Famous Theorems of Mathematics/L'Hôpital's rule

Consider some limit

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

which cannot be evaluated directly, that is, either f(a) = g(a) = 0 or

$$ \lim_{x\to a} f(x) = \lim_{x\to a} g(x) = \infty$$

L'Hopital's rule says that

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

Suppose both are equal to zero. The Newton definition of the derivative is

$$ f'(a) = \lim_{x\to a} \frac{f(x) - f(a)}{x-a} $$

Therefore,

$$ \frac{f'(a)}{g'(a)} = \lim_{x\to a} \frac{\frac{f(x) - f(a)}{x-a}}{\frac{g(x) - g(a)}{x-a}} = \lim_{x\to a} \frac{f(x) - f(a)}{g(x) - g(a)}$$

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

Now suppose that both $$ f $$ and $$ g$$ diverge to positive or negative infinity. Another way to define the derivative is by

$$ f'(a) = \lim_{x\to 0} \frac{f(x+a)}{x} $$

Then

$$ \frac{f'(a)}{g'(a)} = \lim_{x\to 0} \frac{xf(x+a)}{xg(x+a)} = \lim_{x\to 0} \frac{f(x+a)}{g(x+a)} = \frac{f(x+a)}{g(x+a)} $$

QED