L'Hôpital's Rule

Some limits of quotients can't be found by substituting, because the numerator and the denominator both approach 00.
In the simplest case the reason is short. If f(a)=g(a)=0f(a) = g(a) = 0, both are differentiable at aa, and g′(a)≠0g'(a) \ne 0, then for x≠ax \ne a near aa (where g(x)≠0g(x) \ne 0, because g(x)x−a→g′(a)≠0\frac{g(x)}{x - a} \to g'(a) \ne 0)
f(x)g(x)=f(x)−f(a)x−ag(x)−g(a)x−a  →  f′(a)g′(a).\frac{f(x)}{g(x)} = \frac{\dfrac{f(x) - f(a)}{x - a}}{\dfrac{g(x) - g(a)}{x - a}} \;\to\; \frac{f'(a)}{g'(a)}.
If f′f' and g′g' are also continuous at aa, then f′(a)g′(a)=lim⁡x→af′(x)g′(x)\frac{f'(a)}{g'(a)} = \lim_{x \to a} \frac{f'(x)}{g'(x)}, which is the rule.
On a free-response answer, show the form before using the rule: write both limits, lim⁡x→af(x)=0\lim_{x \to a} f(x) = 0 and lim⁡x→ag(x)=0\lim_{x \to a} g(x) = 0, in limit notation. Writing "00\frac{0}{0}" as if it were a value isn't enough.

Using the rule

When the form isn't 0/0

From a table

From a graph