# 4.8 L’hôpital’s rule

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• Recognize when to apply L’Hôpital’s rule.
• Identify indeterminate forms produced by quotients, products, subtractions, and powers, and apply L’Hôpital’s rule in each case.
• Describe the relative growth rates of functions.

In this section, we examine a powerful tool for evaluating limits. This tool, known as L’Hôpital’s rule    , uses derivatives to calculate limits. With this rule, we will be able to evaluate many limits we have not yet been able to determine. Instead of relying on numerical evidence to conjecture that a limit exists, we will be able to show definitively that a limit exists and to determine its exact value.

## Applying l’hôpital’s rule

L’Hôpital’s rule can be used to evaluate limits involving the quotient of two functions. Consider

$\underset{x\to a}{\text{lim}}\frac{f\left(x\right)}{g\left(x\right)}.$

If $\underset{x\to a}{\text{lim}}f\left(x\right)={L}_{1}\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}\underset{x\to a}{\text{lim}}g\left(x\right)={L}_{2}\ne 0,$ then

$\underset{x\to a}{\text{lim}}\frac{f\left(x\right)}{g\left(x\right)}=\frac{{L}_{1}}{{L}_{2}}.$

However, what happens if $\underset{x\to a}{\text{lim}}f\left(x\right)=0$ and $\underset{x\to a}{\text{lim}}g\left(x\right)=0?$ We call this one of the indeterminate forms    , of type $\frac{0}{0}.$ This is considered an indeterminate form because we cannot determine the exact behavior of $\frac{f\left(x\right)}{g\left(x\right)}$ as $x\to a$ without further analysis. We have seen examples of this earlier in the text. For example, consider

$\underset{x\to 2}{\text{lim}}\frac{{x}^{2}-4}{x-2}\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}\underset{x\to 0}{\text{lim}}\frac{\text{sin}\phantom{\rule{0.1em}{0ex}}x}{x}.$

For the first of these examples, we can evaluate the limit by factoring the numerator and writing

$\underset{x\to 2}{\text{lim}}\frac{{x}^{2}-4}{x-2}=\underset{x\to 2}{\text{lim}}\frac{\left(x+2\right)\left(x-2\right)}{x-2}=\underset{x\to 2}{\text{lim}}\left(x+2\right)=2+2=4.$

For $\underset{x\to 0}{\text{lim}}\frac{\text{sin}\phantom{\rule{0.1em}{0ex}}x}{x}$ we were able to show, using a geometric argument, that

$\underset{x\to 0}{\text{lim}}\frac{\text{sin}\phantom{\rule{0.1em}{0ex}}x}{x}=1.$

Here we use a different technique for evaluating limits such as these. Not only does this technique provide an easier way to evaluate these limits, but also, and more important, it provides us with a way to evaluate many other limits that we could not calculate previously.

The idea behind L’Hôpital’s rule can be explained using local linear approximations. Consider two differentiable functions $f$ and $g$ such that $\underset{x\to a}{\text{lim}}f\left(x\right)=0=\underset{x\to a}{\text{lim}}g\left(x\right)$ and such that ${g}^{\prime }\left(a\right)\ne 0$ For $x$ near $a,$ we can write

$f\left(x\right)\approx f\left(a\right)+{f}^{\prime }\left(a\right)\left(x-a\right)$

and

$g\left(x\right)\approx g\left(a\right)+{g}^{\prime }\left(a\right)\left(x-a\right).$

Therefore,

$\frac{f\left(x\right)}{g\left(x\right)}\approx \frac{f\left(a\right)+{f}^{\prime }\left(a\right)\left(x-a\right)}{g\left(a\right)+{g}^{\prime }\left(a\right)\left(x-a\right)}.$

Since $f$ is differentiable at $a,$ then $f$ is continuous at $a,$ and therefore $f\left(a\right)=\underset{x\to a}{\text{lim}}f\left(x\right)=0.$ Similarly, $g\left(a\right)=\underset{x\to a}{\text{lim}}g\left(x\right)=0.$ If we also assume that ${f}^{\prime }$ and ${g}^{\prime }$ are continuous at $x=a,$ then ${f}^{\prime }\left(a\right)=\underset{x\to a}{\text{lim}}{f}^{\prime }\left(x\right)$ and ${g}^{\prime }\left(a\right)=\underset{x\to a}{\text{lim}}{g}^{\prime }\left(x\right).$ Using these ideas, we conclude that

$\underset{x\to a}{\text{lim}}\frac{f\left(x\right)}{g\left(x\right)}=\underset{x\to a}{\text{lim}}\frac{{f}^{\prime }\left(x\right)\left(x-a\right)}{{g}^{\prime }\left(x\right)\left(x-a\right)}=\underset{x\to a}{\text{lim}}\frac{{f}^{\prime }\left(x\right)}{{g}^{\prime }\left(x\right)}.$

Note that the assumption that ${f}^{\prime }$ and ${g}^{\prime }$ are continuous at $a$ and ${g}^{\prime }\left(a\right)\ne 0$ can be loosened. We state L’Hôpital’s rule formally for the indeterminate form $\frac{0}{0}.$ Also note that the notation $\frac{0}{0}$ does not mean we are actually dividing zero by zero. Rather, we are using the notation $\frac{0}{0}$ to represent a quotient of limits, each of which is zero.

## L’hôpital’s rule (0/0 case)

Suppose $f$ and $g$ are differentiable functions over an open interval containing $a,$ except possibly at $a.$ If $\underset{x\to a}{\text{lim}}f\left(x\right)=0$ and $\underset{x\to a}{\text{lim}}g\left(x\right)=0,$ then

$\underset{x\to a}{\text{lim}}\frac{f\left(x\right)}{g\left(x\right)}=\underset{x\to a}{\text{lim}}\frac{{f}^{\prime }\left(x\right)}{{g}^{\prime }\left(x\right)},$

assuming the limit on the right exists or is $\infty$ or $\text{−}\infty .$ This result also holds if we are considering one-sided limits, or if $a=\infty \phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}-\infty .$

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thank you ladies and gentlemen I appreciate the help!
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