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 $\lim _{x \to a} \frac {f'(x)}{g'(x)}$ exists, and $g'(x) \neq 0$ for all $x$ near $a$ , then

$\lim _{x \to a} \frac {f(x)}{g(x)} = \lim _{x \to a} \frac {f'(x)}{g'(x)}.$

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!

1 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 $-\infty$ .

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

3 Exponential Indeterminate Forms

There is a standard trick for dealing with the indeterminate forms

$\ensuremath {\boldsymbol {1^\infty }},\quad \ensuremath {\boldsymbol {0^0}},\quad \ensuremath {\boldsymbol {\infty ^0}}.$

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

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, $\begin{align*} \lim _{x\to \infty }\left (1 + \frac {1}{x}\right )^x &=e^{ \bigg [ \lim _{x\to \infty }x\ln \left (1 + \frac {1}{x}\right ) \bigg ]} \\ &=e^{ \bigg [ \lim _{x\to \infty } \frac {\ln \left (1 + \frac {1}{x}\right )}{x^{-1}} \bigg ]} \\ &=e^{1} \\ &= e. \end{align*}$

2025-01-06 19:43:20

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1 Basic indeterminant forms

2 Indeterminate forms involving subtraction

3 Exponential Indeterminate Forms

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