3.2: Limit and Rate of Change as a Limit

The phrase

\[\text { as } b \text { approaches } a \quad \frac{F(b)-F(a)}{b-a} \quad \text { approaches } \quad m_{a} \label{3.4}\]

introduced in the previous section to define slope of a tangent to the graph of F at a point \((a, F(a))\) can be made more explicit. Almost 200 years after Isaac Newton and Gottfried Leibniz introduced calculus (about 1665), Karl Weierstrass introduced (about 1850) a concise statement of limit that clarifies the phrase. Phrase \ref{3.4} would be rewritten

\[\text { The limit as } b \text { approaches } a \text { of } \frac{F(b)-F(a)}{b-a} \text { is } m_{a} \text { . } \label{3.5}\]

We introduce notation separate from that of the phrase \ref{3.4} because ’limit’ has application well beyond its use in defining tangents.

Definition 3.2.1 Definition of limit

Suppose \(G\) is a function, a is a number, and every open interval that contains a contains a point of the domain of \(G\) distinct from \(a\). Also suppose \(L\) is a number. The statement that

\[\text { as } x \text { approaches } a \quad \mathrm{G}(\mathrm{x}) \text { approaches } L\]

means that if \(\epsilon\) is a positive number, there is a positive number \(\delta\) such that if \(x\) is in the domain of \(G\) and \(0<|x-a|<\delta\) then \(|G(x)-L|<\epsilon\). This relation is denoted by

\[\lim _{x \rightarrow a} G(x)=L \label{3.6}\]

and is read, ’The limit as \(x\) approaches \(a\) of \(G(x)\) is \(L\).’

The statement that ’\(\lim _{x \rightarrow a} G(x)\) exists’ means that \(G\) is a function, a is a number, and every open interval that contains a contains a number in the domain of \(G\) distinct from \(a\), and for some number \(L\), \(\lim _{x \rightarrow a} G(x)=L\).

The use of the word ’limit’ in Definition 3.2.1 is different from its usual use as a bound, as in ’speed limit’ or in "1. the point, line or edge where something ends or must end; · · ·" (Webster’s new College Dictionary, Fourth Edition). Perhaps ’goal’ (something aimed at or striven for) would be a better word, but we and you have no choice: assume a new and well defined meaning for ’limit.’

Important. In Definition 3.2.1, the question of whether \(G(a)\) is defined is irrelevant, and if \(G(a)\) is defined, the value of \(G(a)\) is irrelevant. The inequality \(0 < |x − a| < \delta\) specifically excludes consideration of \(x = a\). Two illustrative examples appear in Figure 3.2.1. They are the graphs of \(F_1\) and \(F_2\), defined by

\[F_{1}(x)=\left\{\begin{array}{cc}
\frac{1}{2} * x & \text { for } x \neq 1 \\
\text { not defined } & \text { for } x=1
\end{array} \quad F_{2}(x)=\left\{\begin{array}{cc}
\frac{1}{2} x & \text { for } x \neq 1 \\
1 & \text { for } x=1
\end{array}\right.\right. \label{3.7}\]

Figure \(\PageIndex{1}\): A. Graphs of \(F_1\) and \(F_2\) as defined in Equation \ref{3.7}.

Intuitively: If x is close to 1 and different from 1, \(F_{1}(x)=\frac{1}{2} x\) is close to 0.5 and \(F_{2}(x)\) is close to 0.5.

\[\lim _{x \rightarrow 1} F_{1}(x)=0.5, \quad \text { and } \quad \lim _{x \rightarrow 1} F_{2}(x)=0.5\]

By Definition 3.2.1: Suppose \(\epsilon\) is a positive number; choose \(\delta = \epsilon\).

\[\begin{aligned}
&\text { If } x \text { is in the domain of } F_{1} \text { and } \quad 0<|x-1|<\delta,\\
&\text { then } \quad\left|F_{1}(x)-\frac{1}{2}\right|=\left|\frac{1}{2} x-\frac{1}{2}\right|=\frac{1}{2}|x-1|<\frac{1}{2} \delta<\epsilon .
\end{aligned}\]

The same argument applies to \(F_2\). That \(F_{2}(1) = 1\) does not change the argument because \(0 < |x − 1| < \delta\) excludes \(x = 1\).

Example 3.2.1 Practice with \(\epsilon\)’s and \(\delta\)’s.

1. Suppose administration of \(a = 3.5\) mg of growth hormone produces the optimum serum hormone level \(L = 8.3\) \(\mu g\) in a 24 kg boy. Suppose further that an amount \(x\) mg of growth hormone produces serum hormone level \(G(x)\) \(\mu g\). You may wish to require \(\epsilon = 0.25\) \(\mu g\) accuracy in serum hormone levels, and need to know what specification \(\delta\) mg accuracy to require in preparation of the growth hormone to be administered. That is, if the actual amount administered is between \(3.5 − \delta\) mg and \(3.5 + \delta\) mg then the resulting serum hormone level will be between \(8.3 − 0.25 = 8.05\) \(\mu g\) and \(8.3 + 0.15 = 8.45\) \(\mu g\). If instead of \(\epsilon = 0.25\) \(\mu g\), your tolerance is \(\epsilon = 0.05\) \(\mu g\), the specified \(\delta\) mg would be smaller.
2. Show that if a is a positive number, then \[\lim _{x \rightarrow a} \sqrt{x}=\sqrt{a} \label{3.8}\] Suppose \(\epsilon > 0\). Let \(\delta=\epsilon \sqrt{a}\). Suppose \(x\) is in the domain of \(\sqrt{x}\) and \(0 < |x − a| < \delta\). Then \[|\sqrt{x}-\sqrt{a}| \stackrel{(i)}{=} \frac{|x-a|}{\sqrt{x}+\sqrt{a}} \stackrel{(i i)}{\leq} \frac{|x-a|}{\sqrt{a}} \quad \stackrel{(i i i)}{<} \frac{\delta}{\sqrt{a}} \stackrel{(i v)}{=} \frac{\epsilon \sqrt{a}}{\sqrt{a}} \quad \stackrel{(v)}{=} \quad \epsilon\] Explore 3.2.1 Our choice of \(\delta=\epsilon \sqrt{a}\) is a bit of a fortuitous lightning bolt³, Step (i) is an algebraic trick, you may well wonder why Step (ii) is even correct or not strictly <, you may wonder where \(\delta\) came from in Step (iii), the fortuitous lightning bolt is the reason for Step (iv) and explains why we chose \(\delta=\epsilon \sqrt{a}\), but by this time you may even doubt Step (v). You would be wise to check each step.
3. Find a number \(\delta > 0\) so that \[\text { if } \quad 0<|x-1|<\delta \quad \text { then } \quad\left|\frac{1}{1+x^{2}}-0.5\right|<0.01 \text { . }\] A graph of \(y = 1/(1 + x^{2})\) is shown in Figure 3.2.2 along with lines a distance 0.01 above and below \(y = 0.5\). We solve for \(x\) at the intersections of the lines with the graph. \[\frac{1}{1+x^{2}}=0.49, \quad x=1.0202 ; \quad \text { and } \quad \frac{1}{1+x^{2}}=0.51, \quad x=0.98019 .\] From the graph it is clear that \[\text { if } \quad 0.981<x<1.02 \quad \text { then } \quad 0.49<\frac{1}{1+x^{2}}<0.51 \text { . }\] We choose \(\delta = 0.019\), the smaller of \(1.0 − 0.981\) and \(1.02 − 1.0\). \[\text { Then if } \quad 0<|x-1|<\delta, \quad\left|\frac{1}{1+x^{2}}-0.5\right|<0.01 \text { . }\]

Figure \(\PageIndex{2}\): A. Graph of \(F(x) = 1/(1 + x^{2})\), and B a magnification of A. In both graphs, segments of the lines y = 0.49 and y = 0.51 are drawn.

4. Find a number \(\delta > 0\) so that \[\text { if } \quad 0<|x-4|<\delta \quad \text { then } \quad\left|\frac{\frac{1}{\sqrt{x}}-\frac{1}{2}}{x-4}+\frac{1}{16}\right|<0.01 \text { . }\] This algebraic challenge springs from the problem of showing that the slope of \[f(x)=\frac{1}{\sqrt{x}} \text { at the point }\left(4, \frac{1}{2}\right) \text { is } \frac{-1}{16}\] We first do some algebra. \[\frac{\frac{1}{\sqrt{x}}-\frac{1}{2}}{x-4}+\frac{1}{16}=\frac{\frac{2-\sqrt{x}}{2 \sqrt{x}}}{(\sqrt{x}-2)(\sqrt{x}+2)}+\frac{1}{16}=\frac{-1}{2 \sqrt{x}(\sqrt{x}+2)}+\frac{1}{16}=\frac{-8+\sqrt{x}(\sqrt{x}+2)}{16 \sqrt{x}(\sqrt{x}+2)}\] It is sufficient to find \(\delta > 0\) so that \[\text { if } \quad 0<|x-4|<\delta \text { then }\left|\frac{-8+\sqrt{x}(\sqrt{x}+2)}{16 \sqrt{x}(\sqrt{x}+2)}\right|<0.01 \quad \text { or } \quad\left|\frac{-8+\sqrt{x}(\sqrt{x}+2)}{\sqrt{x}(\sqrt{x}+2)}\right|<0.16\] Assume that \(x\) is ’close to’ 4 and certainly bigger than 1 so that the denominator, \(\sqrt{x}(\sqrt{x}+2)\) is greater than 1. Then it is sufficient to find \(\delta > 0\) so that \[|-8+\sqrt{x}(\sqrt{x}+2)|=|-8+x+2 \sqrt{x}|<0.16\] Let \(z = \sqrt{x}\) so that \(z^{2} = x\) and solve \[\begin{array}{rlrl}
   -8+z^{2}+2 z & =-0.16 & \text { and } -8+z^{2}+2 z & =0.16 \\
   z_{1} & =1.97321 & z_{2} & =2.026551 .97321 \\
   x_{1} & =3.89356 & x_{2} & =4.10690
   \end{array}\] We will choose \(\delta = 0.1\) so that \[\text { if } 0<|x-4|<0.1 \quad \text { then } x \text { is between } x_{1} \text { and } x_{2} \text { and } \quad\left|\frac{\frac{1}{\sqrt{x}}-\frac{1}{2}}{x-4}+\frac{1}{16}\right|<0.01 \text { . }\]
5. Suppose you try to 'square the circle.' That is, suppose you try to construct a square the area of which is exactly the area, \(\pi\), of a circle of radius 1. A famous problem from antiquity is whether one can construct such a square using only a straight edge and a compass. The answer is no. But can you get close?

Suppose you will be satisfied if the area of your square is within 0.01 of \(\pi\) (think, \(\epsilon = 0.01\)). It helps to know that

\[\pi=3.14159265 \cdots, \quad \text { and } \quad 1.77^{2}=3.1329\]

and that every interval of rational length (for example, 1.77) can be constructed with straight edge and compass. Because \(1.77^{2} = 3.1329\) is within 0.01 of \(\pi\), you can ’almost’ square the circle with straight edge and compass.

Explore 3.2.2 Suppose you are only satisfied if the area of your square is within 0.001 of \(\pi\). What length should the edges of your square be in order to achieve that accuracy?

Now suppose \(\epsilon\) is a positive number and ask, how close (think, \(\delta\)) must your edge be to \(\sqrt{\pi}\) in order to insure that

\[\mid \text { the area of your square }-\pi \mid<\epsilon \text { ? }\]

Let \(x\) be the length of the side of your square. Your problem is to find \(\delta\) so that

\[\text { if } \quad|x-\sqrt{\pi}|<\delta, \quad \text { then }\left|x^{2}-\pi\right|<\epsilon\]

Danger: This may fry your brain. Look at the target \(\left|x^{2}-\pi\right|<\epsilon\) and write

\[\begin{aligned}
\left|x^{2}-\pi\right| &<\epsilon \\
|x-\sqrt{\pi}||x+\sqrt{\pi}| &<\epsilon
\end{aligned}\]

Lets require first of all that \(0 < x < 5\). (We get to write the specifications for \(x\).) Then \(x + \sqrt{\pi} < 10\). (Actually, \(\sqrt{\pi} \doteq 1.7725\) so that \(x + \sqrt{\pi}\) is less than 7, but we can be generous.) Now look at our target,

\[\left|x^{2}-\pi\right|=|x-\sqrt{\pi}||x+\sqrt{\pi}| \quad<\epsilon\]

Choose \(\delta\): Insist that \(|x-\sqrt{\pi}|<\delta=\operatorname{Minimum}(\epsilon / 10,1)\). Then \(0 < x < 5\), and

\[\left|x^{2}-\pi\right|=|x-\sqrt{\pi}||x+\sqrt{\pi}| \quad<\quad \frac{\epsilon}{10} \cdot 10=\epsilon\]

If \(|x-\sqrt{\pi}|<\delta=\operatorname{Minimum}(\epsilon / 10,1)\), then \(\left|x^{2}-\pi\right|<\epsilon\).

Explore 3.2.3 We required that \(0 < x < 5\). (We get to write the specifications for \(x\).) Would it work to require that \(0 < x < 1\)?

Using the definition of limit we rewrite the definitions of tangent to the graph of a function and the rate of change of a function.

Definition 3.2.2 Definition of tangent and rate of change, revisited

Suppose \(F\) is a function, a is a number in the domain of \(F\), and every open interval that contains a contains a point of the domain of \(F\) distinct from \(a\), and \(m_a\) is a number. The statement that the slope of the tangent to the graph of \(F\) at \((a, F(a))\) is \(m_a\) and that the rate of change of \(F\) at \(a\) is \(m_a\) means that

\[\lim _{b \rightarrow a} \frac{F(b)-F(a)}{b-a}=m_{a} \label{3.9}\]

In the previous section we found that the slope of the tangent to the graph of \(F(t) = t^2\) at a point \((a, a^{2})\) was \(2a\). Using the limit notation we would write that development as

\[\begin{aligned}
\lim _{b \rightarrow a} \frac{F(b)-F(a)}{b-a} &=\lim _{b \rightarrow a} \frac{b^{2}-a^{2}}{b-a} \\
&=\lim _{b \rightarrow a} \frac{(b-a)(b+a)}{b-a} \\
&=\lim _{b \rightarrow a}(b+a) \\
&=a+a=2 a
\end{aligned}\]

There is some algebra of the limit symbol, \(\lim _{x \rightarrow a}\), that is important. Suppose each of \(F_1\) and \(F_2\) is a function and a is a number and \(\lim _{x \rightarrow a} F_{1}(x)\) and \(\lim _{x \rightarrow a} F_{2}(x)\) both exist. Suppose further that \(C\) is a number. Then

\[\lim _{x \rightarrow a} C=C \label{3.10}\]

\[\lim _{x \rightarrow a} x=a \label{3.11}\]

\[\text { If } a \neq 0, \text { then } \quad \lim _{x \rightarrow a} \frac{1}{x}=\frac{1}{a} \label{3.12}\]

\[\lim _{x \rightarrow a} C F_{1}(x)=C \lim _{x \rightarrow a} F_{1}(x) \label{3.13}\]

\[\lim _{x \rightarrow a}\left(F_{1}(x)+F_{2}(x)\right)=\lim _{x \rightarrow a} F_{1}(x)+\lim _{x \rightarrow a} F_{2}(x) \label{3.14}\]

\[\lim _{x \rightarrow a}\left(F_{1}(x) \times F_{2}(x)\right)=\left(\lim _{x \rightarrow a} F_{1}(x)\right) \times\left(\lim _{x \rightarrow a} F_{2}(x)\right) \label{3.15}\]

Probably all of these equations are sufficiently intuitive that proofs of them seem superfluous. Equations \ref{3.10} and \ref{3.11} could be more accurately expressed. For Equation \ref{3.10} one might say,

\[" \text {If } F_{1}(x)=C \text { for all } x \neq a \text { then } \quad \lim _{x \rightarrow a} F_{1}(x)=C . " \label{3.16}\]

Explore 3.2.4 Suppose \(F_1\) is defined by

\[\begin{array}{l}
F_{1}(x)=C \quad \text { if } \quad x \neq a \\
F_{1}(a)=C+1
\end{array}\]

Is it true that

\[\lim _{x \rightarrow a} F_{1}(x)=C \quad ?\]

Explore 3.2.5 Write a more accurate expression for Equation \ref{3.11} similar to the more accurate expression \ref{3.16} for \ref{3.10}.

Proof of Equations 3.12, 3.13 and 3.14 are rather easy and are included to illustrate the process.

Proof of Equation 3.12. Suppose \(a \neq 0\) and \(\epsilon > 0\). Let

\[\delta=\operatorname{Minimum}\left(|a| / 2, \epsilon a^{2} / 2\right) . \quad \text { Suppose } \quad 0<|x-a|<\delta .\]

Note: Because \(|x-a|<\delta \leq|a| / 2,|x|>|a| / 2\) and \(|x a|>\left|a^{2}\right| / 2\).

\[\begin{aligned}
\text { Then } \quad\left|\frac{1}{x}-\frac{1}{a}\right| &=\left|\frac{x-a}{x a}\right| \\
&<\frac{\delta}{|x a|} \\
&<\frac{\delta}{a^{2} / 2} \\
& \leq \frac{\epsilon a^{2} / 2}{a^{2} / 2} \quad=\quad \epsilon
\end{aligned}\]

Proof of Equation 3.13. Let \(\lim _{x \rightarrow a} F_{1}(x)=L_{1}\). Suppose \(\epsilon\) is a positive number.

Case \(C = 0\). Let \(\delta = 1\). Then if \(x\) is in the domain of \(F_1\) and \(0 < |x − a| < \delta\),

\[\left|C F_{1}(x)-C L_{1}\right|=|0-0|=0<\epsilon\]

Case \(C \neq 0\). Let

\[\epsilon_{0}=\frac{\epsilon}{|C|}\]

Because \(\lim _{x \rightarrow a} F_{1}(x)=L_{1}\), there is \(\delta>0\) such that if \(0<|x-a|<\delta\) then \(\left|F_{1}(x)-L_{1}\right|<\epsilon_{0}\). Then if \(x\) is in the domain of \(F\) and \(0 < |x − a| < \delta\),

\[\left|C F_{1}(x)-C L_{1}\right|=|C|\left|F_{1}(x)-L_{1}\right|<|C| \epsilon_{0}=|C| \frac{\epsilon}{|C|}=\epsilon\]

Thus

\[\lim _{x \rightarrow a} C F_{1}(x)=C L_{1}=C \lim _{x \rightarrow a} F_{1}(x)\]

Proof of Equation 3.14. Suppose \(\lim _{x \rightarrow a} F_{1}(x)=L_{1}\) and \(\lim _{x \rightarrow a} F_{2}(x)=L_{2}\). Suppose \(\epsilon > 0\). There are numbers \(\delta_1\) and \(\delta_2\) such that

\[\text { if } 0<|x-a|<\delta_{1} \text { then }\left|F_{1}(x)-L_{1}\right|<\frac{\epsilon}{2} \quad \text { and } \quad \text { if } 0<|x-a|<\delta_{2} \text { then } \left|F_{2}(x)-L_{2}\right|<\frac{\epsilon}{2}\]

Let \(\delta=\operatorname{Minimum}\left(\delta_{1}, \delta_{2}\right)\). Suppose \(0<|x-a|<\delta\). Then

\[\begin{aligned}
\left|F_{1}(x)+F_{2}(x)-\left(L_{1}+L_{2}\right)\right| & \leq\left|F_{1}(x)-L_{1}\right|+\left|F_{2}(x)-L_{2}\right| \\
&<\frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon
\end{aligned}\]

Honest Exposition. Almost always in evaluating

\[\lim _{x \rightarrow a} F(x)\]

if \(F\) is a familiar function and \(F(a)\) is defined then

\[\lim _{x \rightarrow a} F(x)=F(a)\]

If this happens, \(F\) is said to be continuous at \(a\) (more about continuity later). It follows from Equations \ref{3.10} through \ref{3.15}, for example, that

\[\text { if } P(x) \text { is any polynomial, then } \quad \lim _{x \rightarrow a} P(x)=P(a) \text {. } \quad \text { See Exercise 3.2.7. }\]

Moreover, even if \(F(a)\) is not meaningful but there is an expression \(E(x)\) for which

\[F(x)=E(x) \text { for } x \neq a, \text { and } E(a) \text { is defined, }\]

then almost always

\[\lim _{x \rightarrow a} F(x)=E(a)\]

For example, for

\[F(x)=\frac{x^{4}-a^{4}}{x-a}, \quad F(a) \quad \text { is meaningless. }\]

But,

\[\begin{array}{l}
\text { for } x \neq a, \quad F(x)=\frac{(x-a)\left(x^{3}+a x^{2}+a^{2} x+a^{3}\right)}{x-a} \\
=x^{3}+x^{2} a+x a^{2}+a^{3} \\
=E(x) . \\
\text { and } \quad E(a)=4 a^{3}, \quad \text { and } \quad \lim _{x \rightarrow a} F(x)=4 a^{3} \text { . }
\end{array}\]

The simple procedure just discussed fails when considering more complex limits such as

\[\lim _{x \rightarrow 0} \frac{\sin x}{x}, \quad \lim _{x \rightarrow 1} \frac{\log x}{x-1}, \quad \text { and } \quad \lim _{x \rightarrow 1} \frac{2^{x}-2}{x-1}\]

Limit of the composition of two functions. A formula for the limit of the composition of two functions has very broad application.

Note

Suppose \(u\) is a function, \(L\) is a number, \(a\) is in the domain of \(u\), and there is an open interval \((p, q)\) in the domain of \(u\) that contains \(a\) and if \(x\) is in \((p, q)\) and is not a then \(u(x) \neq L\). Suppose further that

\[\lim _{x \rightarrow a} u(x)=L\]

Suppose \(F\) is a function and for \(x\) in \((p, q)\), \(u(x)\) is in the domain of \(F\), and

\[\lim _{s \rightarrow L} F(s)=\lambda\]

Then

\[\lim _{x \rightarrow a} F(u(x))=\lambda \label{3.17}\]

Intuitively, if \(x\) is close to but distinct from \(a\), then \(u(x)\) is close to but distinct from \(L\) and \(F(u(x))\) is close to \(\lambda\). The formal argument is perhaps easier than the statement of the property, but is omitted.

Some formulas that follow from Equation \ref{3.17} and previous formulas include

\[\begin{array}{III}
\text{ If } \lim _{x \rightarrow a} F(x)=L>0, & \text{ then } & \lim _{x \rightarrow a} \sqrt{F(x)}=\sqrt{L} \\
\text { If } \lim _{x \rightarrow a} F(x)=L \neq 0, & \text { then } & \lim _{x \rightarrow a} \frac{1}{F(x)}=\frac{1}{L} \\
\text { If } \lim _{x \rightarrow a} F_{2}(x) \neq 0, & \text { then } & \lim _{x \rightarrow a} \frac{F_{1}(x)}{F_{2}(x)}=\frac{\lim _{x \rightarrow a} F_{1}(x)}{\lim _{x \rightarrow a} F_{2}(x)}
\end{array} \label{3.18}\]

 

Exercises for Section 3.2, Limit and rate of change as a limit.

Exercise 3.2.1

1. Find a number, \(\delta > 0\), so that if \(x\) is a number and \(0 < |x − 2| < \delta\) then \(|2x − 4| < 0.01\).
2. Find a number, \(\delta\), so that if \(x\) is a number and \(0 < |x − 2| < \delta\) then \(|x^{2} − 4| < 0.01\).
3. Find a number, \(\delta\), so that if \(x\) is a number and \(0 < |x − 2| < \delta\) then \(\left|\frac{1}{x}-\frac{1}{2}\right|<0.01\).
4. Find a number, \(\delta\), so that if \(x\) is a number and \(0 < |x − 2| < \delta\) then \(|x^{3} − 8| < 0.01\).
5. Find a number, \(\delta\), so that if \(x\) is a number and \(0 < |x − 2| < \delta\) then \(\left|\frac{x}{x+1}-\frac{2}{3}\right|<0.01\).
6. Find a number, \(\delta > 0\), so that if \(x\) is a number and \(0 < |x − 9| < \delta\) then \(| \sqrt{x} − 3| < 0.01\).
7. Find a number, \(\delta\), so that if \(x\) is a number and \(0 < |x − 8| < \delta\) then \(|\sqrt[3]{x}-2|<0.01\).
8. Find a number, \(\delta > 0\), so that if \(x\) is a number and \(0 < |x − 2| < \delta\) then \(\left|x^{4}-3 x-13\right|<0.01\).

Exercise 3.2.2

1. Find a number, \(\delta>0\), so that if \(x\) is a number and \(|x − 3| < \delta\) then \(\left|\frac{x^{2}-9}{x-3}-6\right|<0.01\).
2. Find a number, \(\delta>0\), so that if \(x\) is a number and \(|x − 4| < \delta\) then \(\left|\frac{\sqrt{x}-2}{x-4}-\frac{1}{4}\right|<0.01\).
3. Find a number, \(\delta>0\), so that if \(x\) is a number and \(|x − 2| < \delta\) then \(\left|\frac{\frac{1}{x}-\frac{1}{2}}{x-2}+\frac{1}{4}\right|<0.01\).
4. Find a number, \(\delta>0\), so that if \(x\) is a number and \(|x − 1| < \delta\) then \(\left|\frac{x^{2}+x-2}{x-1}-3\right|<0.01\).

Exercise 3.2.3

1. Use Equations \ref{3.11} and \ref{3.15}, \[\lim _{x \rightarrow a} x=a \quad \text { and } \quad \lim _{x \rightarrow a} F_{1}(x) \times F_{2}(x)=\left(\lim _{x \rightarrow a} F_{1}(x)\right) \times\left(\lim _{x \rightarrow a} F_{2}(x)\right),\] to show that \[\lim _{x \rightarrow a} x^{2}=a^{2}\].
2. Show that \[\lim _{x \rightarrow a} x^{3}=a^{3}\].
3. Show by induction that if \(n\) is a positive integer, \[\lim _{x \rightarrow a} x^{n}=a^{n} \label{3.19}\]

Exercise 3.2.4 Prove the following theorem.

Theorem 3.2.1 : Limit is Unique Theorem. Suppose \(G\) is a function and

\[\lim _{x \rightarrow a} G(x)=L_{1} \quad \text { and } \quad \lim _{x \rightarrow a} G(x)=L_{2} .\]

Then \(L_{1} = L_{2}\).

Begin your proof with,

1. Suppose that \(L_{1} < L_{2}\).
2. Let \(\epsilon=\left(L_{2}-L_{1}\right) / 2\).

Exercise 3.2.5 Evaluate the limits.

1. \(\lim _{x \rightarrow 5} 3 x^{2}-15 x\)
2. \(\lim _{x \rightarrow 0} 3 x\)
3. \(\lim _{x \rightarrow 50} \pi\)
4. \(\lim _{x \rightarrow-2} 3 x^{2}-15 x\)
5. \(\lim _{x \rightarrow-\pi} x\)
6. \(\lim _{x \rightarrow \pi} 50\)
7. \(\lim _{x \rightarrow-5} \frac{x-1}{x-1}\)
8. \(\lim _{x \rightarrow 1} \frac{x^{2}-2 x+1}{x-1}\)
9. \(\lim _{x \rightarrow 1} \frac{x^{4}-1}{x-1}\)
10. \(\lim _{x \rightarrow 4} \frac{x}{\sqrt{x+2 x^{2}}}\)
11. \(\lim _{x \rightarrow 1} \sqrt{\frac{x-2}{x^{3}+1}}\)
12. \(\lim _{x \rightarrow 1} \sqrt{\frac{x-3}{x^{3}-27}}\)

Exercise 3.2.6 Sketch the graph of \(y=\sqrt[3]{x}\) for \(-1 \leq x \leq 1\). Does the graph have a tangent at \((0,0)\)? Remember, Your vote counts.

Exercise 3.2.7 Suppose \(P(x)=a_{n} x^{n}+a_{n-1} x^{n-1}+\cdots a_{1} x+a_{0}\) is a polynomial and a is a number. For each step (i) through (vii), identify the equation of Equations \ref{3.10} - \ref{3.14} and \ref{3.19} that justify the step.

\[\begin{array}{l}
\begin{aligned}
\lim _{x \rightarrow a} P(x) &=\lim _{x \rightarrow a}\left(a_{n} x^{n}+a_{n-1} x^{n-1}+\cdots a_{1} x+a_{0}\right) \\
&=\lim _{x \rightarrow a}\left(a_{n} x^{n}\right)+\lim _{x \rightarrow a}\left(a_{n-1} x^{n-1}+\cdots a_{1} x+a_{0}\right)  &(i)\\
&=\lim _{x \rightarrow a}\left(a_{n} x^{n}\right)+\lim _{x \rightarrow a}\left(a_{n-1} x^{n-1}\right)+\lim _{x \rightarrow a}\left(a_{n-2} x^{n-2} \cdots a_{1} x+a_{0}\right) &(ii)\\
& \vdots \vdots \\
&=\lim _{x \rightarrow a}\left(a_{n} x^{n}\right)+\cdots+\lim _{x \rightarrow a}\left(a_{2} x^{2}\right)+\lim _{x \rightarrow a}\left(a_{1} x+a_{0}\right) \\
&=\lim _{x \rightarrow a}\left(a_{n} x^{n}\right)+\cdots+\lim _{x \rightarrow a}\left(a_{2} x^{2}\right)+\lim _{x \rightarrow a}\left(a_{1} x\right)+\lim _{x \rightarrow a}\left(a_{0}\right) &(iii)\\
&=a_{n} \lim _{x \rightarrow a}\left(x^{n}\right)+\cdots+a_{2} \lim _{x \rightarrow a}\left(x^{2}\right)+a_{1} \lim _{x \rightarrow a}(x)+\lim _{x \rightarrow a}\left(a_{0}\right) &(iv)\\
&=a_{n} a^{n}+\cdots+a_{2} a^{2}+a_{1} \lim _{x \rightarrow a}(x)+\lim _{x \rightarrow a}\left(a_{0}\right) &(v)\\
&=a_{n} a^{n}+a_{n-1} a^{n-1}+\cdots+a_{2} a^{2}+a_{1} a+\lim _{x \rightarrow a}\left(a_{0}\right) &(vi)\\
&=a_{n} a^{n}+a_{n-1} a^{n-1}+\cdots+a_{2} a^{2}+a_{1} a+a_{0} &(vii)\\
&=P(a) \quad \quad \text { Whew! }\\
\end{aligned}
\end{array}\]

 

Exercise 3.2.8 Show that Equation \ref{3.13},

\[\lim _{x \rightarrow a} C F_{1}(x)=C\left(\lim _{x \rightarrow a} F_{1}(x)\right)\]

follows from Equations \ref{3.10} and \ref{3.15},

\[\lim _{x \rightarrow a} C=C \quad \text { and } \quad \lim _{x \rightarrow a}\left(F_{1}(x) F_{2}(x)\right)=\left(\lim _{x \rightarrow a} F_{1}(x)\right)\left(\lim _{x \rightarrow a} F_{2}(x)\right)\]

Exercise 3.2.9 Some of the following statements are true and some are false. For those that are true, provide proofs using Equations \ref{3.10} - {3.15}. For those that are false, provide functions \(F_1\) and \(F_2\) to show that they are false. Assume the limits shown do exist.

Suppose \(F_1\) and \(F_2\) are functions defined for all numbers, \(x\).

1. \(\text { If } \lim _{x \rightarrow a}\left(F_{1}(x)-F_{2}(x)\right)=0,quad \text { then } \lim _{x \rightarrow a} F_{1}(x)=\lim _{x \rightarrow a} F_{2}(x) \text { . }\)
2. \(\text { If } \lim _{x \rightarrow a}\left(F_{1}(x) F_{2}(x)\right)=0, \quad \text { then } \lim _{x \rightarrow a} F_{1}(x)=\lim _{x \rightarrow a} F_{2}(x)=0 .\)
3. \(\text { If } \lim _{x \rightarrow a} \frac{F_{1}(x)}{F_{2}(x)}=0, \quad \text { then } \lim _{x \rightarrow a} F_{1}(x)=0 .\)
4. \(\text { If } \lim _{x \rightarrow a} F_{1}(x)=0, \quad \text { then } \lim _{x \rightarrow a} \frac{F_{1}(x)}{F_{2}(x)}=0 \text { . }\)
5. \(\text { If } \lim _{x \rightarrow a}\left(F_{1}(x) F_{2}(x)\right)=0, \quad \text { then } \lim _{x \rightarrow a} F_{1}(x)=0 \quad \text { or } \quad \lim _{x \rightarrow a} F_{2}(x)=0\) .

Exercise 3.2.10

\(\text { Evaluate } \lim _{x \rightarrow a} \frac{F(x)-F(a)}{x-a} \text { for }\)

1. \(\quad F(x)=x^{2} \quad a=-2\)
2. \(\quad F(x)=17 \quad a=0\)
3. \(\quad F(x)=2 x^{3} \quad a=2\)
4. \(\quad F(x)=x^{2}+2 x \quad a=1\)
5. \(\quad F(x)=\frac{1}{x} \quad a=\frac{1}{2}\)
6. \(\quad F(x)=3 x^{2}-5 x \quad a=7\)
7. \(\quad F(x)=3 \sqrt{x} \quad a=4\)
8. \(\quad F(x)=x^{2}+2 x+1 \quad a=-1\)
9. \(\quad F(x)=\frac{4}{x}+5 \quad a=2\)
10. \(\quad F(x)=x^{6} \quad a=2\)
11. \(\quad F(x)=\frac{1}{x^{3}} \quad a=2\)
12. \(\quad F(x)=x^{10} \quad a=2\)
13. \(\quad F(x)=\frac{4}{x^{5}} \quad a=2\)
14. \(\quad F(x)=x^{67} \quad a=1\)

Exercise 3.2.11 Suppose \(F(x)\) is a function and

\[\lim _{b \rightarrow 2} \frac{F(b)-F(2)}{b-2} \quad \text { exists. }\]

What is \(\lim _{b \rightarrow 2} F(b)\)?

 

 

³ Actually found by working the problem backwards. What must \(\delta\) be in order to insure step (iv) will be correct?