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