# 4.8 L’hôpital’s rule  (Page 2/7)

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## Proof

We provide a proof of this theorem in the special case when $f,g,{f}^{\prime },$ and ${g}^{\prime }$ are all continuous over an open interval containing $a.$ In that case, since $\underset{x\to a}{\text{lim}}f\left(x\right)=0=\underset{x\to a}{\text{lim}}g\left(x\right)$ and $f$ and $g$ are continuous at $a,$ it follows that $f\left(a\right)=0=g\left(a\right).$ Therefore,

$\begin{array}{ccccc}\hfill \underset{x\to a}{\text{lim}}\frac{f\left(x\right)}{g\left(x\right)}& =\underset{x\to a}{\text{lim}}\frac{f\left(x\right)-f\left(a\right)}{g\left(x\right)-g\left(a\right)}\hfill & & & \text{since}\phantom{\rule{0.2em}{0ex}}f\left(a\right)=0=g\left(a\right)\hfill \\ & =\underset{x\to a}{\text{lim}}\frac{\frac{f\left(x\right)-f\left(a\right)}{x-a}}{\frac{g\left(x\right)-g\left(a\right)}{x-a}}\hfill & & & \text{algebra}\hfill \\ & =\frac{\underset{x\to a}{\text{lim}}\frac{f\left(x\right)-f\left(a\right)}{x-a}}{\underset{x\to a}{\text{lim}}\frac{g\left(x\right)-g\left(a\right)}{x-a}}\hfill & & & \text{limit of a quotient}\hfill \\ & =\frac{{f}^{\prime }\left(a\right)}{{g}^{\prime }\left(a\right)}\hfill & & & \text{definition of the derivative}\hfill \\ & =\frac{\underset{x\to a}{\text{lim}}{f}^{\prime }\left(x\right)}{\underset{x\to a}{\text{lim}}{g}^{\prime }\left(x\right)}\hfill & & & \text{continuity of}\phantom{\rule{0.2em}{0ex}}{f}^{\prime }\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}{g}^{\prime }\hfill \\ & =\underset{x\to a}{\text{lim}}\frac{{f}^{\prime }\left(x\right)}{{g}^{\prime }\left(x\right)}.\hfill & & & \text{limit of a quotient}\hfill \end{array}$

Note that L’Hôpital’s rule states we can calculate the limit of a quotient $\frac{f}{g}$ by considering the limit of the quotient of the derivatives $\frac{{f}^{\prime }}{{g}^{\prime }}.$ It is important to realize that we are not calculating the derivative of the quotient $\frac{f}{g}.$

## Applying l’hôpital’s rule (0/0 case)

Evaluate each of the following limits by applying L’Hôpital’s rule.

1. $\underset{x\to 0}{\text{lim}}\frac{1-\text{cos}\phantom{\rule{0.1em}{0ex}}x}{x}$
2. $\underset{x\to 1}{\text{lim}}\frac{\text{sin}\left(\pi x\right)}{\text{ln}\phantom{\rule{0.1em}{0ex}}x}$
3. $\underset{x\to \infty }{\text{lim}}\frac{{e}^{1\text{/}x}-1}{1\text{/}x}$
4. $\underset{x\to 0}{\text{lim}}\frac{\text{sin}\phantom{\rule{0.1em}{0ex}}x-x}{{x}^{2}}$
1. Since the numerator $1-\text{cos}\phantom{\rule{0.1em}{0ex}}x\to 0$ and the denominator $x\to 0,$ we can apply L’Hôpital’s rule to evaluate this limit. We have
$\begin{array}{cc}\hfill \underset{x\to 0}{\text{lim}}\frac{1-\text{cos}\phantom{\rule{0.1em}{0ex}}x}{x}& =\underset{x\to 0}{\text{lim}}\frac{\frac{d}{dx}\left(1-\text{cos}\phantom{\rule{0.1em}{0ex}}x\right)}{\frac{d}{dx}\left(x\right)}\hfill \\ & =\underset{x\to 0}{\text{lim}}\frac{\text{sin}\phantom{\rule{0.1em}{0ex}}x}{1}\hfill \\ & =\frac{\underset{x\to 0}{\text{lim}}\left(\text{sin}\phantom{\rule{0.1em}{0ex}}x\right)}{\underset{x\to 0}{\text{lim}}\left(1\right)}\hfill \\ & =\frac{0}{1}=0.\hfill \end{array}$
2. As $x\to 1,$ the numerator $\text{sin}\left(\pi x\right)\to 0$ and the denominator $\text{ln}\left(x\right)\to 0.$ Therefore, we can apply L’Hôpital’s rule. We obtain
$\begin{array}{cc}\hfill \underset{x\to 1}{\text{lim}}\frac{\text{sin}\left(\pi x\right)}{\text{ln}\phantom{\rule{0.1em}{0ex}}x}& =\underset{x\to 1}{\text{lim}}\frac{\pi \phantom{\rule{0.1em}{0ex}}\text{cos}\left(\pi x\right)}{1\text{/}x}\hfill \\ & =\underset{x\to 1}{\text{lim}}\left(\pi x\right)\text{cos}\left(\pi x\right)\hfill \\ & =\left(\pi ·1\right)\left(-1\right)=\text{−}\pi .\hfill \end{array}$
3. As $x\to \infty ,$ the numerator ${e}^{1\text{/}x}-1\to 0$ and the denominator $\left(\frac{1}{x}\right)\to 0.$ Therefore, we can apply L’Hôpital’s rule. We obtain
$\underset{x\to \infty }{\text{lim}}\frac{{e}^{1\text{/}x}-1}{\frac{1}{x}}=\underset{x\to \infty }{\text{lim}}\frac{{e}^{1\text{/}x}\left(\frac{-1}{{x}^{2}}\right)}{\left(\frac{-1}{{x}^{2}}\right)}=\underset{x\to \infty }{\text{lim}}{e}^{1\text{/}x}={e}^{0}=1·\underset{x\to \infty }{\text{lim}}\frac{{e}^{1\text{/}x}-1}{\text{ln}\phantom{\rule{0.1em}{0ex}}x}.$
4. As $x\to 0,$ both the numerator and denominator approach zero. Therefore, we can apply L’Hôpital’s rule. We obtain
$\underset{x\to 0}{\text{lim}}\frac{\text{sin}\phantom{\rule{0.1em}{0ex}}x-x}{{x}^{2}}=\underset{x\to 0}{\text{lim}}\frac{\text{cos}\phantom{\rule{0.1em}{0ex}}x-1}{2x}.$

Since the numerator and denominator of this new quotient both approach zero as $x\to 0,$ we apply L’Hôpital’s rule again. In doing so, we see that
$\underset{x\to 0}{\text{lim}}\frac{\text{cos}\phantom{\rule{0.1em}{0ex}}x-1}{2x}=\underset{x\to 0}{\text{lim}}\frac{\text{−}\text{sin}\phantom{\rule{0.1em}{0ex}}x}{2}=0.$

Therefore, we conclude that
$\underset{x\to 0}{\text{lim}}\frac{\text{sin}\phantom{\rule{0.1em}{0ex}}x-x}{{x}^{2}}=0.$

Evaluate $\underset{x\to 0}{\text{lim}}\frac{x}{\text{tan}\phantom{\rule{0.1em}{0ex}}x}.$

$1$

We can also use L’Hôpital’s rule to evaluate limits of quotients $\frac{f\left(x\right)}{g\left(x\right)}$ in which $f\left(x\right)\to \text{±}\infty$ and $g\left(x\right)\to \text{±}\infty .$ Limits of this form are classified as indeterminate forms of type $\infty \text{/}\infty .$ Again, note that we are not actually dividing $\infty$ by $\infty .$ Since $\infty$ is not a real number, that is impossible; rather, $\infty \text{/}\infty .$ is used to represent a quotient of limits, each of which is $\infty$ or $\text{−}\infty .$

## L’hôpital’s rule $\left(\infty \text{/}\infty$ Case)

Suppose $f$ and $g$ are differentiable functions over an open interval containing $a,$ except possibly at $a.$ Suppose $\underset{x\to a}{\text{lim}}f\left(x\right)=\infty$ (or $\text{−}\infty \right)$ and $\underset{x\to a}{\text{lim}}g\left(x\right)=\infty$ (or $\text{−}\infty \right).$ 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 the limit is infinite, if $a=\infty$ or $\text{−}\infty ,$ or the limit is one-sided.

## Applying l’hôpital’s rule $\left(\infty \text{/}\infty$ Case)

Evaluate each of the following limits by applying L’Hôpital’s rule.

1. $\underset{x\to \infty }{\text{lim}}\frac{3x+5}{2x+1}$
2. $\underset{x\to {0}^{+}}{\text{lim}}\frac{\text{ln}\phantom{\rule{0.1em}{0ex}}x}{\text{cot}\phantom{\rule{0.1em}{0ex}}x}$
1. Since $3x+5$ and $2x+1$ are first-degree polynomials with positive leading coefficients, $\underset{x\to \infty }{\text{lim}}\left(3x+5\right)=\infty$ and $\underset{x\to \infty }{\text{lim}}\left(2x+1\right)=\infty .$ Therefore, we apply L’Hôpital’s rule and obtain
$\underset{x\to \infty }{\text{lim}}\frac{3x+5}{2x+1}=\underset{x\to \infty }{\text{lim}}\frac{3+5\text{/}x}{2x+1}=\underset{x\to \infty }{\text{lim}}\frac{3}{2}=\frac{3}{2}.$

Note that this limit can also be calculated without invoking L’Hôpital’s rule. Earlier in the chapter we showed how to evaluate such a limit by dividing the numerator and denominator by the highest power of $x$ in the denominator. In doing so, we saw that
$\underset{x\to \infty }{\text{lim}}\frac{3x+5}{2x+1}=\underset{x\to \infty }{\text{lim}}\frac{3+5\text{/}x}{2x+1}=\frac{3}{2}.$

L’Hôpital’s rule provides us with an alternative means of evaluating this type of limit.
2. Here, $\underset{x\to {0}^{+}}{\text{lim}}\text{ln}\phantom{\rule{0.1em}{0ex}}x=\text{−}\infty$ and $\underset{x\to {0}^{+}}{\text{lim}}\text{cot}\phantom{\rule{0.1em}{0ex}}x=\infty .$ Therefore, we can apply L’Hôpital’s rule and obtain
$\underset{x\to {0}^{+}}{\text{lim}}\frac{\text{ln}\phantom{\rule{0.1em}{0ex}}x}{\text{cot}\phantom{\rule{0.1em}{0ex}}x}=\underset{x\to {0}^{+}}{\text{lim}}\frac{1\text{/}x}{\text{−}{\text{csc}}^{2}x}=\underset{x\to {0}^{+}}{\text{lim}}\frac{1}{\text{−}x\phantom{\rule{0.1em}{0ex}}{\text{csc}}^{2}x}.$

Now as $x\to {0}^{+},$ ${\text{csc}}^{2}x\to \infty .$ Therefore, the first term in the denominator is approaching zero and the second term is getting really large. In such a case, anything can happen with the product. Therefore, we cannot make any conclusion yet. To evaluate the limit, we use the definition of $\text{csc}\phantom{\rule{0.1em}{0ex}}x$ to write
$\underset{x\to {0}^{+}}{\text{lim}}\frac{1}{\text{−}x\phantom{\rule{0.1em}{0ex}}{\text{csc}}^{2}x}=\underset{x\to {0}^{+}}{\text{lim}}\frac{{\text{sin}}^{2}x}{\text{−}x}.$

Now $\underset{x\to {0}^{+}}{\text{lim}}{\text{sin}}^{2}x=0$ and $\underset{x\to {0}^{+}}{\text{lim}}x=0,$ so we apply L’Hôpital’s rule again. We find
$\underset{x\to {0}^{+}}{\text{lim}}\frac{{\text{sin}}^{2}x}{\text{−}x}=\underset{x\to {0}^{+}}{\text{lim}}\frac{2\phantom{\rule{0.1em}{0ex}}\text{sin}\phantom{\rule{0.1em}{0ex}}x\phantom{\rule{0.1em}{0ex}}\text{cos}\phantom{\rule{0.1em}{0ex}}x}{-1}=\frac{0}{-1}=0.$

We conclude that
$\underset{x\to {0}^{+}}{\text{lim}}\frac{\text{ln}\phantom{\rule{0.1em}{0ex}}x}{\text{cot}\phantom{\rule{0.1em}{0ex}}x}=0.$

f(x) = x-2 g(x) = 3x + 5 fog(x)? f(x)/g(x)
fog(x)= f(g(x)) = x-2 = 3x+5-2 = 3x+3 f(x)/g(x)= x-2/3x+5
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