)>\mu\) if \(c-\epsilon<x\le c\) (Exercise~). Therefore, \(c-\epsilon\) is an upper bound for \(S\), which contradicts the definition of \(c\) as the supremum of \(S\). If \(f(c) <\mu\), then \(c<b\) and there is an \(\epsilon>0\) such that \(f(x)<\mu\) for \(c\le x<c+\epsilon\), so \(c\) is not an upper bound for \(S\). This is also a contradiction. Therefore, \(f(c)=\mu\).

The proof for the case where \(f(b)<\mu<f(a)\) can be obtained by applying this result to \(-f\).

Theorem~ and Definition~ imply that a

function \(f\) is continuous on a subset \(S\) of its domain if for each \(\epsilon>0\) and each \(x_0\) in \(S\), there is a \(\delta>0\), {}, such that \[ |f(x)-f(x_0)|<\epsilon\mbox{\quad if \quad} |x-x_0|<\delta \mbox{\quad and \quad} x\in D_f. \nonumber \]

The next definition introduces another kind of continuity on a set \(S\).

We emphasize that in this definition \(\delta\) depends only on \(\epsilon\) and \(S\) and not on the particular choice of \(x\) and \(x'\), provided that they are both in \(S\).

Often a concept is clarified by considering its negation: a function \(f\) is {} uniformly continuous on \(S\) if there is an \(\epsilon_0>0\) such that if \(\delta\) is any positive number, there are points \(x\) and \(x'\) in \(S\) such that \[ |x-x'|<\delta\mbox{\quad but\quad} |f(x)-f(x')|\ge\epsilon_0. \nonumber \]

Examples~ and show that a function may be continuous but not uniformly continuous on an interval. The next theorem shows that this cannot happen if the interval is closed and bounded, and therefore compact.

Suppose that \(\epsilon>0\). Since \(f\) is continuous on \([a,b]\), for each \(t\) in \([a,b]\) there is a positive number \(\delta_{t}\) such that \[\begin{equation}\label{eq:2.2.10} |f(x)-f(t)|<\frac{\epsilon}{2} \mbox{\quad if \quad} |x-t|<2\delta_{t} \mbox{\quad and \quad} x\in[a,b]. \end{equation} \] If \(I_{t}=(t-\delta_{t },t+\delta_{t})\), the collection \[ {\mathcal H}=\set{I_{t}}{t\in [a,b]} \nonumber \] is an open covering of \([a,b]\). Since \([a,b]\) is compact, the Heine–Borel theorem implies that there are finitely many points \(t_1\), \(t_2\), , \(t_n\) in \([a,b]\) such that \(I_{t_1}\), \(I_{t_2}\), , \(I_{t_n}\) cover \([a,b]\). Now define \[\begin{equation}\label{eq:2.2.11} \delta=\min\{\delta_{t_1},\delta_{t_2}, \dots,\delta_{t_n}\}. \end{equation} \] We will show that if \[\begin{equation} \label{eq:2.2.12} |x-x'|<\delta \mbox{\quad and \quad}x,x'\in [a,b], \end{equation} \] then \(|f(x)-f(x')|<\epsilon\).

From the triangle inequality, \[\begin{equation} \label{eq:2.2.13} \begin{array}{rcl} |f(x)-f(x')|\ar = |\left(f(x)-f(t_r)\right)+\left(f(t_r)-f(x')\right)|\\ \ar\le |f(x)-f(t_r)|+|f(t_r)-f(x')|. \end{array} \end{equation} \] Since \(I_{t_1}\), \(I_{t_2}\), , \(I_{t_n}\) cover \([a,b]\), \(x\) must be in one of these intervals. Suppose that \(x\in I_{t_r}\); that is, \[\begin{equation} \label{eq:2.2.14} |x-t_r|<\delta_{t_r}. \end{equation} \] From with \(t=t_r\), \[\begin{equation} \label{eq:2.2.15} |f(x)-f(t_r)|<\frac{\epsilon}{2}. \end{equation} \] From , , and the triangle inquality, \[ |x'-t_r|=|(x'-x)+(x-t_r)|\le |x'-x|+|x-t_r|<\delta+\delta_{t_r}\le2\delta_{t_r}. \nonumber \] Therefore, with \(t=t_r\) and \(x\) replaced by \(x'\) implies that \[ |f(x')-f(t_r)|<\frac{\epsilon}{2}. \nonumber \] This, , and imply that \(|f(x)-f(x')|<\epsilon\).

This proof again shows the utility of the Heine–Borel theorem, which allowed us to define \(\delta\) in as the smallest of a {} set of positive numbers, so that \(\delta\) is sure to be positive. (An infinite set of positive numbers may fail to have a smallest positive member; for example, consider the open interval \((0,1)\).)

Applied to Example~, Corollary~ implies that the function \(g(x)=\cos1/x\) is uniformly continuous on \([\rho,1]\) if \(0<\rho<1\).

Theorem~ implies that if \(f\) is monotonic on an interval \(I\), then \(f\) is either continuous or has a jump discontinuity at each \(x_0\) in \(I\). This and Theorem~ provide the key to the proof of the following theorem.

We assume that \(f\) is nondecreasing, and leave the case where \(f\) is nonincreasing to you (Exercise_{). Theorem}

implies that the set \(\widetilde R_f=\set{f(x)}{x\in(a,b)}\) is a subset of the open interval \((f(a+),f(b-))\). Therefore, \[\begin{equation} \label{eq:2.2.16} R_f=\{f(a)\}\cup\widetilde R_f\cup\{f(b)\}\subset\{f(a)\}\cup(f(a+),f(b-))\cup\{f(b)\}. \end{equation} \] Now suppose that \(f\) is continuous on \([a,b]\). Then \(f(a)=f(a+)\), \(f(b-)=f(b)\), so implies that \(R_f\subset[f(a),f(b)]\). If \(f(a)<\mu<f(b)\), then Theorem~ implies that \(\mu=f(x)\) for some \(x\) in \((a,b)\). Hence, \(R_f=[f(a),f(b)]\).

For the converse, suppose that \(R_f=[f(a),f(b)]\). Since \(f(a)\le f(a+)\) and \(f(b-)\le f(b)\), implies that \(f(a)=f(a+)\) and \(f(b-)=f(b)\). We know from Theorem~

that if \(f\) is nondecreasing and \(a<x_0<b\), then \[ f(x_0-)\le f(x_0)\le f(x_0+). \nonumber \] If either of these inequalities is strict, \(R_f\) cannot be an interval. Since this contradicts our assumption, \(f(x_0-)=f(x_0)=f(x_0+)\). Therefore, \(f\) is continuous at \(x_0\) (Exercise~). We can now conclude that \(f\) is continuous on \([a,b]\).

Theorem~ implies the following theorem.

We first show that there is a function \(g\) satisfying and . Since \(f\) is continuous, Theorem~ implies that for each \(y_0\) in \([c,d]\) there is an \(x_0\) in \([a,b]\) such that \[\begin{equation}\label{eq:2.2.19} f(x_0)=y_0, \end{equation} \]

and, since \(f\) is increasing, there is only one such \(x_0\). Define \[\begin{equation}\label{eq:2.2.20} g(y_0)=x_0. \end{equation} \] The definition of \(x_0\) is illustrated in Figure~: with \([c,d]\) drawn on the \(y\)-axis, find the intersection of the line \(y=y_0\) with the curve \(y=f(x)\) and drop a vertical from the intersection to the \(x\)-axis to find \(x_0\).

6pt

12pt

Substituting into yields \[ f(g(y_0))=y_0, \nonumber \] and substituting into yields \[ g(f(x_0))=x_0. \nonumber \] Dropping the subscripts in these two equations yields and .

The uniqueness of \(g\) follows from our assumption that \(f\) is increasing, and therefore only one value of \(x_0\) can satisfy for each \(y_0\).

To see that \(g\) is increasing, suppose that \(y_1<y_2\) and let \(x_1\) and \(x_2\) be the points in \([a,b]\) such that \(f(x_1)=y_1\) and \(f(x_2)=y_2\). Since \(f\) is increasing, \(x_1<x_2\). Therefore, \[ g(y_1)=x_1<x_2=g(y_2), \nonumber \] so \(g\) is increasing. Since \(R_g=\set{g(y)}{y\in[c,d]}\) is the interval \([g(c),g(d)]=[a,b]\), Theorem~ with \(f\) and \([a,b]\) replaced by \(g\) and \([c,d]\) implies that \(g\) is continuous on \([c,d]\).

The function \(g\) of Theorem~ is the {} of \(f\), denoted by \(f^{-1}\). Since and are symmetric in \(f\) and \(g\), we can also regard \(f\) as the inverse of \(g\), and denote it by \(g^{-1}\).

In calculus you studied differentiation, emphasizing rules for calculating derivatives. Here we consider the theoretical properties of differentiable functions. In doing this, we assume that you know how to differentiate elementary functions such as \(x^n\), \(e^x\), and \(\sin x\), and we will use such functions in examples.

If \(f\) is defined on an open set \(S\), we say that \(f\) is {} \(S\) if \(f\) is differentiable at every point of \(S\). If \(f\) is differentiable on \(S\), then \(f'\) is a function on \(S\). We say that \(f\) is {} on \(S\) if \(f'\) is continuous on \(S\). If \(f\) is differentiable on a neighborhood of \(x_0\), it is reasonable to ask if \(f'\) is differentiable at \(x_0\). If so, we denote the derivative of \(f'\) at \(x_0\) by \(f''(x_0)\). This is the {}, and it is also denoted by \(f^{(2)}(x_0)\). Continuing inductively, if \(f^{(n-1)}\) is defined on a neighborhood of \(x_0\), then the \(n\)th {}, denoted by \(f^{(n)}(x_0)\), is the derivative of \(f^{(n-1)}\) at \(x_0\). For convenience we define the {} of \(f\) to be \(f\) itself; thus \[ f^{(0)}=f. \nonumber \]

We assume that you are familiar with the other standard notations for derivatives; for example, \[ f^{(2)}=f'',\quad f^{(3)}=f''', \nonumber \]

and so on, and \[ \frac{d^nf}{ dx^n}=f^{(n)}. \nonumber \]

To derive differentiation formulas for elementary functions such as \(\sin x\), \(\cos x\), and \(e^x\) directly from Definition~ requires estimates based on the properties of these functions. Since this is done in calculus, we will not repeat it here.

If \(f(x)\) is the position of a particle at time \(x\ne x_0\), the difference quotient \[ \frac{f(x)-f(x_0)}{ x-x_0} \nonumber \] is the average velocity of the particle between times \(x_0\) and \(x\). As \(x\) approaches \(x_0\), the average applies to shorter and shorter intervals. Therefore, it makes sense to regard the limit , if it exists, as the particle’s {}. This interpretation may be useful even if \(x\) is not time, so we often regard \(f'(x_0)\) as the {}, regardless of the specific nature of the variable \(x\). The derivative also has a geometric interpretation. The equation of the line through two points \((x_0,f(x_0))\) and \((x_1,f(x_1))\) on the curve \(y=f(x)\) (Figure~) is \[ y=f(x_0)+\frac{f(x_1)-f(x_0)}{ x_1-x_0} (x-x_0). \nonumber \]

Varying \(x_1\) generates lines through \((x_0,f(x_0))\) that rotate into the line \[\begin{equation}\label{eq:2.3.2} y=f(x_0)+f'(x_0)(x-x_0) \end{equation} \]

as \(x_1\) approaches \(x_0\). This is the {} to the curve \(y=f(x)\) at the point \((x_0,f(x_0))\). Figure~ depicts the situation for various values of \(x_1\).

6pt

12pt

Here is a less intuitive definition of the tangent line: If the function \[ T(x)=f(x_0)+m(x-x_0) \nonumber \] approximates \(f\) so well near \(x_0\) that \[ \lim_{x\to x_0}\frac{f(x)-T(x)}{ x-x_0}=0, \nonumber \] we say that the line \(y=T(x)\) is {}.

This tangent line exists if and only if \(f'(x_0)\) exists, in which case \(m\) is uniquely determined by \(m=f'(x_0)\) (Exercise~). Thus, is the equation of the tangent line.

We will use the following lemma to study differentiable functions.

Define \[\begin{equation} \label{eq:2.3.4} E(x)=\left\{\casespace\begin{array}{ll} \dst\frac{f(x)-f(x_0)}{ x-x_0}- f'(x_0),&x\in D_f\mbox{ and }x\ne x_0,\\[2\jot] 0,&x=x_0. \end{array}\right. \end{equation} \] Solving for \(f(x)\) yields if \(x\ne x_0\), and is obvious if \(x=x_0\). Definition~ implies that \(\lim_{x\to x_0}E(x)=0\). We defined \(E(x_0)=0\) to make \(E\) continuous at \(x_0\).

Since the right side of is continuous at \(x_0\), so is the left. This yields the following theorem.

The converse of this theorem is false, since a function may be continuous at a point without being differentiable at the point.

The following theorem should be familiar from calculus.

The proof is accomplished by forming the appropriate difference quotients and applying Definition~ and Theorem~. We will prove

and leave the rest to you (Exercises~, , and ).

The trick is to add and subtract the right quantity in the numerator of the difference quotient for \((fg)'(x_0)\); thus, \[\begin{eqnarray*} \frac{f(x)g(x)-f(x_0)g(x_0)}{ x-x_0}\ar= \frac{f(x)g(x)-f(x_0)g(x)+f(x_0)g(x)-f(x_0)g(x_0)}{ x-x_0}\\ \ar=\frac{f(x)-f(x_0)}{ x-x_0} g(x)+f(x_0)\frac{g(x)-g(x_0)}{ x-x_0}. \end{eqnarray*} \nonumber \] The difference quotients on the right approach \(f'(x_0)\) and \(g'(x_0)\) as \(x\) approaches \(x_0\), and \(\lim_{x\to x_0}g(x)=g(x_0)\) (Theorem~). This proves

.

Here is the rule for differentiating a composite function.

Since \(f\) is differentiable at \(g(x_0)\), Lemma~ implies that \[ f(t)-f(g(x_0))=[f'(g(x_0))+E(t)][t-g(x_0)], \nonumber \] where \[\begin{equation}\label{eq:2.3.9} \lim_{t\to g(x_0)} E(t)=E(g(x_0))=0. \end{equation} \] Letting \(t=g(x)\) yields \[ f(g(x))-f(g(x_0))=[f'(g(x_0))+E(g(x))][g(x)-g(x_0)]. \nonumber \] Since \(h(x)=f(g(x))\), this implies that \[\begin{equation}\label{eq:2.3.10} \frac{h(x)-h(x_0)}{ x-x_0}=[f'(g(x_0))+E(g(x))] \frac{g(x)-g(x_0)}{ x-x_0}. \end{equation} \] Since \(g\) is continuous at \(x_0\) (Theorem~), and Theorem~ imply that \[ \lim_{x\to x_0} E(g(x))=E(g(x_0))=0. \nonumber \] Therefore, implies that \[ h'(x_0)=\lim_{x\to x_0}\frac{h(x)-h(x_0)}{ x-x_0}=f'(g(x_0)) g'(x_0), \nonumber \] as stated.

It may seem reasonable to justify the chain rule by writing \[\begin{eqnarray*} \frac{h(x)-h(x_0)}{ x-x_0}\ar=\frac{f(g(x))-f(g(x_0))}{x-x_0}\\ \ar=\frac{f(g(x))-f(g(x_0))}{ g(x)-g(x_0)}\, \frac{g(x)-g(x_0)}{ x-x_0} \end{eqnarray*} \nonumber \] and arguing that \[ \lim_{x\to x_0} \frac{f(g(x))-f(g(x_0))}{ g(x)-g(x_0)}=f'(g(x_0)) \nonumber \]

(because \(\lim_{x\to x_0} g(x)=g(x_0))\) and \[ \lim_{x\to x_0}\frac{g(x)-g(x_0)}{ x-x_0}=g'(x_0). \nonumber \] However, this is not a valid proof (Exercise~).

One-sided limits of difference quotients such as and in Example~ are called {} or {}. That is, if \(f\) is defined on \([x_0,b)\), the {} is defined to be \[ f_+'(x_0)=\lim_{x\to x_0+}\frac{f(x)-f(x_0)}{ x-x_0} \nonumber \] if the limit exists, while if \(f\) is defined on \((a,x_0]\), the {} is defined to be \[ f_-'(x_0)=\lim_{x\to x_0-}\frac{f(x)-f(x_0)}{ x-x_0} \nonumber \] if the limit exists. Theorem~ implies that \(f\) is differentiable at \(x_0\) if and only if \(f_+'(x_0)\) and \(f_-'(x_0)\) exist and are equal, in which case \[ f'(x_0)=f_+'(x_0)=f_-'(x_0). \nonumber \]

In Example~, \(f_+'(0)=1\) and \(f_-'(0)=-1\).

This example shows that there is a difference between a one-sided derivative and a one-sided limit of a derivative, since \(f_+'(0)=0\), but, from , \(f'(0+)=\lim_{x\to 0+}f'(x)\) does not exist. It also shows that a derivative may exist in a neighborhood of a point \(x_0\) (\(=0\) in this case), but be discontinuous at \(x_0\).

Exercise~ justifies the method used in Example~ to compute \(f'(x)\) for \(x\ne0\).

We say that \(f(x_0)\) is a {} of \(f\) if there is a \(\delta>0\) such that \(f(x)-f(x_0)\) does not change sign on \[\begin{equation}\label{eq:2.3.13} (x_0-\delta,x_0+\delta)\cap D_f. \end{equation} \] More specifically, \(f(x_0)\) is a {} of \(f\) if \[\begin{equation}\label{eq:2.3.14} f(x) \le f(x_0) \end{equation} \] or a {} of \(f\) if \[\begin{equation}\label{eq:2.3.15} f(x)\ge f(x_0) \end{equation} \] for all \(x\) in the set . The point \(x_0\) is called a {} of \(f\), or, more specifically, a {} or {} of \(f\).

6pt

12pt

4pt

It is geometrically plausible that if the curve \(y=f(x)\) has a tangent at a local extreme point of \(f\), then the tangent must be horizontal; that is, have zero slope. (For example, in Figure~, see \(x=1\), \(x=3\), and every \(x\) in \((-1,-1/2)\).) The following theorem shows that this must be so. 4pt

4pt

We will show that \(x_0\) is not a local extreme point of \(f\) if \(f'(x_0)\ne0\). From Lemma~, \[\begin{equation}\label{eq:2.3.16} \frac{f(x)-f(x_0)}{ x-x_0}=f'(x_0)+E(x), \end{equation} \] where \(\lim_{x\to x_0} E(x)=0\). Therefore, if \(f'(x_0)\ne0\), there is a \(\delta>0\) such that \[ |E(x)|<|f'(x_0)|\mbox{\quad if\quad} |x-x_0|<\delta, \nonumber \] and the right side of must have the same sign as \(f'(x_0)\) for \(|x-x_0|<\delta\). Since the same is true of the left side, \(f(x)-f(x_0)\) must change sign in every neighborhood of \(x_0\) (since \(x-x_0\) does). Therefore, neither nor can hold for all \(x\) in any interval about \(x_0\). 4pt

If \(f'(x_0)=0\), we say that \(x_0\) is a {} of \(f\). Theorem~ says that every local extreme point of \(f\) at which \(f\) is differentiable is a critical point of \(f\). The converse is false. For example, \(0\) is a critical point of \(f(x)=x^3\), but not a local extreme point. 4pt

The use of Theorem~ for finding local extreme points is covered in calculus, so we will not pursue it here. However, we will use Theorem~ to prove the following fundamental theorem, which says that if a curve \(y=f(x)\) intersects a horizontal line at \(x=a\) and \(x=b\) and has a tangent at \((x,f(x))\) for every \(x\) in \((a,b)\), then there is a point \(c\) in \((a,b)\) such that the tangent to the curve at \((c,f(c))\) is horizontal (Figure~).

6pt

12pt

Since \(f\) is continuous on \([a,b]\), \(f\) attains a maximum and a minimum value on \([a,b]\) (Theorem~). If these two extreme values are the same, then \(f\) is constant on \((a,b)\), so \(f'(x)=0\) for all \(x\) in \((a,b)\). If the extreme values differ, then at least one must be attained at some point \(c\) in the open interval \((a,b)\), and \(f'(c)=0\), by Theorem~.

A derivative may exist on an interval \([a,b]\) without being continuous on \([a,b]\). Nevertheless, an intermediate value theorem similar to Theorem~ applies to derivatives.

Suppose first that \[\begin{equation}\label{eq:2.3.17} f'(a)<\mu<f'(b) \end{equation} \] and define \[ g(x)=f(x)-\mu x. \nonumber \] Then \[\begin{equation}\label{eq:2.3.18} g'(x)=f'(x)-\mu,\quad a\le x\le b, \end{equation} \] and implies that \[\begin{equation}\label{eq:2.3.19} g'(a)<0\mbox{\quad and\quad} g'(b)>0. \end{equation} \] Since \(g\) is continuous on \([a,b]\), \(g\) attains a minimum at some point \(c\) in \([a,b]\). Lemma~ and imply that there is a \(\delta>0\) such that \[ g(x)<g(a),\quad a<x<a+\delta,\mbox{\quad and\quad} g(x)<g(b),\quad b-\delta<x<b \nonumber \]

(Exercise~), and therefore \(c\ne a\) and \(c\ne b\). Hence, \(a<c<b\), and therefore \(g'(c)=0\), by Theorem~. From , \(f'(c)=\mu\).

The proof for the case where \(f'(b)<\mu<f'(a)\) can be obtained by applying this result to \(-f\).

The function \[ h(x)=[g(b)-g(a)]f(x)-[f(b)-f(a)]g(x) \nonumber \] is continuous on \([a,b]\) and differentiable on \((a,b)\), and \[ h(a)=h(b)=g(b)f(a)-f(b)g(a). \nonumber \] Therefore, Rolle’s theorem implies that \(h'(c)=0\) for some \(c\) in \((a,b)\). Since \[ h'(c)=[g(b)-g(a)]f'(c)-[f(b)-f(a)]g'(c), \nonumber \] this implies .

The following special case of Theorem~ is important enough to be stated separately.

Apply Theorem~ with \(g(x)=x\).

Theorem~ implies that the tangent to the curve \(y=f(x)\) at \((c,f(c))\) is parallel to the line connecting the points \((a,f(a))\) and \((b,f(b))\) on the curve (Figure~, page ).

If \(f\) is differentiable on \((a,b)\) and \(x_1\), \(x_2\in (a,b)\) then \(f\) is continuous on the closed interval with endpoints \(x_1\) and \(x_2\) and differentiable on its interior. Hence, the mean value theorem implies that \[ f(x_2)-f(x_1)=f'(c)(x_2-x_1) \nonumber \] for some \(c\) between \(x_1\) and \(x_2\). (This is true whether $x_1<x_2 $ or \(x_2<x_1\).) The next three theorems follow from this.

A function that satisfies an inequality like for all \(x\) and \(x'\) in an interval is said to satisfy a {} on the interval.

6pt

12pt

The method of Theorem~ for finding limits of the sum, difference, product, and quotient of functions breaks down in connection with indeterminate forms. The generalized mean value theorem (Theorem~) leads to a method for evaluating limits of indeterminate forms.

We prove the theorem for finite \(L\) and leave the case where \(L=\pm\infty\) to you (Exercise~).

Suppose that \(\epsilon >0\). From , there is an \(x_0\) in \((a,b)\) such that \[\begin{equation}\label{eq:2.4.5} \left|\frac{f'(c)}{g'(c)}-L\right|<\epsilon\mbox{\quad if\quad} x_0 <c<b. \end{equation} \] Theorem~ implies that if \(x\) and \(t\) are in \([x_0, b)\), then there is a \(c\) between them, and therefore in \((x_0,b)\), such that \[\begin{equation}\label{eq:2.4.6} [g(x)-g(t)]f'(c)=[f(x)-f(t)]g'(c). \end{equation} \] Since \(g'\) has no zeros in \((a,b)\), Theorem~ implies that \[ g(x)-g(t)\ne0\mbox{\quad if\quad} x, t\in (a,b). \nonumber \] This means that \(g\) cannot have more than one zero in \((a,b)\). Therefore, we can choose \(x_0\) so that, in addition to , \(g\) has no zeros in \([x_0,b)\). Then can be rewritten as \[ \frac{f(x)-f(t)}{ g(x)-g(t)}=\frac{f'(c)}{ g'(c)}, \nonumber \] so implies that \[\begin{equation}\label{eq:2.4.7} \left|\frac{f(x)-f(t)}{ g(x)-g(t)}-L\right|<\epsilon \mbox{\quad if\quad} x, t\in [x_0,b). \end{equation} \]

If holds, let \(x\) be fixed in \([x_0,b)\), and consider the function \[ G(t)=\frac{f(x)-f(t)}{ g(x)-g(t)}-L. \nonumber \] From , \[ \lim_{t\to b-}f(t)=\lim_{t\to b-}g(t)=0, \nonumber \] so \[\begin{equation}\label{eq:2.4.8} \lim_{t\to b-}G(t)=\frac{f(x)}{g(x)}-L. \end{equation} \]

Since \[ |G(t)|<\epsilon\mbox{\quad if\quad} x_0<t<b, \nonumber \] because of , implies that \[ \left|\frac{f(x)}{ g(x)}-L\right|\le\epsilon. \nonumber \] This holds for all \(x\) in \((x_0,b)\), which implies .

The proof under assumption is more complicated. Again choose \(x_0\) so that holds and \(g\) has no zeros in \([x_0,b)\). Letting \(t=x_0\) in , we see that \[\begin{equation} \label{eq:2.4.9} \left|\frac{f(x)-f(x_0)}{ g(x)-g(x_0)}-L\right|<\epsilon \mbox{\quad if\quad} x_0\le x<b. \end{equation} \] Since \(\lim_{x\to b-}f(x)=\pm\infty\), we can choose \(x_1>x_0\) so that \(f(x)\ne0\) and \(f(x)\ne f(x_0)\) if \(x_1<x<b\). Then the function \[ u(x)=\frac{1-g(x_0)/g(x)}{1-f(x_0)/f(x)} \nonumber \] is defined and nonzero if \(x_1<x<b\), and \[\begin{equation}\label{eq:2.4.10} \lim_{x\to b-} u(x)=1, \end{equation} \] because of .

Since \[ \frac{f(x)-f(x_0)}{ g(x)-g(x_0)}=\frac{f(x)}{ g(x)}\ \frac{1-f(x_0)/f(x)}{ 1-g(x_0)/g(x)}=\frac{f(x)}{ g(x)u(x)}, \nonumber \] implies that \[ \left|\frac{f(x)}{ g(x) u(x)}-L\right|<\epsilon\mbox{\quad if\quad} x_1<x<b, \nonumber \] which can be rewritten as \[\begin{equation} \label{eq:2.4.11} \left|\frac{f(x)}{ g(x)}-Lu(x)\right|<\epsilon |u(x)|\mbox{\quad if \quad}x_1<x<b. \end{equation} \] From this and the triangle inequality, \[\begin{equation}\label{eq:2.4.12} \left|\frac{f(x)}{ g(x)}-L\right|\le\left|\frac{f(x)}{ g(x)}-Lu(x)\right| +|Lu(x)-L|\le\epsilon|u(x)|+|L|\,|u(x)-1|. \end{equation} \] Because of , there is a point \(x_2\) in \((x_1,b)\) such that \[ |u(x)-1|<\epsilon\mbox{\quad and therefore \quad} |u(x)|<1+\epsilon \mbox{\quad if\quad} x_2<x<b. \nonumber \] This, , and imply that \[ \left|\frac{f(x)}{ g(x)}-L\right|<\epsilon (1+\epsilon)+|L|\epsilon \mbox{\quad if\quad} x_2<x<b, \nonumber \]

which proves under assumption .

Theorem~ and the proof given here remain valid if \(b=\infty\) and $x\to b-$'' is replaced by\(x\to\infty\)’’ throughout. Only minor changes in the proof are required to show that similar theorems are valid for limits from the right, limits at \(-\infty\), and ordinary (two-sided) limits. We will take these as given.

-.4em We say that \(f/g\) {} \(x\to b-\) if \[ \lim_{x\to b-}f(x)=\lim_{x\to b-}g(x)=0, \nonumber \] or {} if \[ \lim_{x\to b-} f(x)=\pm\infty \nonumber \] and \[ \lim_{x\to b-} g(x)=\pm\infty. \nonumber \] The corresponding definitions for \(x\to b+\) and \(x\to\pm\infty\) are similar. If \(f/g\) is of one of these forms as \(x\to b-\) and as \(x\to b+\), then we say that it is of that form as \(x\to b\).

In using L’Hospital’s rule we usually write, for example, \[\begin{equation}\label{eq:2.4.13} \lim_{x\to b}\frac{f(x)}{ g(x)}=\lim_{x\to b}\frac{f'(x)}{ g'(x)} \end{equation} \] and then try to find the limit on the right. This is convenient, but technically incorrect, since is true only if the limit on the right exists in the extended reals. It may happen that the limit on the left exists but the one on the right does not. In this case, is incorrect.

We say that a product \(fg\) {} if one of the factors approaches \(0\) and the other approaches \(\pm\infty\) as \(x\to b-\). In this case, it may be useful to apply L’Hospital’s rule after writing \[ f(x)g(x)=\frac{f(x)}{1/g(x)}\mbox{\quad or\quad} f(x)g(x)=\frac{g(x)}{1/f(x)}, \nonumber \] since one of these ratios is of the form \(0/0\) and the other is of the form \(\infty/\infty\) as \(x\to b-\).

Similar statements apply to limits as \(x\to b+\), \(x\to b\), and \(x\to \pm\infty\).

    \enlargethispage{100pt}

A difference \(f-g\) {} if \[ \lim_{x\to b-} f(x)=\lim_{x\to b-} g(x)=\pm\infty. \nonumber \] In this case, it may be possible to manipulate \(f-g\) into an expression that is no longer indeterminate, or is of the form \(0/0\) or \(\infty/\infty\) as \(x\to b-\). Similar remarks apply to limits as \(x\to b+\), \(x\to b\), or \(x\to\pm\infty\).

The function \(f^g\) is defined by \[ f(x)^{g(x)}=e^{g(x)\log f(x)}=\exp(g(x)\log f(x)) \nonumber \] for all \(x\) such that \(f(x)>0\). Therefore, if \(f\) and \(g\) are defined and \(f(x)>0\) on an interval \((a,b)\), Exercise~ implies that \[\begin{equation}\label{eq:2.4.14} \lim_{x\to b-}[f(x)]^{g(x)}=\exp\left(\lim_{x\to b-} g(x)\log f(x)\right) \end{equation} \] if \(\lim_{x\to b-}g(x)\log f(x)\) exists in the extended reals. (If this limit is \(\pm\infty\) then is valid if we define \(e^{-\infty}=0\) and \(e^\infty=\infty\).) The product \(g\log f\) can be of the form \(0\cdot\infty\) in three ways as \(x\to b-\): 10pt

10pt In these three cases, we say that \(f^g\) {}. Similar definitions apply to limits as \(x\to b+\), \(x\to b\), and \(x\to\pm\infty\).

10pt

10pt

\begin{exerciselist}

Prove Theorem~ for the case where \(\lim_{x\to b-}f'(x)/g'(x)=\pm\infty\).

\(\dst\lim_{x\to 0}\frac{\tan^{-1}x}{\sin^{-1}x}\) & \(\dst\lim_{x\to 0}\frac{1-\cos x}{\log(1+x^2)}\)&

\(\dst\lim_{x\to 0+}\frac{1+\cos x}{e^x-1}\)\ \end{tabular}

\(\dst\lim_{x\to\pi}\frac{\sin nx}{\sin x}\)& \(\dst\lim_{x\to 0}\frac{\log(1+x)}{ x}\)&

\(\dst\lim_{x\to\infty}e^x\sin e^{-x^2}\)\ \end{tabular}

\(\dst\lim_{x\to\infty} x\sin(1/x)\)& \(\dst\lim_{x\to\infty}\sqrt{x}(e^{-1/x}-1)\)&

\(\dst\lim_{x\to 0+}\tan x\log x\) \end{tabular}

\(\dst\lim_{x\to\pi}\sin x\log (|\tan x|)\)&

\(\dst\lim_{x\to 0+} {\left[\frac{1}{x}+\log(\tan x)\right]}\)\ \end{tabular}

\(\dst\lim_{x\to\infty} (\sqrt{x+1}-\sqrt{x})\)&

\(\dst\lim_{x\to 0} \left(\frac{1}{ e^x-1}-\frac{1}{ x}\right)\)\ \end{tabular}

\(\dst\lim_{x\to 0} (\cot x-\csc x)\)&

\(\dst\lim_{x\to 0} {\left(\frac{1}{\sin x}-\frac{1}{ x}\right)}\)\ \end{tabular}

\(\dst\lim_{x\to\pi} |\sin x|^{\tan x}\)&

\(\dst\lim_{x\to\pi/2} |\tan x|^{\cos x}\)\ \end{tabular}

\(\dst\lim_{x\to 0} |\sin x|^x\)&

\(\dst\lim_{x\to 0}(1+x)^{1/x}\)\ \end{tabular}

\(\dst\lim_{x\to\infty} x^{\sin(1/x)}\)&

\(\dst\lim_{x\to 0} {\left(\frac{x}{1-\cos x}-\frac{2}{ x}\right)}\)\ \end{tabular}

\(\dst\lim_{x\to 0+} x^\alpha\log x\)&

\(\dst\lim_{x\to e} \frac{\log(\log x)}{\sin(x-e)}\)\ \end{tabular}

\(\dst\lim_{x\to\infty}{\left(\frac{x+1}{ x-1}\right)}^{\sqrt{x^2-1}}\)&

\(\dst\lim_{x\to 1+}{\left( \frac{x+1}{ x-1}\right)}^{\sqrt{x^2-1}}\)\ \end{tabular}

\(\dst\lim_{x\to\infty}\frac{(\log x)^\beta}{ x}\)&

\(\dst\lim_{x\to\infty}(\cosh x-\sinh x)\)\ \end{tabular}

\(\dst\lim_{x\to\infty}(x^\alpha-\log x)\)&

\(\dst\lim_{x\to-\infty}e^{x^2}\sin(e^x)\)\ \end{tabular}

\(\dst\lim_{x\to\infty} x(x+1)\left[\log(1+1/x)\right]^2\)&

\(\dst\lim_{x\to 0}\frac{\sin x-x+x^3/6}{x^5}\)\ \end{tabular}

\(\dst\lim_{x\to\infty}\frac{e^x}{ x^\alpha}\)&

\(\dst\lim_{x\to 3\pi/2-} e^{\tan x}\cos x\)\ \end{tabular}

\(\dst\lim_{x\to1+}\dst{(\log x)}^\alpha\log(\log x)\)&

\(\dst\lim_{x\to\infty}\frac{x^x}{ x\log x}\) \end{tabular}

\(\dst\lim_{x\to\pi/2}(\sin x)^{\tan x}\)

\(\dst\lim_{x\to 0} \frac{e^x-\dst\sum_{r=0}^n{x^r}{ r!}}{ x^n}\quad (n=\mbox{ integer}\ge1)\)

\(\dst\lim_{x\to 0} \frac{\sin x-\dst{\sum_{r=0}^n (-1)^r\frac{x^{2r+1} }{(2r+1)!}}}{ x^{2n+1}}\quad (n=\mbox{ integer}\ge0)\)

\(\dst{\lim_{x\to 0}\frac{e^{-1/x^2}}{ x^n}=0}\)(\(n=\) integer)

The {} are defined by \(L_0(x)=x\) and \[ L_n(x)=\log(L_{n-1}(x)),\quad x>a_n,\quad n\ge1, \nonumber \] where \(a_1=0\) and \(a_n=e^{a_{n-1}}, n\ge1\). Show that

Let \(f\) be positive and differentiable on \((0,\infty)\), and suppose that \[ \lim_{x\to\infty}\frac{f'(x)}{ f(x)}=L,\mbox{\quad where \quad} 0<L\le\infty. \nonumber \] Define \(f_0(x)=x\) and \[ f_n(x)=f\left(f_{n-1}(x)\right),\quad n\ge1. \nonumber \] Use L’Hospital’s rule to show that \[ \lim_{x\to\infty}\frac{(f_n(x))^\alpha}{ f_{n-1}(x)}=\infty \mbox{\quad if \quad}\alpha>0\mbox{\quad and \quad} n\ge1. \nonumber \]

Let \(f\) be differentiable on some deleted neighborhood \(N\) of \(x_0\), and suppose that \(f\) and \(f'\) have no zeros in \(N\). Find

Suppose that \(f\) and \(g\) are differentiable and \(g'\) has no zeros on \((a,b)\). Suppose also that \(\lim_{x\to b-}f'(x)/g'(x)=L\) and either \[ \lim_{x\to b-}f(x)=\lim_{x\to b-}g(x)=0 \nonumber \] or \[ \lim_{x\to b-}f(x)=\infty\mbox{\quad and \quad}\lim_{x\to b-}g(x)=\pm\infty. \nonumber \] Find \(\lim_{x\to b-}(1+f(x))^{1/g(x)}\).

We distinguish between \(\infty\cdot\infty\) \((=\infty)\) and \((-\infty)\infty\) \((=-\infty)\) and between \(\infty+\infty\) \((=\infty)\) and \(-\infty-\infty\) \((=-\infty)\). Why don’t we distinguish between \(0\cdot\infty\) and \(0\cdot (-\infty)\), \(\infty-\infty\) and \(-\infty+\infty\), \(\infty/\infty\) and \(-\infty/\infty\), and \(1^\infty\) and \(1^{-\infty}\)?

\end{exerciselist}

A {} is a function of the form \[\begin{equation}\label{eq:2.5.1} p(x)=a_0+a_1(x-x_0)+\cdots+a_n(x-x_0)^n, \end{equation} \] where \(a_0\), , \(a_n\) and \(x_0\) are constants. Since it is easy to calculate the values of a polynomial, considerable effort has been devoted to using them to approximate more complicated functions. Taylor’s theorem is one of the oldest and most important results on this question.

The polynomial is said to be written {} \(x-x_0\), and is {} \(n\) if \(a_n\ne0\). If we wish to leave open the possibility that \(a_n=0\), we say that \(p\) is of degree \(\le n\). In particular, a constant polynomial \(p(x)=a_0\) is of degree zero if \(a_0\ne0\). If \(a_0=0\), so that \(p\) vanishes identically, then \(p\) has no degree according to our definition, which requires at least one coefficient to be nonzero. For convenience we say that the identically zero polynomial \(p\) has degree \(-\infty\). (Any negative number would do as well as \(-\infty\). The point is that with this convention, the statement that \(p\) is a polynomial of degree \(\le n\) includes the possibility that \(p\) is identically zero.)

We saw in Lemma~ that if \(f\) is differentiable at \(x_0\), then \[ f(x)=f(x_0)+f'(x_0)(x-x_0)+E(x)(x-x_0), \nonumber \]

where \[ \lim_{x\to x_0} E(x)=0. \nonumber \] To generalize this result, we first restate it: the polynomial \[ T_1(x)=f(x_0)+f'(x_0)(x-x_0), \nonumber \] which is of degree \(\le1\) and satisfies \[ T_1(x_0)=f(x_0),\quad T'_1(x_0)=f'(x_0), \nonumber \] approximates \(f\) so well near \(x_0\) that \[\begin{equation} \label{eq:2.5.2} \lim_{x\to x_0}\frac{f(x)-T_1(x)}{ x-x_0}=0. \end{equation} \]

Now suppose that \(f\) has \(n\) derivatives at \(x_0\) and \(T_n\) is the polynomial of degree~\(\le~n\) such that \[\begin{equation}\label{eq:2.5.3} T^{(r)}_n (x_0)=f^{(r)}(x_0),\quad 0\le r\le n. \end{equation} \] How well does \(T_n\) approximate \(f\) near \(x_0\)?

To answer this question, we must first find \(T_n\). Since \(T_n\) is a polynomial of degree~\(\le~n\), it can be written as \[\begin{equation}\label{eq:2.5.4} T_n(x)=a_0+a_1(x-x_0)+\cdots+a_n(x-x_0)^n, \end{equation} \] where \(a_0\), , \(a_n\) are constants. Differentiating yields \[ T^{(r)}_n(x_0)=r!a_r,\quad 0\le r\le n, \nonumber \] so determines \(a_r\) uniquely as \[ a_r=\frac{f^{(r)}(x_0)}{ r!},\quad 0\le r\le n. \nonumber \] Therefore, \[\begin{eqnarray*} T_n(x)\ar=f(x_0)+\frac{f'(x_0)}{1!} (x-x_0)+\cdots+\frac{f^{(n)}(x_0) }{ n!} (x-x_0)^n\\ \ar=\sum_{r=0}^n\frac{f^{(r)} (x_0)}{ r!} (x-x_0)^r. \end{eqnarray*} \nonumber \] We call \(T_n\) {} {}.

The following theorem describes how \(T_n\) approximates \(f\) near \(x_0\).

The proof is by induction. Let \(P_n\) be the assertion of the theorem. From we know that is true if \(n=1\); that is, \(P_1\) is true. Now suppose that \(P_n\) is true for some integer \(n\ge1\), and \(f^{(n+1)}\) exists. Since the ratio \[ \frac{f(x)-T_{n+1}(x)}{(x-x_0)^{n+1}} \nonumber \] is indeterminate of the form \(0/0\) as \(x\to x_0\), L’Hospital’s rule implies that \[\begin{equation}\label{eq:2.5.6} \lim_{x\to x_0}\frac{f(x)-T_{n+1}(x)}{(x-x_0)^{n+1}}=\frac{1}{ n+1}\lim_{x\to x_0}\frac{f'(x)-T'_{n+1}(x)}{(x-x_0)^n} \end{equation} \] if the limit on the right exists. But \(f'\) has an \(n\)th derivative at \(x_0\), and \[ T'_{n+1}(x)=\sum_{r=0}^n\frac{f^{(r+1)}(x_0)}{ r!}(x-x_0)^r \nonumber \] is the \(n\)th Taylor polynomial of \(f'\) about \(x_0\). Therefore, the induction assumption, applied to \(f'\), implies that \[ \lim_{x\to x_0}\frac{f'(x)-T'_{n+1}(x)}{(x-x_0)^n}=0. \nonumber \] This and imply that \[ \lim_{x\to x_0}\frac{f(x)-T_{n+1}(x)}{(x-x_0)^{n+1}}=0, \nonumber \] which completes the induction.

It can be shown (Exercise~) that if \[ p_n=a_0+a_1(x-x_0)+\cdots+a_n(x-x_0)^n \nonumber \] is a polynomial of degree \(\le n\) such that \[ \lim_{x\to x_0}\frac{f(x)-p_n(x)}{(x-x_0)^n}=0, \nonumber \] then \[ a_r=\frac{f^{(r)}(x_0)}{ r!}; \nonumber \] that is, \(p_n=T_n\). Thus, \(T_n\) is the only polynomial of degree \(\le n\) that approximates \(f\) near \(x_0\) in the manner indicated in .

Theorem~ can be restated as a generalization of Lemma~.

Define \[ E_n(x)= \left\{\casespace\begin{array}{ll} \dst\frac{f(x)-T_n(x)}{(x-x_0)^n},&x\in D_f-\{x_0\},\\ 0,&x=x_0.\end{array}\right. \nonumber \] Then implies that \(\lim_{x\to x_0}E_n(x)=E_n(x_0)=0\), and it is straightforward to verify .

If \(q\) is a nonnegative integer, then \(\dst\binom{q}{n}\) is the binomial coefficient defined in Exercise~. In this case, we see from that \[ T_n(x)=(1+x)^q=f(x),\quad n\ge q. \nonumber \]

Lemma~ yields the following theorem.

Since \(f^{(r)}(x_0)=0\) for \(1\le r\le n-1\), implies that \[\begin{equation}\label{eq:2.5.10} f(x)-f(x_0)=\left[\frac{f^{(n)}(x_0)}{ n!}+E_n(x)\right] (x-x_0)^n \end{equation} \] in some interval containing \(x_0\). Since \(\lim_{x\to x_0} E_n(x)=0\) and \(f^{(n)}(x_0)\ne0\), there is a \(\delta>0\) such that \[ |E_n(x)|<\left|\frac{f^{(n)}(x_0)}{ n!}\right|\mbox{\quad if\quad} |x-x_0| <\delta. \nonumber \]

This and imply that \[\begin{equation}\label{eq:2.5.11} \frac{f(x)-f(x_0)}{(x-x_0)^n} \end{equation} \] has the same sign as \(f^{(n)}(x_0)\) if \(0<|x-x_0|<\delta\). If \(n\) is odd the denominator of changes sign in every neighborhood of \(x_0\), and therefore so must the numerator (since the ratio has constant sign for \(0<|x-x_0|<\delta\)). Consequently, \(f(x_0)\) cannot be a local extreme value of \(f\). This proves . If \(n\) is even, the denominator of is positive for \(x\ne x_0\), so \(f(x)-f(x_0)\) must have the same sign as \(f^{(n)}(x_0)\) for \(0<|x-x_0|<\delta\). This proves

.

For \(n=2\),

is called the {} for local extreme points.

Theorem~ implies that the error in approximating \(f(x)\) by \(T_n(x)\) approaches zero faster than \((x-x_0)^n\) as \(x\) approaches \(x_0\); however, it gives no estimate of the error in approximating \(f(x)\) by \(T_n(x)\) for a {} \(x\). For instance, it provides no estimate of the error in the approximation \[\begin{equation}\label{eq:2.5.12} e^{0.1}\approx T_2(0.1)=1+\frac{0.1}{1!}+\frac{(0.1)^2}{2!}=1.105 \end{equation} \] obtained by setting \(n=2\) and \(x=0.1\) in . The following theorem provides a way of estimating errors of this kind under the additional assumption that \(f^{(n+1)}\) exists in a neighborhood of \(x_0\).

This theorem follows from an extension of the mean value theorem that we will prove below. For now, let us assume that Theorem~ is correct, and apply it.

We now consider the extended mean value theorem, which implies Theorem~ (Exercise~). In the following theorem, \(a\) and \(b\) are the endpoints of an interval, but we do not assume that \(a<b\).

The proof is by induction. The mean value theorem (Theorem~) implies the conclusion for \(n=0\). Now suppose that \(n\ge1\), and assume that the assertion of the theorem is true with \(n\) replaced by \(n-1\). The left side of can be written as \[\begin{equation}\label{eq:2.5.18} f(b)-\sum_{r=0}^n\frac{f^{(r)}(a)}{ r!}(b-a)^r=K\frac{(b-a)^{n+1}}{(n+1)!} \end{equation} \] for some number \(K\). We must prove that \(K=f^{(n+1)}(c)\) for some \(c\) in \(I^0\). To this end, consider the auxiliary function \[ h(x)=f(x)-\sum_{r=0}^n\frac{f^{(r)}(a)}{ r!}(x-a)^r-K\frac{(x-a)^{n+1}}{ (n+1)!}, \nonumber \] which satisfies \[ h(a)=0,\quad h(b)=0, \nonumber \] (the latter because of ) and is continuous on the closed interval \(I\) and differentiable on \(I^0\), with \[\begin{equation}\label{eq:2.5.19} h'(x)=f'(x)-\sum_{r=0}^{n-1}\frac{f^{(r+1)}(a)}{ r!}(x-a)^r-K\frac{(x-a)^n}{n!}. \end{equation} \] Therefore, Rolle’s theorem (Theorem~) implies that \(h'(b_1)=0\) for some \(b_1\) in \(I^0\); thus, from , \[ f'(b_1)-\sum_{r=0}^{n-1}\frac{f^{(r+1)}(a)}{ r!}(b_1-a)^r-K\frac{(b_1-a)^n}{n!}=0. \nonumber \] If we temporarily write \(f'=g\), this becomes \[\begin{equation}\label{eq:2.5.20} g(b_1)-\sum_{r=0}^{n-1}\frac{g^{(r)}(a)}{ r~}(b_1-a)^r-K\frac{(b_1-a)^n}{n!}=0. \end{equation} \]

Since \(b_1\in I^0\), the hypotheses on \(f\) imply that \(g\) is continuous on the closed interval \(J\) with endpoints \(a\) and \(b_1\), \(g^{(n)}\) exists on \(J^0\), and, if \(n\ge1\), \(g'\), , \(g^{(n-1)}\) exist and are continuous at \(a\) (also at \(b_1\), but this is not important). The induction hypothesis, applied to \(g\) on the interval \(J\), implies that \[ g(b_1)-\sum_{r=0}^{n-1}\frac{g^{(r)}(a)}{ r!} (b_1-a)^r=\frac{g^{(n)}(c)}{n!}(b_1-a)^n \nonumber \] for some \(c\) in \(J^0\). Comparing this with and recalling that \(g=f'\) yields \[ K=g^{(n)}(c)=f^{(n+1)}(c). \nonumber \] Since \(c\) is in \(I^0\), this completes the induction.