
## Section3.6Using Derivatives to Evaluate Limits

###### Motivating Questions
• How can derivatives be used to help us evaluate indeterminate limits of the form $\frac{0}{0}\text{?}$

• What does it mean to say that $\lim\limits_{x \to \infty} f(x) = L$ and $\lim\limits_{x \to a} f(x) = \infty\text{?}$

• How can derivatives assist us in evaluating indeterminate limits of the form $\frac{\infty}{\infty}\text{?}$

Because differential calculus is based on the definition of the derivative, and the definition of the derivative involves a limit, there is a sense in which all of calculus rests on limits. Limits in which separately evaluating the numerator and denominator at the limit point results in the fraction $\frac{0}{0}$ are called indeterminate form. When we come across a limit like this, we have some work to do. Remember, saying that a limit has an indeterminate form only means that we don't yet know its value and have more work to do: indeed, limits of the form $\frac{0}{0}$ can take on any value.

We have learned many techniques for evaluating the limits that arise from applying the definition of the derivative, including a large number of shortcut rules that seem to bypass the use of limits entirely. In this section, we will explore how to use derivatives to evaluate certain limits that we previously could not compute.

###### Example3.57

Let $h$ be the function given by $h(x) = \frac{x^5 + x - 2}{x^2 - 1}\text{.}$

1. Evaluate $f(x) = x^5 + x - 2$ and $g(x) = x^2 - 1$ at $x=1\text{.}$ Can you evaluate $\lim\limits_{x \rightarrow 1}\frac{x^5 + x - 2}{x^2 - 1}$ by just plugging $x=1$ into the quotient?

2. Next we will investigate the behavior of both the numerator and denominator of $h$ near the point where $x = 1\text{.}$ Let $f(x) = x^5 + x - 2$ and $g(x) = x^2 - 1\text{.}$ Find the local linearizations of $f$ and $g$ at $a = 1\text{,}$ and call these functions $L_f(x)$ and $L_g(x)\text{,}$ respectively.

3. Explain why $h(x) \approx \frac{L_f(x)}{L_g(x)}$ for $x$ near $a = 1\text{.}$

4. Using your work from (b), evaluate

\begin{equation*} \lim_{x \to 1} \frac{L_f(x)}{L_g(x)}\text{.} \end{equation*}

What do you think your result tells us about $\lim\limits_{x \to 1} h(x)\text{?}$

5. Investigate the function $h(x)$ graphically and numerically near $x = 1\text{.}$ What do you think is the value of $\lim\limits_{x \to 1} h(x)\text{?}$

Hint
1. Is $h$ continuous at $x=1\text{?}$ Why (not)?

2. Remember that the local linearization of a function at a point is just the tangent line to the function at that point.

3. What can you say about the relationship between $f$ and $L_f$ near $a=1\text{?}$

4. Based on (c), how should $\lim\limits_{x\to1}\frac{L_f(x)}{L_g(x)}$ compare to $\lim\limits_{x\to1}h(x)\text{?}$

5. You can also investigate $h(x)$ algebraically by factoring the numerator and denominator (with a common factor of $x-1$).

1. $f(1)=0=g(1)\text{.}$ $\lim\limits_{x\to1}h(x)$ has indeterminate form $\frac00\text{.}$

2. $L_f(x)=6(x-1)\text{;}$ $L_g(x)=2(x-1)\text{.}$

3. Near $a=1\text{,}$ we have $L_f(x)\approx f(x)$ and $L_g(x)\approx g(x)\neq0\text{.}$ Hence, near $a=1\text{,}$ $\frac{L_f(x)}{L_g(x)}\approx\frac{f(x)}{g(x)}=h(x)\text{.}$

4. $\lim\limits_{x\to1}\frac{L_f(x)}{L_g(x)}=3\text{.}$

5. $\lim\limits_{x\to1}h(x)=3\text{.}$

Solution
1. Since $f(x)=x^5+x-2\text{,}$ we have $f(1)=1+1-2=0\text{.}$ Likewise, $g(x)=x^2-1\text{,}$ so $g(1)=1-1=0\text{.}$ As both $f(1)=0$ and $g(1)=0\text{,}$ we can not say $\lim\limits_{x\to1}\frac{f(x)}{g(x)}=\frac{f(1)}{g(1)}\text{,}$ because the right side of this equation is not a well-defined number. If only the denominator were $0\text{,}$ we could say the limit was infinite; if only the numerator were $0\text{,}$ we could say the limit was also $0\text{.}$ Instead, we are left with an indeterminate form of $\frac00\text{,}$ which could be zero or infinite or anywhere in between.

2. Recall that the local linearization of a function $y(x)$ at a point $x=a$ is given by the formula

\begin{equation*} L_y(x)=y'(a)(x-a)+y(a)\text{.} \end{equation*}

Since $f(1)=0$ and $f'(x)=5x^4+1\text{,}$ then $f'(1)=6$ and

\begin{equation*} L_f(x)=6(x-1)\text{.} \end{equation*}

Since $g(1)=0$ and $g'(x)=2x\text{,}$ then $g'(1)=2$ and

\begin{equation*} L_g(x)=2(x-1)\text{.} \end{equation*}
3. $L_f$ and $L_g$ are the tangent line approximations for $f$ and $g\text{,}$ respectively, near the point $x=1\text{.}$ Because $f$ and $g$ are each differentiable and hence locally linear, this means $L_f(x)\approx f(x)$ for $x$ near $1\text{,}$ and $L_g(x)\approx g(x)$ for $x$ near $1\text{.}$ Thus for values of $x$ that are close to $1$ and are not roots of $g(x)\text{,}$ we can approximate the quotient $h(x)=\frac{f(x)}{g(x)}$ by $\frac{L_f(x)}{L_g(x)}\text{.}$ Finally, since the only roots of $g$ are $\pm1\text{,}$ we can thus say: for $x\neq1$ near $a=1\text{,}$

\begin{equation*} h(x)\approx\frac{L_f(x)}{L_g(x)}\text{.} \end{equation*}
4. Notice that

\begin{equation*} \lim_{x\to1}\frac{L_f(x)}{L_g(x)}=\lim_{x\to1}\frac{6(x-1)}{2(x-1)}=\lim_{x\to1}\frac62=3\text{.} \end{equation*}

By our argument in (c), we know $h(x)\approx\frac{L_f(x)}{L_g(x)}$ near $x=1\text{.}$ It should thus be the case that $\lim\limits_{x\to1}h(x)=\lim\limits_{x\to1}\frac{L_f(x)}{L_g(x)}=3\text{.}$

5. Looking at a graph of $y=h(x)\text{,}$ we note that there is a hole at $(1,3)\text{,}$ implying that $x=1$ is a removable discontinuity, and $\lim\limits_{x\to1}h(x)=3\text{.}$ With a numerical approach, we see that $h(0.99)\approx2.965\text{,}$ $h(0.999)\approx2.997\text{,}$ $h(1.01)\approx3.035\text{,}$ and $h(1.001)\approx3.0035\text{.}$ These values again suggest that $\lim\limits_{h\to1}h(x)=3\text{.}$ Finally, we note that $h$ is a rational function and that $1$ is a root of both its numerator and its denominator. It follows that $x-1$ is a factor of both the numerator and denominator, and in fact

\begin{equation*} h(x)=\frac{(x^4+x^3+x^2+x+2)(x-1)}{(x+1)(x-1)}\text{.} \end{equation*}

Hence, we can actually compute $\lim\limits_{x\to1}h(x)$ algebraically (even though it wasn't clear before that we could), and thus confirm that

\begin{equation*} \lim_{x\to1}h(x)=\lim_{x\to1}\frac{(x^4+x^3+x^2+x+2)}{(x+1)}=\frac62=3\text{.} \end{equation*}

### SubsectionUsing Derivatives to Evaluate Indeterminate Limits of the Form $\frac{0}{0}$

The idea demonstrated in Example3.57 that we can evaluate an indeterminate limit of the form $\frac{0}{0}$ by replacing each of the numerator and denominator with their local linearizations at the point of interest can be generalized in a way that enables us to evaluate a wide range of limits. Given a function $h(x)$ that can be written as a quotient $h(x) = \frac{f(x)}{g(x)}\text{,}$ where $f$ and $g$ are both differentiable at $x=a$ and for which $f(a) = g(a) = 0\text{,}$ we would like to evaluate the indeterminate limit given by $\lim\limits_{x \to a} h(x)\text{.}$ Below, Figure3.58 illustrates this situation.

In this figure, we see that both $f$ and $g$ have an $x$-intercept at $x = a\text{.}$ Their respective tangent line approximations $L_f$ and $L_g$ at $x = a$ are also shown in the figure. We can take advantage of the fact that a differentiable function and its tangent line approximation become indistinguishable as $x$ approaches $a\text{.}$

First, let's recall that the local linearizations of $f$ and $g$ at $x=a$ are, respectively, $L_f(x) = f'(a)(x-a) + f(a)$ and $L_g(x) = g'(a)(x-a) +g(a)\text{.}$ Because $x$ is getting arbitrarily close to $a$ when we take the limit, we can replace $f$ with $L_f$ and replace $g$ with $L_g\text{,}$ and thus we observe that

\begin{align*} \lim_{x \to a} \frac{f(x)}{g(x)} =\mathstrut \amp \lim_{x \to a} \frac{L_f(x)}{L_g(x)}\\ =\mathstrut \amp \lim_{x \to a} \frac{f'(a)(x-a) + f(a)}{g'(a)(x-a) + g(a)}\text{.} \end{align*}

Next, we remember that both $f(a) = 0$ and $g(a) = 0\text{,}$ which is precisely what makes the original limit indeterminate. Substituting these values for $f(a)$ and $g(a)$ in the limit above, we now have

\begin{align*} \lim_{x \to a} \frac{f(x)}{g(x)} =\mathstrut \amp \lim_{x \to a} \frac{f'(a)(x-a)}{g'(a)(x-a)}\\ =\mathstrut \amp \lim_{x \to a} \frac{f'(a)}{g'(a)}\text{,} \end{align*}

where the latter equality holds because $\frac{x-a}{x-a} = 1$ when $x$ is approaching (but not equal to) $a\text{.}$ Finally, we note that $\frac{f'(a)}{g'(a)}$ is constant with respect to $x\text{,}$ and thus

\begin{equation*} \lim_{x \to a} \frac{f(x)}{g(x)} = \frac{f'(a)}{g'(a)}\text{.} \end{equation*}

This result holds as long as $g'(a)$ is not equal to zero. The formal name of the result is L'Hopital's Rule.

###### L'Hopital's Rule

Let $f$ and $g$ be differentiable at $x=a\text{,}$ and suppose that $f(a) = g(a) = 0$ and that $g'(a) \neq 0\text{.}$ Then

\begin{equation*} \lim_{x \to a} \frac{f(x)}{g(x)} = \frac{f'(a)}{g'(a)}. \end{equation*}

In practice, we typically work with a slightly more general version of L'Hopital's Rule, which states that (under the identical assumptions as the boxed rule above and the extra assumption that $g'$ is continuous at $x=a$)

\begin{equation*} \lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)}\text{,} \end{equation*}

provided the righthand limit exists. This form reflects the basic idea of L'Hopital's Rule: if $\frac{f(x)}{g(x)}$ produces an indeterminate limit of form $\frac{0}{0}$ as $x$ tends to $a\text{,}$ that limit is equivalent to the limit of the quotient of the two functions' derivatives, $\frac{f'(x)}{g'(x)}\text{.}$

For example, we can revisit the limit from Example3.57:

\begin{equation*} \lim_{x \to 1} \frac{x^5 + x - 2}{x^2 - 1}\text{.} \end{equation*}

Applying L'Hopital's Rule to this limit, we again see that

\begin{equation*} \lim_{x \to 1} \frac{x^5 + x - 2}{x^2 - 1} = \lim_{x \to 1} \frac{5x^4 + 1}{2x} = \frac{6}{2} = 3. \\ \end{equation*}

As was the case here, by replacing the numerator and denominator with their respective derivatives, we often replace an indeterminate limit with one whose value we can easily determine.

###### Example3.59

Evaluate each of the following limits. If you use L'Hopital's Rule, indicate where it was used, and be certain its hypotheses are met before you apply it.

1. $\lim\limits_{x \to 0} \frac{\ln(1 + x)}{x}$

2. $\lim\limits_{x \to \pi} \frac{\cos(x)}{x}$

3. $\lim\limits_{x \to 1} \frac{2 \ln(x)}{1-e^{x-1}}$

4. $\lim\limits_{x \to 0} \frac{\sin(x) - x}{\cos(2x)-1}$

Hint
1. Remember that $\ln(1) = 0\text{.}$

2. Note that this limit has $x$ approaching $\pi$ rather than $x$ going to $0\text{.}$

3. Observe that $e^{x-1}$ tends to $1$ as $x$ approaches $1\text{.}$

4. If necessary, L'Hopital's Rule can be applied more than once.

1. $\lim\limits_{x \to 0} \frac{\ln(1 + x)}{x} = 1\text{.}$

2. $\lim\limits_{x \to \pi} \frac{\cos(x)}{x} = -\frac{1}{\pi}\text{.}$

3. $\lim\limits_{x \to 1} \frac{2 \ln(x)}{1-e^{x-1}} = -2\text{.}$

4. $\lim\limits_{x \to 0} \frac{\sin(x) - x}{\cos(2x)-1} = 0\text{.}$

Solution
1. As $x$ tends to $0\text{,}$ we see that $\ln(1+x)$ approaches $\ln(1) = 0\text{,}$ thus this limit has an indeterminate form. By L'Hopital's Rule, we have

\begin{equation*} \lim_{x \to 0} \frac{\ln(1 + x)}{x} = \lim_{x \to 0} \frac{\big(\frac{1}{1 + x}\big)}{1}\text{.} \end{equation*}

As this limit is no longer indeterminate, we may simply allow $x$ to go to $0\text{,}$ and thus we find that

\begin{equation*} \lim_{x \to 0} \frac{\ln(1 + x)}{x} = \frac{\big(\frac{1}{1 + 0}\big)}{1} = 1\text{.} \end{equation*}
2. Observe that

\begin{equation*} \lim_{x \to \pi} \frac{\cos(x)}{x} = \frac{\cos(\pi)}{\pi} = -\frac{1}{\pi}\text{.} \end{equation*}

This limit is not indeterminate because the function $\frac{\cos(x)}{x}$ is continuous at $x = \pi\text{.}$

3. Since $\ln(x)$ goes to $0$ and $e^{x-1}$ goes to $1$ as $x$ approaches $1\text{,}$ this limit is indeterminate with form $\frac{0}{0}\text{.}$ Hence, by L'Hopital's Rule,

\begin{equation*} \lim_{x \to 1} \frac{2 \ln(x)}{1-e^{x-1}} = \lim_{x \to 1} \frac{\big(\frac{2}{x}\big)}{-e^{x-1}}\text{.} \end{equation*}

The updated limit is not indeterminate, so we allow $x$ to approach $1$ and find

\begin{equation*} \lim_{x \to 1} \frac{2 \ln(x)}{1-e^{x-1}} = \frac{\big(\frac{2}{1}\big)}{-e^{0}} = -2\text{.} \end{equation*}
4. Since the given limit is indeterminate of form $\frac{0}{0}\text{,}$ by L'Hopital's Rule we have

\begin{equation*} \lim_{x \to 0} \frac{\sin(x) - x}{\cos(2x)-1} = \lim_{x \to 0} \frac{\cos(x) - 1}{-2\sin(2x)}\text{.} \end{equation*}

Now, as $x$ nears $0\text{,}$ we see that $\cos(x)$ approaches $1$ and $\sin(2x)$ tends to $0\text{,}$ which makes the latest limit also indeterminate in form $\frac{0}{0}\text{.}$ Applying L'Hopital's Rule a second time, we now have

\begin{equation*} \lim_{x \to 0} \frac{\sin(x) - x}{\cos(2x)-1} = \lim_{x \to 0} \frac{-\sin(x)}{-4\cos(2x)}\text{.} \end{equation*}

In the newest limit, we note that $\sin(x)$ tends to $0$ but $\cos(2x)$ tends to $1$ as $x$ approaches $0\text{,}$ so the numerator is tending to 0 while the denominator is approaching $-4\text{.}$ Thus, the value of the limit is determined to be

\begin{equation*} \lim_{x \to 0} \frac{\sin(x) - x}{\cos(2x)-1} = 0\text{.} \end{equation*}

While L'Hopital's Rule can be applied in an entirely algebraic way, it is important to remember that the justification of the rule is graphical: the main idea is that the slopes of the tangent lines to $f$ and $g$ at $x = a$ determine the value of the limit of $\frac{f(x)}{g(x)}$ as $x$ tends to $a\text{.}$ We see this in Figure3.60 below, where we can see from the grid that $f'(a) = 2$ and $g'(a) = -1\text{,}$16This is assuming that the grid has a $1\times1$ scale, which is not clear from the diagram. In fact, as long as the scales on the two axes in Figure3.60 are the same (but not necessarily having one unit per box), then the ratio of $f'(a)$ and $g'(a)$ will be $-2$ as stated below. In general though, it is always important to be sure of the scale of a graph before assuming that of course it's a $1\times1$ grid, because that will not always be the case. hence by L'Hopital's Rule,

\begin{equation*} \lim_{x \to a}\frac{f(x)}{g(x)} = \frac{f'(a)}{g'(a)} = \frac{2}{-1} = -2\text{.} \end{equation*}

It's not the fact that $f$ and $g$ both approach zero that matters most, but rather the rate at which each approaches zero that determines the value of the limit. This is a good way to remember what L'Hopital's Rule says: if $f(a) = g(a) = 0\text{,}$ the the limit of $\frac{f(x)}{g(x)}$ as $x$ approaches $a$ is given by the ratio of the slopes of $f$ and $g$ at $x = a\text{.}$

###### Example3.61

In this example, we reason graphically from the following figure to evaluate limits of ratios of functions about which some information is known.

1. Use the left-hand graph to determine the values of $f(2)\text{,}$ $f'(2)\text{,}$ $g(2)\text{,}$ and $g'(2)\text{.}$ Then, evaluate $\lim\limits_{x \to 2} \frac{f(x)}{g(x)}\text{.}$

2. Use the middle graph to find $p(2)\text{,}$ $p'(2)\text{,}$ $q(2)\text{,}$ and $q'(2)\text{.}$ Then, determine the value of $\lim\limits_{x \to 2} \frac{p(x)}{q(x)}\text{.}$

3. Assume that $r$ and $s$ are functions whose for which $r''(2) \ne 0$ and $s''(2) \ne 0$ Use the right-hand graph to compute $r(2)\text{,}$ $r'(2)\text{,}$ $s(2)\text{,}$ $s'(2)\text{.}$ Explain why you cannot determine the exact value of $\lim\limits_{x \to 2} \frac{r(x)}{s(x)}$ without further information being provided, but that you can determine the sign of $\lim\limits_{x \to 2} \frac{r(x)}{s(x)}\text{.}$ In addition, state what the sign of the limit will be, with justification.

Hint
1. Don't forget that $f'(a)$ measures the slope of the tangent line to $y = f(x)$ at the point $(a,f(a))\text{.}$

2. Do the functions $p$ and $q$ meet the criteria of L'Hopital's Rule?

3. Remember that L'Hopital's Rule can be applied more than once to a particular limit.

1. $\lim\limits_{x \to 2} \frac{f(x)}{g(x)} = \frac{1}{8}\text{.}$

2. $\lim\limits_{x \to 2} \frac{p(x)}{q(x)} = 1\text{.}$

3. $\lim\limits_{x \to 2} \frac{r(x)}{s(x)} \lt 0\text{.}$

Solution
1. From the given graph, we observe that $f(2) = 0\text{,}$ $f'(2) = \frac{1}{2}\text{,}$ $g(2)=0\text{,}$ and $g'(2) = 4\text{.}$ By L'Hopital's Rule,

\begin{equation*} \lim_{x \to 2} \frac{f(x)}{g(x)} = \frac{f'(2)}{g'(2)} = \frac{\frac{1}{2}}{4} = \frac{1}{8}\text{.} \end{equation*}
2. The given graph tells us that $p(2) = 1.5\text{,}$ $p'(2)=-1\text{,}$ $q(2)=1.5\text{,}$ and $q'(2)=0\text{.}$ Note well that the given limit,

\begin{equation*} \lim_{x \to 2} \frac{p(x)}{q(x)}\text{,} \end{equation*}

is not indeterminate, and thus L'Hopital's Rule does not apply. Rather, since $p(x) \to 1.5$ and $q(x) \to 1.5$ as $x \to 2\text{,}$ we have that

\begin{equation*} \lim_{x \to 2} \frac{p(x)}{q(x)} = \frac{p(2)}{q(2)} = \frac{1.5}{1.5} = 1\text{.} \end{equation*}
3. From the third graph, $r(2)=0\text{,}$ $r'(2)=0\text{,}$ $s(2)=0\text{,}$ $s'(2)=0\text{.}$ By L'Hopital's Rule,

\begin{equation*} \lim_{x \to 2} \frac{r(x)}{s(x)} = \lim_{x \to 2} \frac{r'(x)}{s'(x)}\text{,} \end{equation*}

but this limit is still indeterminate, so by L'Hopital's Rule again,

\begin{equation*} \lim_{x \to 2} \frac{r(x)}{s(x)} = \lim_{x \to 2} \frac{r''(x)}{s''(x)} = \frac{r''(2)}{s''(2)}\text{,} \end{equation*}

provided that $s''(2) \ne 0\text{.}$ Since we do not know the values of $r''(2)$ and $s''(2)\text{,}$ we can't determine the actual value of the limit, but from the graph it appears that $r''(2) \gt 0$ (since $r$ is concave up) and that $s''(2) \lt 0$ (because $s$ is concave down), and therefore

\begin{equation*} \lim_{x \to 2} \frac{r(x)}{s(x)} \lt 0\text{.} \end{equation*}

### SubsectionLimits Involving $\infty$

The concept of infinity, denoted $\infty\text{,}$ arises naturally in calculus, as it does in much of mathematics. It is important to note from the outset that $\infty$ is a concept, but not a number itself. Indeed, the notion of $\infty$ naturally invokes the idea of limits. Consider, for example, the function $f(x) = \frac{1}{x}\text{,}$ whose graph is pictured in Figure3.63.

We note that $x = 0$ is not in the domain of $f\text{,}$ so we may naturally wonder what happens as $x \to 0\text{.}$ As $x \to 0^+\text{,}$ we observe that $f(x)$ increases without bound. That is, we can make the value of $f(x)$ as large as we like by taking $x$ closer and closer (but not equal) to 0, while keeping $x \gt 0\text{.}$ This is a good way to think about what infinity represents: a quantity is tending to infinity if there is no single number that the quantity is always less than.

Recall that the statement $\lim\limits_{x \to a} f(x) = L\text{,}$ means that can make $f(x)$ as close to $L$ as we'd like by taking $x$ sufficiently close (but not equal) to $a\text{.}$ We now expand this notation and language to include the possibility that either $L$ or $a$ can be $\infty\text{.}$ For instance, for $f(x) = \frac{1}{x}\text{,}$ we now write

\begin{equation*} \lim_{x \to 0^+} \frac{1}{x} = \infty\text{,} \end{equation*}

by which we mean that we can make $\frac{1}{x}$ as large as we like by taking $x$ sufficiently close (but not equal) to 0. In a similar way, we write

\begin{equation*} \lim_{x \to \infty} \frac{1}{x} = 0\text{,} \end{equation*}

since we can make $\frac{1}{x}$ as close to 0 as we'd like by taking $x$ sufficiently large (i.e., by letting $x$ increase without bound).

In general, the notation $\lim\limits_{x \to a} f(x) = \infty$ means that we can make $f(x)$ as large as we like by taking $x$ sufficiently close (but not equal) to $a\text{,}$ and the notation $\lim\limits_{x \to \infty} f(x) = L$ means that we can make $f(x)$ as close to $L$ as we like by taking $x$ sufficiently large. This notation also applies to left- and right-hand limits, and to limits involving $-\infty\text{.}$ For example, returning to Figure3.63 and $f(x) = \frac{1}{x}\text{,}$ we can say that

\begin{equation*} \lim_{x \to 0^-} \frac{1}{x} = -\infty \ \ \text{and} \ \ \lim_{x \to -\infty} \frac{1}{x} = 0\text{.} \end{equation*}

Finally, we write

\begin{equation*} \lim_{x \to \infty} f(x) = \infty \end{equation*}

if we can make the value of $f(x)$ as large as we'd like by taking $x$ sufficiently large. For example,

\begin{equation*} \lim_{x \to \infty} x^2 = \infty\text{.} \end{equation*}

Limits involving infinity identify vertical and horizontal asymptotes of a function. If $\lim\limits_{x \to a} f(x) = \infty\text{,}$ then $x = a$ is a vertical asymptote of $f\text{,}$ while if $\lim\limits_{x \to \infty} f(x) = L\text{,}$ then $y = L$ is a horizontal asymptote of $f\text{.}$ Similar statements can be made using $-\infty\text{,}$ and with left- and right-hand limits as $x \to a^-$ or $x \to a^+\text{.}$

In precalculus classes, it is common to study the end behavior of certain families of functions, by which we mean the behavior of a function as $x \to \infty$ and as $x \to -\infty\text{.}$ Here we briefly examine some familiar functions and note the values of several limits involving $\infty\text{.}$

For the natural exponential function $e^x\text{,}$ we note that $\lim\limits_{x \to \infty} e^x = \infty$ and $\lim\limits_{x \to -\infty} e^x = 0\text{.}$ For the exponential decay function $e^{-x}\text{,}$ these limits are reversed, with $\lim\limits_{x \to \infty} e^{-x} = 0$ and $\lim\limits_{x \to -\infty} e^{-x} = \infty\text{.}$ Turning to the natural logarithm function, we have $\lim\limits_{x \to 0^+} \ln(x) = -\infty$ and $\lim\limits_{x \to \infty} \ln(x) = \infty\text{.}$ While both $e^x$ and $\ln(x)$ grow without bound as $x \to \infty\text{,}$ the exponential function does so much more quickly than the logarithm function does. We'll soon use limits to quantify what we mean by quickly.

For polynomial functions of the form

\begin{equation*} p(x) = a_n x^n + a_{n-1}x^{n-1} + \cdots a_1 x + a_0\text{,} \end{equation*}

the end behavior depends on the sign of $a_n$ and whether the highest power $n$ is even or odd. If $n$ is even and $a_n$ is positive, then $\lim\limits_{x \to \infty} p(x) = \infty$ and $\lim\limits_{x \to -\infty} p(x) = \infty\text{,}$ as in the plot of $g$ in Figure3.64. If instead $a_n$ is negative, then $\lim\limits_{x \to \infty} p(x) = -\infty$ and $\lim\limits_{x \to -\infty} p(x) = -\infty\text{.}$ In the situation where $n$ is odd, then either $\lim\limits_{x \to \infty} p(x) = \infty$ and $\lim\limits_{x \to -\infty} p(x) = -\infty$ (which occurs when $a_n$ is positive, as in the graph of $f$ in Figure3.64), or $\lim\limits_{x \to \infty} p(x) = -\infty$ and $\lim\limits_{x \to -\infty} p(x) = \infty$ (when $a_n$ is negative).

A function can fail to have a limit as $x \to \infty\text{.}$ For example, consider the plot of the sine function at right in Figure3.64. Because the function continues oscillating between $-1$ and $1$ as $x \to \infty\text{,}$ we say that $\lim\limits_{x \to \infty} \sin(x)$ does not exist.

Finally, it is straightforward to analyze the behavior of any rational function as $x \to \infty\text{.}$

###### Example3.65

Determine the limit of the function

\begin{equation*} q(x) = \frac{3x^2 - 4x + 5}{7x^2 + 9x - 10} \end{equation*}

as $x \to \infty\text{.}$

Note that both $(3x^2 - 4x + 5) \to \infty$ as $x \to \infty$ and $(7x^2 + 9x - 10) \to \infty$ as $x \to \infty\text{.}$ Here we say that $\lim\limits_{x \to \infty} q(x)$ has indeterminate form $\frac{\infty}{\infty}\text{.}$ We can determine the value of this limit through a standard algebraic approach. Multiplying the numerator and denominator each by $\frac{1}{x^2}\text{,}$ we find that

\begin{align*} \lim_{x \to \infty} q(x) =\mathstrut \amp \lim_{x \to \infty} \frac{3x^2 - 4x + 5}{7x^2 + 9x - 10} \cdot \frac{\frac{1}{x^2}}{\frac{1}{x^2}}\\ =\mathstrut \amp \lim_{x \to \infty} \frac{3 - 4\frac{1}{x} + 5\frac{1}{x^2}}{7 + 9\frac{1}{x} - 10\frac{1}{x^2}} = \frac{3}{7} \end{align*}

since $\frac{1}{x^2} \to 0$ and $\frac{1}{x} \to 0$ as $x \to \infty\text{.}$ This shows that the rational function $q$ has a horizontal asymptote at $y = \frac{3}{7}\text{.}$ A similar approach can be used to determine the limit of any rational function as $x \to \infty\text{.}$

But how should we handle a limit such as

\begin{equation*} \lim_{x \to \infty} \frac{x^2}{e^x}? \end{equation*}

Here, both $x^2 \to \infty$ and $e^x \to \infty\text{,}$ but there is not an obvious algebraic approach that enables us to find the limit's value. Fortunately, it turns out that L'Hopital's Rule extends to cases involving infinity.

###### L'Hopital's Rule ($\infty$)

If $f$ and $g$ are differentiable and both approach zero or both approach $\pm \infty$ as $x \to a$ (where $a$ is allowed to be $\infty$) , then

\begin{equation*} \lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)}\text{.} \end{equation*}

(To be technically correct, we need to add the additional hypothesis that $g'(x) \ne 0$ on an open interval that contains $a$ or in every neighborhood of infinity if $a$ is $\infty\text{;}$ this is almost always met in practice.)

To evaluate $\lim\limits_{x \to \infty} \frac{x^2}{e^x}\text{,}$ we can apply L'Hopital's Rule, since both $x^2 \to \infty$ and $e^x \to \infty\text{.}$ Doing so, it follows that

\begin{equation*} \lim_{x \to \infty} \frac{x^2}{e^x} = \lim_{x \to \infty} \frac{2x}{e^x}\text{.} \end{equation*}

This updated limit is still indeterminate and of the form $\frac{\infty}{\infty}\text{,}$ but it is simpler since $2x$ has replaced $x^2\text{.}$ Hence, we can apply L'Hopital's Rule again, and find that

\begin{equation*} \lim_{x \to \infty} \frac{x^2}{e^x} = \lim_{x \to \infty} \frac{2x}{e^x} = \lim_{x \to \infty} \frac{2}{e^x}\text{.} \end{equation*}

Now, since $2$ is constant and $e^x \to \infty$ as $x \to \infty\text{,}$ it follows that $\frac{2}{e^x} \to 0$ as $x \to \infty\text{,}$ which shows that

\begin{equation*} \lim_{x \to \infty} \frac{x^2}{e^x} = 0\text{.} \end{equation*}
###### Example3.66

Evaluate each of the following limits. If you use L'Hopital's Rule, indicate where it was used, and be certain its hypotheses are met before you apply it.

1. $\lim\limits_{x \to \infty} \frac{x}{\ln(x)}$

2. $\lim\limits_{x \to \infty} \frac{e^{x} + x}{2e^{x} + x^2}$

3. $\lim\limits_{x \to 0^+} \frac{\ln(x)}{\frac{1}{x}}$

4. $\lim\limits_{x \to \frac{\pi}{2}^-} \frac{\tan(x)}{x-\frac{\pi}{2}}$

Hint
1. Remember that $\ln(x) \to \infty$ as $x \to \infty\text{.}$

2. Both the numerator and denominator tend to $\infty$ as $x \to \infty\text{.}$

3. Note that $x \to 0^+\text{,}$ not $\infty\text{.}$

4. As $x \to \frac{\pi}{2}^-\text{,}$ $\tan(x) \to \infty\text{.}$

1. $\lim\limits_{x \to \infty} \frac{x}{\ln(x)} = \infty\text{.}$

2. $\lim\limits_{x \to \infty} \frac{e^{x} + x}{2e^{x} + x^2} = \frac{1}{2}\text{.}$

3. $\lim\limits_{x \to 0^+} \frac{\ln(x)}{\frac{1}{x}} = 0\text{.}$

4. $\lim\limits_{x \to \frac{\pi}{2}^-} \frac{\tan(x)}{x-\frac{\pi}{2}} = -\infty\text{.}$

Solution
1. As both numerator and denominator tend to $\infty$ as $x \to \infty\text{,}$ by L'Hopital's Rule followed by some elementary algebra,

\begin{equation*} \lim_{x \to \infty} \frac{x}{\ln(x)} = \lim_{x \to \infty} \frac{1}{\frac{1}{x}} = \lim_{x \to \infty} x = \infty\text{.} \end{equation*}
2. Because this limit has indeterminate form $\frac{\infty}{\infty}\text{,}$ L'Hopital's Rule tells us that

\begin{equation*} \lim_{x \to \infty} \frac{e^{x} + x}{2e^{x} + x^2} = \lim_{x \to \infty} \frac{e^{x} + 1}{2e^{x} + 2x}\text{.} \end{equation*}

The latest limit is indeterminate for the same reason, and a second application of the rule shows

\begin{equation*} \lim_{x \to \infty} \frac{e^{x} + x}{2e^{x} + x^2} = \lim_{x \to \infty} \frac{e^{x}}{2e^{x} + 2}\text{.} \end{equation*}

Note how each application of the rule produces a simpler numerator and denominator. With one more use of L'Hopital's Rule, followed by a simple algebraic simplification, we have

\begin{equation*} \lim_{x \to \infty} \frac{e^{x} + x}{2e^{x} + x^2} = \lim_{x \to \infty} \frac{e^{x}}{2e^{x}} = \lim_{x \to \infty} \frac{1}{2} = \frac{1}{2}\text{.} \end{equation*}
3. As $x \to 0^+\text{,}$ $\ln(x) \to -\infty$ and $\frac{1}{x} \to +\infty\text{,}$ thus by L'Hopital's Rule,

\begin{equation*} \lim_{x \to 0^+} \frac{\ln(x)}{\frac{1}{x}} = \lim_{x \to 0^+} \frac{\frac{1}{x}}{-\frac{1}{x^2}}\text{.} \end{equation*}

Reciprocating, multiplying, and simplifying, it follows that

\begin{equation*} \lim_{x \to 0^+} \frac{\ln(x)}{\frac{1}{x}} = \lim_{x \to 0^+} \frac{1}{x}\cdot \frac{x^2}{-1} = \lim_{x \to 0^+} -x = 0\text{.} \end{equation*}
4. Here, the numerator tends to $\infty$ while the denominator tends to $0^-\text{.}$ Note well that this limit is not indeterminate, but rather produces a collection of fractions with large positive numerators and small negative denominators. Hence

\begin{equation*} \lim_{x \to \frac{\pi}{2}^-} \frac{\tan(x)}{x-\frac{\pi}{2}} = -\infty\text{.} \end{equation*}

In particular, we observe that L'Hopital's Rule is not applicable here.

To evaluate the limit of a quotient of two functions $\frac{f(x)}{g(x)}$ that results in an indeterminate form of $\frac{\infty}{\infty}\text{,}$ in essence we are asking which function is growing faster without bound. We say that the function $g$ dominates the function $f$ as $x \to \infty$ provided that

\begin{equation*} \lim_{x \to \infty} \frac{f(x)}{g(x)} = 0\text{,} \end{equation*}

whereas $f$ dominates $g$ provided that $\lim\limits_{x \to \infty} \frac{f(x)}{g(x)} = \infty\text{.}$ Finally, if the value of $\lim\limits_{x \to \infty} \frac{f(x)}{g(x)}$ is finite and nonzero, we say that $f$ and $g$ grow at the same rate. For example, we saw that $\lim\limits_{x \to \infty} \frac{x^2}{e^x} = 0\text{,}$ so $e^x$ dominates $x^2\text{,}$ while $\lim\limits_{x \to \infty} \frac{3x^2 - 4x + 5}{7x^2 + 9x - 10} = \frac{3}{7}\text{,}$ so $f(x) = 3x^2 - 4x + 5$ and $g(x) = 7x^2 + 9x - 10$ grow at the same rate.

### SubsectionOther Indeterminate Forms

While the main two indeterminate forms are $\frac{0}{0}$ and $\frac{\infty}{\infty}\text{,}$ there are other indeterminate forms which can be rewritten in order to use L'Hopital's Rule.

The following example is a classic use of L'Hopital's Rule.

###### Example3.67

Evaluate the following limit. If you use L'Hopital's Rule, indicate where it was used, and be certain its hypotheses are met before you apply it.

\begin{equation*} \lim_{x\to \infty}(1+\frac{1}{x})^x \end{equation*}

Notice that $1+\frac{1}{x}\to 1$ as $x\to\infty\text{.}$ So this limit takes the form $1^{\infty}\text{,}$ which is an indeterminate form.

We need a couple extra steps before we can apply L'Hopital's Rule. First, let's set $y=(1+\frac{1}{x})^x\text{.}$ Note that $\ln(y)=\ln(1+\frac{1}{x})^x=x\ln(1+\frac{1}{x}).$ Then

\begin{equation*} \lim_{x\to\infty}(1+\frac{1}{x})^x=\lim_{x\to\infty}y=\lim_{x\to\infty}e^{\ln(y)}=e^{\lim\limits_{x\to\infty}\ln(y)}. \end{equation*}

Therefore, we need to evaluate $\lim\limits_{x\to\infty}\ln(y)=\lim\limits_{x\to\infty}x\ln(1+\frac{1}{x}).$

As $x\to\infty$ this limit approaches $\infty\cdot 0\text{,}$ which is another indeterminate form. We still have not met the conditions to apply L'Hopital's rule, but are getting closer. We can rewrite the limit in the following way

\begin{equation*} \lim_{x\to\infty}x\ln(1+\frac{1}{x})=\lim_{x\to\infty}\frac{\ln(1+\frac{1}{x})}{1/x}. \end{equation*}

This limit has the indeterminate form $\frac{0}{0}\text{,}$ so we can apply L'Hopital's Rule at this point:

\begin{equation*} \lim_{x\to\infty}\frac{\ln(1+\frac{1}{x})}{\frac{1}{x}}=\lim_{x\to\infty}\frac{\frac{1}{1+\frac{1}{x}}\cdot\frac{-1}{x^2}}{-\frac{1}{x^2}}=\lim_{x\to\infty}\frac{1}{1+\frac{1}{x}}=1. \end{equation*}

So finally, we have

\begin{equation*} \lim_{x\to\infty}(1+\frac{1}{x})^x=e^{1}=e. \end{equation*}

Other types of forms include $0\cdot\infty\text{,}$ $\infty-\infty\text{,}$ $1^{\infty}\text{,}$ $0^0\text{,}$ and $\infty^0\text{.}$ The following examples demonstrate a few of these other forms.

###### Example3.68

Evaluate each of the following limits. If you use L'Hopital's Rule, indicate where it was used, and be certain its hypotheses are met before you apply it.

1. $\lim\limits_{x \to \infty} xe^{-x}$

2. $\lim\limits_{x \to \frac{\pi}{2}^{-}}\left( \sec(x)-\tan(x)\right)$

3. $\lim\limits_{x \to 0^{+}} x^{x}$

Hint
1. Observe that $e^{-x} = \frac{1}{e^x}\text{.}$

2. Recall that $\sec(x)=\frac{1}{\cos(x)}$ and $\tan(x)=\frac{\sin(x)}{\cos(x)}\text{.}$

3. Observe that $x^{x} = e^{x\ln(x)}\text{.}$

1. $\lim\limits_{x \to \infty} xe^{-x} = 0\text{.}$

2. $\lim\limits_{x \to \frac{\pi}{2}^{-}} \sec(x)-\tan(x)=0\text{.}$

3. $\lim\limits_{x \to 0^{+}} x^{x}=1\text{.}$

Solution
1. In its original form, $\lim\limits_{x \to \infty} xe^{-x}\text{,}$ is indeterminate of form $\infty \cdot 0\text{.}$ Rewriting $e^{-x}$ as $\frac{1}{e^x}\text{,}$ a straightforward application of L'Hopital's Rule tells us that

\begin{equation*} \lim_{x \to \infty} xe^{-x} = \lim_{x \to \infty} \frac{x}{e^x} = \lim_{x \to \infty} \frac{1}{e^x}\text{.} \end{equation*}

Since $e^x \to \infty$ as $x \to \infty\text{,}$ we find that

\begin{equation*} \lim_{x \to \infty} xe^{-x} = 0\text{.} \end{equation*}
2. In its original form, $\lim\limits_{x\to \frac{\pi}{2}^{-}}\left(\sec(x)-\tan(x)\right)$ has the indeterminate form $\infty-\infty\text{.}$ Rewriting $\sec(x)$ as $\frac{1}{\cos(x)}$ and $\tan(x)$ as $\frac{\sin(x)}{\cos(x)}$ we have

\begin{equation*} \lim_{x\to \frac{\pi}{2}^{-}}\left(\sec(x)-\tan(x)\right)=\lim_{x\to \frac{\pi}{2}^{-}}\frac{1}{\cos(x)}-\frac{\sin(x)}{\cos(x)}=\lim_{x\to \frac{\pi}{2}^{-}}\frac{1-\sin(x)}{\cos(x)}. \end{equation*}

Since both numerator and demoninator approach 0, we can apply L'Hopital's Rule,

\begin{equation*} \lim_{x\to \frac{\pi}{2}^{-}}\sec(x)-\tan(x)=\lim_{x\to \frac{\pi}{2}^{-}}\frac{1-\sin(x)}{\cos(x)}=\lim_{x\to \frac{\pi}{2}^{-}}\frac{-\cos(x)}{-\sin(x)}=0. \end{equation*}
3. In its original form, $\lim\limits_{x \to 0^+} x^{x}$ has the indeterminate form $0^0\text{.}$ We can rewrite $x^x$ using logarithmic and exponential functions as $e^{x\ln(x)}\text{.}$ Thus,

\begin{equation*} \lim_{x\to 0^+}x^x=\lim_{x\to 0^+}e^{x\ln(x)}=e^{\lim\limits_{x\to 0^+}x\ln(x)}. \end{equation*}

We evaluate $\lim\limits_{x\to 0^+}x\ln(x)$ to get the indeterminate form $0\cdot \infty\text{.}$ Rewriting again, we have

\begin{equation*} \lim_{x\to 0^+}x\ln(x)=\lim_{x\to 0^+}\frac{\ln(x)}{\frac{1}{x}}. \end{equation*}

Since both numerator and denominator approach $\infty\text{,}$ we can apply L'Hopital's Rule,

\begin{equation*} \lim_{x\to 0^+}\frac{\ln(x)}{\frac{1}{x}}=\lim_{x\to 0^+}\frac{\frac{1}{x}}{-\frac{1}{x^2}}=\lim_{x\to 0^+}\frac{-x^2}{x}=\lim_{x\to 0^+}-x=0. \end{equation*}

Finally, we have

\begin{equation*} \lim_{x \to 0^+} x^{x}=e^0=1. \end{equation*}

### SubsectionSummary

• Derivatives can be used to help us evaluate indeterminate limits of the form $\frac{0}{0}$ through L'Hopital's Rule, by replacing the functions in the numerator and denominator with their tangent line approximations. In particular, if $f(a) = g(a) = 0$ and $f$ and $g$ are differentiable at $a\text{,}$ L'Hopital's Rule tells us that

\begin{equation*} \lim_{x \to a}\frac{f(x)}{g(x)} = \lim_{x \to a}\frac{f'(x)}{g'(x)}\text{.} \end{equation*}
• When we write $x \to \infty\text{,}$ this means that $x$ is increasing without bound. Thus, $\lim\limits_{x \to \infty} f(x) = L$ means that we can make $f(x)$ as close to $L$ as we like by choosing $x$ to be sufficiently large. Similarly, $\lim\limits_{x \to a} f(x) = \infty\text{,}$ means that we can make $f(x)$ as large as we like by choosing $x$ sufficiently close to $a\text{.}$

• A version of L'Hopital's Rule also helps us evaluate indeterminate limits of the form $\frac{\infty}{\infty}\text{.}$ If $f$ and $g$ are differentiable and both approach zero or both approach $\pm \infty$ as $x \to a$ (where $a$ is allowed to be $\infty$), then

\begin{equation*} \lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)}\text{.} \end{equation*}

### SubsectionExercises

Let $f$ and $g$ be differentiable functions about which the following information is known: $f(3) = g(3) = 0\text{,}$ $f'(3) = g'(3) = 0\text{,}$ $f''(3) = -2\text{,}$ and $g''(3) = 1\text{.}$ Let a new function $h$ be given by the rule $h(x) = \frac{f(x)}{g(x)}\text{.}$ On the same set of axes, sketch possible graphs of $f$ and $g$ near $x = 3\text{,}$ and use the provided information to determine the value of

\begin{equation*} \lim_{x \to 3} h(x)\text{.} \end{equation*}

Provide explanation to support your conclusion.

Find all vertical and horizontal asymptotes of the function

\begin{equation*} R(x) = \frac{3(x-a)(x-b)}{5(x-a)(x-c)}\text{,} \end{equation*}

where $a\text{,}$ $b\text{,}$ and $c$ are distinct, arbitrary constants. In addition, state all values of $x$ for which $R$ is not continuous. Sketch a possible graph of $R\text{,}$ clearly labeling the values of $a\text{,}$ $b\text{,}$ and $c\text{.}$

Consider the function $g(x) = x^{2x}\text{,}$ which is defined for all $x \gt 0\text{.}$ Observe that $\lim\limits_{x \to 0^+} g(x)$ is indeterminate due to its form of $0^0\text{.}$ (Think about how we know that $0^k = 0$ for all $k \gt 0\text{,}$ while $b^0 = 1$ for all $b \ne 0\text{,}$ but that neither rule can apply to $0^0\text{.}$)

1. Let $h(x) = \ln(g(x))\text{.}$ Explain why $h(x) = 2x \ln(x)\text{.}$

2. Next, explain why it is equivalent to write $h(x) = \frac{2\ln(x)}{\frac{1}{x}}\text{.}$

3. Use L'Hopital's Rule and your work in (b) to compute $\lim\limits_{x \to 0^+} h(x)\text{.}$

4. Based on the value of $\lim\limits_{x \to 0^+} h(x)\text{,}$ determine $\lim\limits_{x \to 0^+} g(x)\text{.}$

Recall we say that the function $g$ dominates the function $f$ provided that $\lim\limits_{x \to \infty} f(x) = \infty\text{,}$ $\lim\limits_{x \to \infty} g(x) = \infty\text{,}$ and $\lim\limits_{x \to \infty} \frac{f(x)}{g(x)} = 0\text{.}$

1. Which function dominates the other: $\ln(x)$ or $\sqrt{x}\text{?}$

2. Which function dominates the other: $\ln(x)$ or $\sqrt[n]{x}\text{?}$ ($n$ can be any positive integer)

3. Explain why $e^x$ will dominate any polynomial function.

4. Explain why $x^n$ will dominate $\ln(x)$ for any positive integer $n\text{.}$

5. Give any example of two nonlinear functions such that neither dominates the other.