L’Hôpital’s Rule

Occasionally, one comes across a limit which results in ${\frac {0}{0}}$ or ${\frac {\infty }{\infty }}$ , which are called indeterminate limits. However, it is still possible to solve these by using L'Hôpital's rule. This rule is vital in explaining how other limits can be derived.

Suppose that for real functions $f$ and $g$ , $\lim _{x\to a}f(x)=\lim _{x\to a}g(x)=0$ and that $\lim _{x\to a}{\frac {f'(x)}{g'(x)}}$ exists. Thus $f'(x)$ and $g'(x)$ exist in an interval $(a-\delta ,a+\delta )$ around $a$ , but maybe not at $a$ itself. Thus, for any $x\in (a-\delta ,a+\delta )$ , in any interval $[x,a]$ or $[a,x]$ , $f$ and $g$ are continuous and differentiable, with the possible exception of $a$ . Define

{\begin{array}{l}F(x)={\begin{cases}f(x)&x\neq a\\\lim \limits _{x\to a}f(x)&x=a\end{cases}}\\G(x)={\begin{cases}g(x)&x\neq a\\\lim \limits _{x\to a}g(x)&x=a\end{cases}}\end{array}}

Note that $\lim _{x\to a}{\frac {f(x)}{g(x)}}=\lim _{x\to a}{\frac {F(x)}{G(x)}}$ , $\lim _{x\to a}{\frac {f'(x)}{g'(x)}}=\lim _{x\to a}{\frac {F'(x)}{G'(x)}}$ , and that $F,G$ are continuous in any interval $[a,x]$ or $[x,a]$ and differentiable in any interval $(a,x)$ or $(x,a)$ when $x\in (a-\delta ,a+\delta )$ .

Cauchy's Mean Value Theorem tells us that ${\frac {F(x)-F(a)}{G(x)-G(a)}}={\frac {F'(c)}{G'(c)}}$ for some $c\in (a,x)$ or $c\in (x,a)$ . Since $F(a)=G(a)=0$ , we have ${\frac {F(x)}{G(x)}}={\frac {F'(c)}{G'(c)}}$ for $x,c\in (a-\delta ,a+\delta )$ .

Since $c\in (x,a)$ or $c\in (a,x)$ , by the squeeze theorem

\lim _{x\to a}x=a\quad \Rightarrow \quad \lim _{x\to a}c=a

This implies

\lim _{x\to a}{\frac {F'(c)}{G'(c)}}=\lim _{x\to a}{\frac {F'(x)}{G'(x)}}

So taking the limit as $x\to a$ of the last equation gives $\lim _{x\to a}{\frac {F(x)}{G(x)}}=\lim _{x\to a}{\frac {F'(x)}{G'(x)}}$ , which is equivalent to the more commonly used form $\lim _{x\to a}{\frac {f(x)}{g(x)}}=\lim _{x\to a}{\frac {f'(x)}{g'(x)}}$ .

Examples

Example 1

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

Since plugging in 0 for x results in ${\frac {0}{0}}$ , use L'Hôpital's rule to take the derivative of the top and bottom, giving:

\lim _{x\to 0}{\frac {{\frac {d}{dx}}\left(\sin(x)\right)}{{\frac {d}{dx}}(x)}}=\lim _{x\to 0}{\frac {\cos(x)}{1}}

Plugging in 0 for x gives 1 here. Note that it is logically incorrect to prove this limit by using L'Hôpital's rule, as the same limit is required to prove that the derivative of the sine function exists: it would be a form of begging the question.

Example 2

Find $\lim _{x\to 0}x\cot(x)$

First, you need to rewrite the function into an indeterminate limit fraction:

\lim _{x\to 0}{\frac {x}{\tan(x)}}

Now it's indeterminate. Take the derivative of the top and bottom:

\lim _{x\to 0}{\frac {1}{\sec ^{2}(x)}}

Plugging in 0 for $x$ once again gives 1.

Example 3

Find $\lim _{x\to \infty }{\frac {4x+22}{5x+9}}$

This time, plugging in $\infty$ for x gives you ${\frac {\infty }{\infty }}$ . So using L'Hôpital's rule gives:

\lim _{x\to \infty }{\frac {4}{5}}

Therefore, ${\frac {4}{5}}$ is the answer.

Example 4

Find $\lim _{x\to \infty }\left(1+{\frac {1}{x}}\right)^{x}$

Plugging the value of x into the limit yields

\lim _{x\to \infty }\left(1+{\frac {1}{x}}\right)^{x}=1^{\infty }

(indeterminate form).

Let $k=\lim _{x\to \infty }\left(1+{\frac {1}{x}}\right)^{x}=1^{\infty }$

$\ln(k)$	$=\lim _{x\to \infty }\ln \left(1+{\frac {1}{x}}\right)^{x}$
	$=\lim _{x\to \infty }x\ln \left(1+{\frac {1}{x}}\right)$
	$=\lim _{x\to \infty }{\frac {\ln \left(1+{\frac {1}{x}}\right)}{\frac {1}{x}}}={\frac {\ln(1)}{\frac {1}{x}}}={\frac {0}{0}}$