Proof of the ${\displaystyle {\frac {0}{0}}}$ case

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

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

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

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

Note that since ${\displaystyle c\in (x,a)}$ or ${\displaystyle c\in (a,x)}$ , by the squeeze theorem

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

This implies

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

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

Example 1

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

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

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

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

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

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

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

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

Plugging in 0 for x once again gives 1.

Example 3

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

This time, plugging in ${\displaystyle \infty }$ for x gives you ${\displaystyle {\frac {\infty }{\infty }}}$ . You know the drill:

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

This time, though, there is no x term left! ${\displaystyle {\frac {4}{5}}}$ is the answer.

Example 4

Sometimes, forms exist where it is not intuitively obvious how to solve them. One might think the value ${\displaystyle 1^{\infty }=1.}$ However, as was noted in the definition of an indeterminate form, this isn't possible to evaluate using the rules learned before now, and we need to use L'Hôpital's rule to solve.

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

Plugging the value of x into the limit yields

${\displaystyle \lim _{x\to \infty }\left(1+{\frac {1}{x}}\right)^{x}=1^{\infty }}$ (indeterminate form).

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

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

We now apply L'Hôpital's rule by taking the derivative of the top and bottom with respect to ${\displaystyle x}$ .

${\displaystyle {\frac {d}{dx}}\left[\ln \left(1+{\frac {1}{x}}\right)\right]={\frac {x}{x+1}}\cdot {\frac {-1}{x^{2}}}=-{\frac {1}{x(x+1)}}}$

${\displaystyle {\frac {d}{dx}}\left({\frac {1}{x}}\right)=-{\frac {1}{x^{2}}}}$

Returning to the expression above

${\displaystyle \ln(k)}$	${\displaystyle =\lim _{x\to \infty }{\frac {-(-x^{2})}{x(x+1)}}}$
	${\displaystyle =\lim _{x\to \infty }{\frac {x}{x+1}}={\frac {\infty }{\infty }}}$ (indeterminate form)

We apply L'Hôpital's rule once again

${\displaystyle \ln(k)=\lim _{x\to \infty }{\frac {1}{1}}=1}$

Therefore

${\displaystyle k=e}$

And

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

Careful: this does not prove that ${\displaystyle 1^{\infty }=e}$ because

${\displaystyle \lim _{x\to \infty }\left(1+{\frac {2}{x}}\right)^{x}=1^{\infty }\neq e}$