L’Hopital’s rule

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We use derivatives to give us a “short-cut” for computing limits.

Derivatives allow us to take problems that were once difficult to solve and convert them to problems that are easier to solve. Let us consider L’Hôpital’s rule:

L’Hôpital’s Rule Let $f(x)$ and $g(x)$ be functions that are differentiable near $a$ . If $\lim _{x \to a} f(x) = \lim _{x \to a}g(x) = 0 \qquad \text {or} \pm \infty ,$ and $g'(x) \neq 0$ for all $x$ near $a$ , then $\lim _{x \to a} \frac {f(x)}{g(x)} = \lim _{x \to a} \frac {f'(x)}{g'(x)},$ provided that $\lim _{x \to a} \frac {f'(x)}{g'(x)}$ exists or $\pm \infty$ .

This theorem is somewhat difficult to prove, in part because it incorporates so many different possibilities, so we will not prove it here.

L’Hôpital’s rule applies even when $\lim _{x\to a}f(x) = \pm \infty$ and $\lim _{x\to a}g(x) = \mp \infty$ .

L’Hôpital’s rule allows us to investigate limits of indeterminate form.

List of Indeterminate Forms

$\relax \boldsymbol {\tfrac {0}{0}}$: This refers to a limit of the form $\lim _{x\to a} \frac {f(x)}{g(x)}$ where $f(x)\to 0$ and $g(x)\to 0$ as $x\to a$ .
$\relax \boldsymbol {\tfrac {\infty }{\infty }}$: This refers to a limit of the form $\lim _{x\to a} \frac {f(x)}{g(x)}$ where $f(x)\to \infty$ and $g(x)\to \infty$ as $x\to a$ .
$\relax \small \boldsymbol {0\cdot \infty }$: This refers to a limit of the form $\lim _{x\to a} \left (f(x)\cdot g(x)\right )$ where $f(x)\to 0$ and $g(x)\to \infty$ as $x\to a$ .
$\relax \small \boldsymbol {\infty - \infty }$: This refers to a limit of the form $\lim _{x\to a}\left ( f(x)-g(x)\right )$ where $f(x)\to \infty$ and $g(x)\to \infty$ as $x\to a$ .
$\relax \boldsymbol {1^\infty }$: This refers to a limit of the form $\lim _{x\to a} f(x)^{g(x)}$ where $f(x)\to 1$ and $g(x)\to \infty$ as $x\to a$ .
$\relax \boldsymbol {0^0}$: This refers to a limit of the form $\lim _{x\to a} f(x)^{g(x)}$ where $f(x)\to 0$ and $g(x)\to 0$ as $x\to a$ .
$\relax \boldsymbol {\infty ^0}$: This refers to a limit of the form $\lim _{x\to a} f(x)^{g(x)}$ where $f(x)\to \infty$ and $g(x)\to 0$ as $x\to a$ .

In each of these cases, the value of the limit is not immediately obvious. Hence, a careful analysis is required!

Basic indeterminant forms

Our first example is the computation of a limit that was somewhat difficult before.

Compute $\lim _{x\to 0} \frac {\sin (x)}{x}.$

Set $f(x) = \sin (x)$ and $g(x) = x$ . Since both $f(x)$ and $g(x)$ are differentiable functions at $0$ , and $\lim _{x \to 0} f(x) = \lim _{x \to 0}g(x) = 0,$ this situation is ripe for L’Hôpital’s Rule. Now $f'(x) = \answer [given]{\cos (x)}$ and $g'(x) = \answer [given]{1}.$ L’Hôpital’s rule tells us that $\lim _{x \to 0} \frac {\sin (x)}{x} = \lim _{x \to 0} \frac {\cos (x)}{1} = 1.$

Note, the astute mathematician will notice that in our example above, we are somewhat cheating. To apply L’Hôpital’s rule, we need to know the derivative of sine; however, to know the derivative of sine we must be able to compute the limit: $\lim _{x\to 0}\frac {\sin (x)}{x}$ Hence using L’Hôpital’s rule to compute this limit is a circular argument! We encourage the gentle reader to view L’Hôpital’s rule a “reminder” as to what is true, not as the formal derivation of the result.

Our next set of examples will run through the remaining indeterminate forms one is likely to encounter.

Compute $\lim _{x\to \pi /2^+} \frac {\sec (x)}{\tan (x)}.$

Set $f(x) = \sec (x)$ and $g(x) = \tan (x)$ . Both $f(x)$ and $g(x)$ are differentiable near $\pi /2$ . Additionally, $\lim _{x \to \pi /2^+} f(x) = \lim _{x \to \pi /2^+}g(x) = -\infty .$ This situation is ripe for L’Hôpital’s Rule. Now $f'(x) = \answer [given]{\sec (x)\tan (x)}$ and $g'(x) = \answer [given]{\sec ^2(x)}.$ L’Hôpital’s rule tells us that $\begin{align*} \lim_{x\to \pi/2^+} \frac{\sec(x)}{\tan(x)} &= \lim_{x\to \pi/2^+} \frac{\sec(x)\tan(x)}{\sec^2(x)} \\ &= \lim_{x\to \pi/2^+} \sin(x)\\ &=\answer[given]{1}. \end{align*}$

Compute $\lim _{x\to 0^+} x\ln x.$

This doesn’t appear to be suitable for L’Hôpital’s Rule. As $x$ approaches zero, $\ln x$ goes to $-\infty$ , so the product looks like $(\text {something very small})\cdot (\text {something very large and negative}).$ This product could be anything. A careful analysis is required. Write $x\ln x = \frac {\ln x}{x^{-1}}.$ Set $f(x) = \ln (x)$ and $g(x) = x^{-1}$ . Since both functions are differentiable near zero and $\lim _{x\to 0+} \ln (x) = -\infty \qquad \text {and}\qquad \lim _{x\to 0+} x^{-1} = \infty ,$ we may apply L’Hôpital’s rule. Write with me $f'(x) = \answer [given]{x^{-1}}$ and $g'(x) = \answer [given]{-x^{-2}},$ so $\begin{align*} \lim_{x\to 0^+} x\ln x &= \lim_{x\to 0^+} \frac{\ln x}{x^{-1}} \\ &= \lim_{x\to 0^+} \frac{x^{-1}}{-x^{-2}}\\ &=\lim_{x\to 0^+} -x \\ &= 0. \end{align*}$

One way to interpret this is that since $\lim _{x\to 0^+}x\ln x = 0$ , the function $x$ approaches zero much faster than $\ln x$ approaches $-\infty$ .

Indeterminate forms involving subtraction

There are two basic cases here, we’ll do an example of each.

Compute $\lim _{x\to 0} \left (\cot (x) - \csc (x)\right ).$

Here we simply need to write each term as a fraction, $\begin{align*} \lim_{x\to 0} \left(\cot(x) - \csc(x)\right) &= \lim_{x\to 0} \left(\frac{\cos(x)}{\sin(x)} - \frac{1}{\sin(x)}\right)\\ &= \lim_{x\to 0} \frac{\cos(x)-1}{\sin(x)} \end{align*}$

Setting $f(x) = \cos (x)-1$ and $g(x)=\sin (x)$ , both functions are differentiable near zero and $\lim _{x\to 0}(\cos (x)-1)=\lim _{x\to 0}\sin (x) = 0.$ We may now apply L’Hôpital’s rule. Write with me $f'(x) = \answer [given]{-\sin (x)}$ and $g'(x) = \answer [given]{\cos (x)},$ so

$\begin{align*} \lim_{x\to 0} \left(\cot(x) - \csc(x)\right) &= \lim_{x\to 0} \frac{\cos(x)-1}{\sin(x)}\\ &= \lim_{x\to 0} \frac{-\sin(x)}{\cos(x)} \\ &=0. \end{align*}$

Sometimes one must be slightly more clever.

Compute $\lim _{x\to \infty }\left (\sqrt {x^2+x}-x\right ).$

Again, this doesn’t appear to be suitable for L’Hôpital’s Rule. A bit of algebraic manipulation will help. Write with me $\begin{align*} \lim_{x\to\infty}\left(\sqrt{x^2+x}-x\right) &= \lim_{x\to\infty}\left(x\left(\sqrt{1+1/x}-1\right)\right)\\ &=\lim_{x\to\infty}\frac{\sqrt{1+1/x}-1}{x^{-1}} \end{align*}$

Now set $f(x) = \sqrt {1+1/x}-1$ , $g(x) = x^{-1}$ . Since both functions are differentiable for large values of $x$ and $\lim _{x\to \infty } (\sqrt {1+1/x}-1) = \lim _{x\to \infty }x^{-1} = 0,$ we may apply L’Hôpital’s rule. Write with me $f'(x) = \answer [given]{(1/2)(1+1/x)^{-1/2}\cdot (-x^{-2})}$ and $g'(x) = \answer [given]{-x^{-2}}$ so

$\begin{align*} \lim_{x\to\infty}\left(\sqrt{x^2+x}-x\right) &= \lim_{x\to\infty}\frac{\sqrt{1+1/x}-1}{x^{-1}} \\ &= \lim_{x\to\infty}\frac{(1/2)(1+1/x)^{-1/2}\cdot(-x^{-2})}{-x^{-2}} \\ &= \lim_{x\to\infty} \frac{1}{2\sqrt{1+1/x}}\\ &= \frac{1}{2}. \end{align*}$

Exponential Indeterminate Forms

There is a standard trick for dealing with the indeterminate forms $\oneToInfty ,\quad \zeroToZero ,\quad \inftyToZero .$ Given $u(x)$ and $v(x)$ such that $\lim _{x\to a}u(x)^{v(x)}$ falls into one of the categories described above, rewrite as $\lim _{x\to a}e^{\ln \left (u(x)^{v(x)}\right )}$ and then we rewrite as $\lim _{x\to a}e^{v(x)\ln {(u(x))}}.$ Recall that the exponential function is continuous. Therefore $\lim _{x\to a}e^{v(x)\ln {(u(x))}}=e^{\left [\lim _{x\to a} v(x)\ln (u(x))\right ]}.$ Now we will focus on the limit $\lim _{x\to a} v(x)\ln (u(x))$ using L’Hôpital’s rule. We will only give a single example.

First determine the form of the limit, then compute the limit. $\lim _{x\to \infty }\left (1 + \frac {1}{x}\right )^x.$ Select the correct choice. The form of the limit is

$\zeroOverZero$ $\relax \boldsymbol {\tfrac {\infty }{\infty }}$ $\zeroTimesInfty$ $\relax \small \boldsymbol {\infty - \infty }$ $\oneToInfty$ $\inftyToZero$ $\zeroToZero$

Write $\lim _{x\to \infty }\left (1 + \frac {1}{x}\right )^x = e^{\left [\lim _{x\to \infty }x\ln \left (1 + \frac {1}{x}\right )\right ]}.$ So now look at the limit of the exponent $\lim _{x\to \infty } x\ln \left (1 + \frac {1}{x}\right ) = \lim _{x\to \infty } \frac {\ln \left (1 + \frac {1}{x}\right )}{x^{-1}}.$ Setting $f(x) = \ln \left (1 + \frac {1}{x}\right )$ and $g(x) = x^{-1}$ , both functions are differentiable for large values of $x$ and $\lim _{x\to \infty }\ln \left (1 + \frac {1}{x}\right )=\lim _{x\to \infty }x^{-1} = 0.$ We may now apply L’Hôpital’s rule. Write $f'(x) = \answer [given]{\frac {-x^{-2}}{1 + \frac {1}{x}}}$ and $g'(x) = \answer [given]{-x^{-2}},$ so $\begin{align*} \lim_{x\to\infty} \frac{\ln\left(1 + \frac{1}{x}\right)}{x^{-1}} &= \lim_{x\to\infty} \frac{\frac{-x^{-2}}{1 + \frac{1}{x}}}{-x^{-2}} \\ &=\lim_{x\to\infty} \frac{1}{1 + \frac{1}{x}}\\ &=1. \end{align*}$

Hence,

$\begin{align*} \lim_{x\to \infty}\left(1 + \frac{1}{x}\right)^x &=e^{ \eval{\lim_{x\to \infty}x\ln\left(1 + \frac{1}{x}\right)}} \\ &=e^{ \eval{\lim_{x\to\infty} \frac{\ln\left(1 + \frac{1}{x}\right)}{x^{-1}}}} \\ &=e^{1} \\ &= e. \end{align*}$