Theorem   (L'Hôpital's Rule for 0/0) Assume $f(x)$ and $g(x)$ have derivatives $f'(x)$ and $g'(x)$ at each point $x$ of an open interval $(a,b)$ and suppose that

$$\lim_{x\to a^+} f(x) = 0 \;\;$$    and $$\;\; \lim_{x\to a^+} g(x) = 0 \;.$$

Assume also that $g'(x)\neq 0$ for each $x$ in $(a,b)$. If the limit

$$\lim_{x\to a^+} \frac{f'(x)}{g'(x)}$$

exists and has the value $L$, then the limit

$$\lim_{x\to a^+} \frac{f(x)}{g(x)}$$

also exists and has value $L$.

Note 1: This statement is for the indeterminate case 0/0. Using this result can you get a corresponding result for the case of $\pm\infty/ \pm\infty$? Try taking reciprocals of both $f$ and $g$ in that case.

Note 2: Also a similar result holds when the limit is two-sided (i.e. $x\to a$ instead of $x \to a^+$), or the limit is at infinity (i.e. $x \to \infty$), or the limit value $L$ is $\pm \infty$.

Example (where L'Hôpital's rule works)  

Find the value of

$$\lim_{t \to 0^+} \frac{t}{1-e^{2t}} \;.$$

This quotient assumes the indeterminate form 0/0 as $t \to 0^+$. So we can apply the L'Hôpital's rule:

$$\lim_{t \to 0^+} \frac{t}{1-e^{2t}} = \lim_{t \to 0^+} \frac{1}{-2e^{2t}} = -\frac{1}{2} \; .$$

Example (this example doesn't work exactly)  

Find the value of

$$\lim_{t \to 0^+} \frac{e^{-1/x}}{x}\;.$$

This quotient also assumes the form 0/0. So consider the quotient of the derivatives:

$$\frac{f'}{g'} = \frac{\frac{1}{x^2} e^{-1/x}}{1} = \frac{e^{-1/x}}{x^2} \;.$$

This quotient still assumes the form 0/0, so we can continue applying the L'Hôpital's rule, but it only gets worse... After $n$-steps, we get to the quotient

$$\frac{e^{-1/x}}{n\! x^{n+1}}\;.$$

The indeterminacy doesn't disappear.

Example (when students don't get enough sleep)  

Find the value of

$$\lim_{x\to 1} \frac{3x^2 - 2x - 1}{x^2 -x} \; .$$

We have the form 0/0, so we apply L'Hôpital's rule:

$$\lim_{x\to 1} \frac{3x^2 - 2x - 1}{x^2 -x} = \lim_{x\to 1} \frac{6x-2}{2x-1} = \lim_{x\to 1} \frac{6}{2} = 3 \; .$$   NOT CORRECT!

The first step is correct, and stop there. You don't have an indeterminacy any more, so plug in $x=1$ and get

$$\lim_{x\to 1} \frac{6x-2}{2x-1} = \frac{6-2}{2-1} = 4\;.$$

Example   What happens if the limit $f'/g'$ doesn't exist? Does that mean the limit $f/g$ does not exist?

Consider the example

$$\lim_{x\to \infty} \frac{x- \sin x}{x}\;.$$

Find limit of $f'/g'$:

$$\lim_{x\to \infty} \frac{1-\cos x}{1}\;.$$

This limit does not exist because $\cos x$ oscillates between -1 and 1. But clearly if we rewrite the original limit, we find

$$\lim_{x\to \infty} \frac{x- \sin x}{x} = \lim_{x\to \infty} 1 - \frac{\sin x}{x} = 1-0=1\;.$$