17.4: Appendix D- Limits

We introduced notation involving limits without carefully defining what was meant. Here, such technical matters are briefly discussed.

The concept of a limit helps us to describe the behavior of a function close to some point of interest. This is useful in the case of functions that are either not continuous, or not defined somewhere. We use the notation

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

to denote the value the function \(f\) approaches as \(x\) gets closer and closer to the value \(a\).

D.1 Limits for continuous functions

If \(x=a\) is a point at which the function is defined and continuous (informally: has no "breaks in its graph") the value of the limit and the value of the function at a point are the same, i.e.

If \(f\) is continuous at \(x=a\) then

\[\lim _{x \rightarrow a} f(x)=f(a) . \nonumber \]

Example D.1

Find \(\lim _{x \rightarrow 0} f(x)\) for the function \(y=f(x)=10\).

Solution

This function is continuous (and constant) everywhere. In fact, the value of the function is independent of \(x\). We conclude immediately that

\[\lim _{x \rightarrow 0} f(x)=\lim _{x \rightarrow 0} 10=10 . \nonumber \]

Example D.2

Find \(\lim _{x \rightarrow 0} f(x)\) for the function \(y=f(x)=\sin (x)\).

Solution

This function is a continuous trigonometric function, and has the value \(\sin (0)=0\) at the origin. Thus

\[\lim _{x \rightarrow 0} f(x)=\lim _{x \rightarrow 0} \sin (x)=0 . \nonumber \]

Power functions are continuous everywhere. This motivates the next example.

Example D.3

Compute the limit \(\lim _{x \rightarrow 0} x^{n}\) where \(n\) is a positive integer.

Solution

The function in question, \(f(x)=x^{n}\), is a simple power function that is continuous everywhere. Further, \(f(0)=0\) for any \(n\) a positive integer. Hence the limit as \(x \rightarrow 0\) coincides with the value of the function at that point, so

\[\lim _{x \rightarrow 0} x^{n}=0 . \nonumber \]

D.2 Properties of limits

Suppose we are given two functions, \(f(x)\) and \(g(x)\). we also assume that both functions have (finite) limits at the point \(x=a\). Then the following statements follow.

1. \(\lim _{x \rightarrow a}(f(x)+g(x))=\lim _{x \rightarrow a} f(x)+\lim _{x \rightarrow a} g(x)\)
2. \(\lim _{x \rightarrow a}(c f(x))=c \lim _{x \rightarrow a} f(x)\)
3. \(\lim _{x \rightarrow a}(f(x) \cdot g(x))=\left(\lim _{x \rightarrow a} f(x)\right) \cdot\left(\lim _{x \rightarrow a} g(x)\right)\)
4. Provided that \(\lim _{x \rightarrow a} g(x) \neq 0\), we also have that

\[\lim _{x \rightarrow a}\left(\frac{f(x)}{g(x)}\right)=\left(\frac{\lim _{x \rightarrow a} f(x)}{\lim _{x \rightarrow a} g(x)}\right) . \nonumber \]

The first two statements are equivalent to linearity of the process of computing a limit.

Example D.4

Find \(\lim _{x \rightarrow 2} f(x)\) for the function \(y=f(x)=2 x^{2}-x^{3}\).

Solution

Since this function is a polynomial, and so continuous everywhere, we can simply plug in the relevant value of \(x\), i.e.

\[\lim _{x \rightarrow 2}\left(2 x^{2}-x^{3}\right)=2 \cdot 2^{2}-2^{3}=0 \nonumber \]

Thus when \(x\) gets closer to 2 , the value of the function gets closer to 0 .

Note: when the function is continuous, the value of the limit is the same as the value of the function at the given point.

D.3 Limits of rational functions

Case 1: Denominator nonzero

We first consider functions that are the quotient of two polynomials, \(y=\) \(f(x) / g(x)\) at points were \(g(x) \neq 0\). This allows us to apply Property 4 of limits together with what we have learned about the properties of power functions and polynomials. Much of this discussion is related to the properties of power functions and dominance of lower (higher) powers at small (large) values of \(x\), as discussed in Chapter 1 . In the examples below, we consider both limits at the origin (at \(x=0\) ) and at infinity (for \(x \rightarrow \infty\) ). The latter means "very large \(x\) ". See Section 1.4 for examples of the informal version of the same reasoning used to reach the same conclusions.

Example D.5

Find the limit as \(x \rightarrow 0\) and as \(x \rightarrow \infty\) of the quotients

1. \(\frac{K x}{k_{n}+x}\)
2. \(\frac{A x^{n}}{a^{n}+x^{n}}\)

Solution

We recognize (a) as an example of the Michaelis-Menten kinetics, found in (1.8) and (b) as a Hill function in (1.7) of Chapter 1. We now compute, first for \(x \rightarrow 0\),

(a) \(\lim _{x \rightarrow 0} \frac{K x}{k_{n}+x}=0\) (b) \(\lim _{x \rightarrow 0} \frac{A x^{n}}{a^{n}+x^{n}}=0\).

This follows from the fact that, provided \(a, k_{n} \neq 0\), both functions are continuous at \(x=0\), so that their limits are the same as the actual values attained by the functions. Now for \(x \rightarrow \infty\) (a) \(\lim _{x \rightarrow \infty} \frac{K x}{k_{n}+x}=\lim _{x \rightarrow \infty} \frac{K x}{x}=K\),

(b) \(\lim _{x \rightarrow \infty} \frac{A x^{n}}{a^{n}+x^{n}}=\lim _{x \rightarrow \infty} \frac{A x^{n}}{x^{n}}=A\).

This follows from the fact that the constants \(k_{n}, a^{n}\) are always "swapped out" by the value of \(x\) as \(x \rightarrow \infty\), allowing us to obtain the result. Other than the formal limit notation, there is nothing new here that we have not already discussed in Sections 1.5.

Below we apply similar reasoning to other examples of rational functions.

Example D.6

Find the limit as \(x \rightarrow 0\) and as \(x \rightarrow \infty\) of the quotients

1. \(\frac{3 x^{2}}{9+x^{2}}\)
2. \(\frac{1+x}{1+x^{3}}\)

Solution

For part (a) we note that as \(x \rightarrow \infty\), the quotient approaches \(3 x^{2} / x^{2}=\) 3. As \(x \rightarrow 0\), both numerator and denominator are defined and the denominator is nonzero, so we can use the 4th property of limits. We thus find that

(a) \(\lim _{x \rightarrow \infty} \frac{3 x^{2}}{9+x^{2}}=3, \quad \lim _{x \rightarrow 0} \frac{3 x^{2}}{9+x^{2}}=0\)

For part (b), we use the fact that as \(x \rightarrow \infty\), the limit approaches \(x / x^{3}=x^{-2} \rightarrow\) 0 . As \(x \rightarrow 0\) we can apply property 4 yet again to compute the (finite) limit, so that

(b) \[\lim _{x \rightarrow \infty} \frac{1+x}{1+x^{3}}=0, \quad \lim _{x \rightarrow 0} \frac{1+x}{1+x^{3}}=1 \text {. } \nonumber \]

Example D.7

Find the limits of the following function at 0 and \(\infty\)

\[y=\frac{x^{4}-3 x^{2}+x-1}{x^{5}+x} . \nonumber \]

Solution

For \(x \rightarrow \infty\) powers with the largest power dominate, whereas for \(x \rightarrow 0\), smaller powers dominate. Hence, we find

\[\begin{gathered} \lim _{x \rightarrow \infty} \frac{x^{4}-3 x^{2}+x-1}{x^{5}+x}=\lim _{x \rightarrow \infty} \frac{x^{4}}{x^{5}}=\lim _{x \rightarrow \infty} \frac{1}{x}=0 . \\ \lim _{x \rightarrow 0} \frac{x^{4}-3 x^{2}+x-1}{x^{5}+x}=\lim _{x \rightarrow 0} \frac{-1}{x}=-\lim _{x \rightarrow 0} \frac{1}{x}=\infty \end{gathered} \nonumber \]

So in the latter case, the limit does not exist.

Case 2: zero in the denominator and "holes" in a graph

In the previous examples, evaluating the limit, where it existed, was as simple as plugging the appropriate value of \(x\) into the function itself. The next example shows that this is not always possible.

Example D.8

Compute the limit as \(x \rightarrow 4\) of the function \(f(x)=1 /(x-4)\)

Solution

This function has a vertical asymptote at \(x=4\). Indeed, the value of the function shoots off to \(+\infty\) if we approach \(x=4\) from above, and \(-\infty\) if we approach the same point from below. We say that the limit does not exist in this case.

Example D.9

Compute the limit as \(x \rightarrow-1\) of the function \(f(x)=x /\left(x^{2}-1\right)\)

Solution

We compute

\[\lim _{x \rightarrow-1} \frac{x}{x^{2}-1}=\lim _{x \rightarrow-1} \frac{x}{(x-1)(x+1)} \nonumber \]

It is evident (even before factoring as we have done) that this function has a vertical asymptote at \(x=-1\) where the denominator approaches zero. Hence, the limit does not exist.

Next, we describe an extremely important example where the function has a "hole" in its graph, but where a finite limit exists. This kind of limit plays a huge role in the definition of a derivative.

Example D.10

Find \(\lim _{x \rightarrow 2} f(x)\) for the function \(y=(x-2) /\left(x^{2}-4\right)\).

Solution

This function is a quotient of two rational expressions \(f(x) / g(x)\) but we note that \(\lim _{x \rightarrow 2} g(x)=\lim _{x \rightarrow 2}\left(x^{2}-4\right)=0\). Thus we cannot use property 4 directly. However, we can simplify the quotient by observing that for \(x \neq 2\) the function \(y=(x-2) /\left(x^{2}-4\right)=(x-2) /(x-2)(x+2)\) takes on the same values as the expression \(1 /(x+2)\). At the point \(x=2\), the function itself is not defined, since we are not allowed division by zero. However, the limit of this function does exist:

\[\lim _{x \rightarrow 2} f(x)=\lim _{x \rightarrow 2} \frac{(x-2)}{\left(x^{2}-4\right)} . \nonumber \]

Provided \(x \neq 2\) we can factor the denominator and cancel:

\[\lim _{x \rightarrow 2} \frac{(x-2)}{\left(x^{2}-4\right)}=\lim _{x \rightarrow 2} \frac{(x-2)}{(x-2)(x+2)}=\lim _{x \rightarrow 2} \frac{1}{(x+2)} \nonumber \]

Now we can substitute \(x=2\) to obtain

\[\lim _{x \rightarrow 2} f(x)=\frac{1}{(2+2)}=\frac{1}{4} \nonumber \]

Figure D.1: The function \(y=\frac{(x-2)}{\left(x^{2}-4\right)}\) has a "hole" in its graph at \(x=2\). The limit of the function as \(x\) approaches 2 does exist, and "supplies the missing point": \(\lim _{x \rightarrow 2} f(x)=\frac{1}{4}\).

Example D.11

Compute the limit

\[\lim _{h \rightarrow 0} \frac{K(x+h)^{2}-K x^{2}}{h} . \nonumber \]

Solution

This is a calculation we would perform to compute the derivative of the function \(y=K x^{2}\) from the definition of the derivative. Details have already been displayed in Example 2.10. The essential idea is that we expand the numerator and simplify algebraically as follows:

\[\lim _{h \rightarrow 0} K \frac{\left(2 x h+h^{2}\right)}{h}=\lim _{h \rightarrow 0} K(2 x+h)=2 K x . \nonumber \]

Even though the quotient is not defined at the value \(h=0\) (as the denominator is zero there), the limit exists, and hence the derivative can be defined.

See also Example \(3.12\) for a similar calculation for the function \(K x^{3}\).

D.4 Right and left sided limits

Some functions are discontinuous at a point, but we may still be able to define a limit that the function attains as we approach that point from the right or from the left. (This is equivalent to gradually decreasing or gradually increasing \(x\) as we get closer to the point of interest.

Consider the function

\[f(x)= \begin{cases}0 & \text { if } x<0 \\ 1 & \text { if } x>0\end{cases} \nonumber \]

This is a step function, whose values is 0 for negative real numbers, and 1 for positive real numbers. The function is not even defined at the point \(x=0\) and has a jump in its graph. However, we can still define a right and a left limit as follows:

\[\lim _{x \rightarrow+0} f(x)=1, \quad \lim _{x \rightarrow-0} f(x)=0 . \nonumber \]

That is, the limit as we approach from the right is 0 whereas from the left it is 1 . We also state the following result:

If \(f(x)\) has a right and a left limit at a point \(x=a\) and if those limits are equal, then we say that the limit at \(x=a\) exists, and we write

\[\lim _{x \rightarrow+\infty} f(x)=\lim _{x \rightarrow-a} f(x)=\lim _{x \rightarrow a} f(x) \nonumber \]

Example D.12

Find \(\lim _{x \rightarrow \pi / 2} f(x)\) for the function \(y=f(x)=\tan (x)\).

Solution

The function \(\tan (x)=\sin (x) / \cos (x)\) cannot be continuous at \(x=\) \(\pi / 2\) because \(\cos (x)\) in the denominator takes on the value of zero at the point \(x=\pi / 2\). Moreover, the value of this function becomes unbounded (grows without a limit) as \(x \rightarrow \pi / 2\). We say in this case that "the limit does not exist". We sometimes use the notation

\[\lim _{x \rightarrow \pi / 2} \tan (x)=\pm \infty . \nonumber \]

(We can distinguish the fact that the function approaches \(+\infty\) as \(x\) approaches \(\pi / 2\) from below, and \(-\infty\) as \(x\) approaches \(\pi / 2\) from higher values.)

D.5 Limits at infinity

We can also describe the behavior "at infinity" i.e. the trend displayed by a function for very large (positive or negative) values of \(x\). We consider a few examples of this sort below.

Example D.13

Find \(\lim _{x \rightarrow \infty} f(x)\) for the function \(y=f(x)=x^{3}-x^{5}+x\).

Solution

All polynomials grow in an unbounded way as \(x\) tends to very large values. We can determine whether the function approaches positive or negative unbounded values by looking at the coefficient of the highest power of \(x\), since that power dominates at large \(x\) values. In this example, we find that the term \(-x^{5}\) is that highest power. Since this has a negative coefficient, the function approaches unbounded negative values as \(x\) gets larger in the positive direction, i.e.

\[\lim _{x \rightarrow \infty} x^{3}-x^{5}+x=\lim _{x \rightarrow \infty}-x^{5}=-\infty . \nonumber \]

Example D.14

Determine the following two limits:

1. \(\lim _{x \rightarrow \infty} e^{-2 x}\)
2. \(\lim _{x \rightarrow-\infty} e^{5 x}\)

Solution

The function \(y=e^{-2 x}\) becomes arbitrarily small as \(x \rightarrow \infty\). The function \(y=e^{5 x}\) becomes arbitrarily small as \(x \rightarrow-\infty\). Thus we have

(a) \(\lim _{x \rightarrow \infty} e^{-2 x}=0\)

(b) \(\lim _{x \rightarrow-\infty} e^{5 x}=0\)

Example D.15

Find the limits below:

1. \(\lim _{x \rightarrow \infty} x^{2} e^{-2 x}\)
2. \(\lim _{x \rightarrow 0} \frac{1}{x} e^{-x}\)

Solution

For part (a) we state here the fact that as \(x \rightarrow \infty\), the exponential function with negative exponent decays to zero faster than any power function increases.

For part (b) we note that for the quotient \(e^{-x} / x\) we have that as \(x \rightarrow 0\) the top satisfies \(e^{-x} \rightarrow e^{0}=1\), while the denominator has \(x \rightarrow 0\). Thus the limit at \(x \rightarrow 0\) cannot exist. We find that (a) \(\lim _{x \rightarrow \infty} x^{2} e^{-2 x}=0\) (b) \(\lim _{x \rightarrow 0} \frac{1}{x} e^{-x}=\infty\)

D.6 Summary of special limits

As a reference, in the table below, we collect some of the special limits that are useful in a variety of situations.

Table D.1: A collection of useful limits.

Function	\(x \rightarrow\)	Limit notation	Value
\(e^{-a x}, a>0\)	\(\infty\)		\(0\)
\(e^{-a x}, a>0\)	\(-\infty\)		\(\infty\)
\(e^{a x}, a>0\)	\(\infty\)		\(\infty\)
\(e^{k x}\)	\(0\)		\(1\)
\(x^n e^{-a x}, a>0\)	\(\infty\)		\(0\)
\(\ln (a x), a>0\)	\(\infty\)		\(\infty\)
\(\ln (a x), a>0\)	\(1\)		\(0\)
\(\ln (a x), a>0\)	\(0\)		\(-\infty\)
\(x \ln (a x), a>0\)	\(0\)		\(0\)
\(\frac{\ln (a x)}{x}, a>0\)	\(\infty\)		\(0\)
\(\frac{\sin (x)}{x}\)	\(0\)		\(1\)

We can summarize the information in this table informally as follows:

1. The exponential function \(e^{x}\) grows faster than any power function as \(x\) increases, and conversely the function \(e^{-x}=1 / e^{x}\) decreases faster than any power of \((1 / x)\) as \(x\) grows. The same is true for \(e^{a x}\) provided \(a>0\).
2. The logarithm \(\ln (x)\) is an increasing function that keeps growing without bound as \(x\) increases, but it does not grow as rapidly as the function \(y=x\). The same is true for \(\ln (a x)\) provided \(a>0\). The logarithm is not defined for negative values of its argument and as \(x\) approaches zero, this function becomes unbounded and negative. However, it approaches \(-\infty\) more slowly than \(x\) approaches 0 . For this reason, the expression \(x \ln (x)\) has a limit of 0 as \(x \rightarrow 0\).