# Famous Theorems of Mathematics/L'Hôpital's rule

Consider some limit

${\displaystyle \lim _{x\to a}{\frac {f(x)}{g(x)}}}$

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

${\displaystyle \lim _{x\to a}f(x)=\lim _{x\to a}g(x)=\infty }$

L'Hopital's rule says that

${\displaystyle \lim _{x\to a}{\frac {f(x)}{g(x)}}=\lim _{x\to a}{\frac {f'(x)}{g'(x)}}}$

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

${\displaystyle f'(a)=\lim _{x\to a}{\frac {f(x)-f(a)}{x-a}}}$

Therefore,

${\displaystyle {\frac {f'(a)}{g'(a)}}=\lim _{x\to a}{\frac {\frac {f(x)-f(a)}{x-a}}{\frac {g(x)-g(a)}{x-a}}}=\lim _{x\to a}{\frac {f(x)-f(a)}{g(x)-g(a)}}}$

${\displaystyle {\frac {f'(a)}{g'(a)}}={\frac {f(a)}{g(a)}}=\lim _{x\to a}{\frac {f'(a)}{g'(a)}}}$

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

${\displaystyle f'(a)=\lim _{x\to 0}{\frac {f(x+a)}{x}}}$

Then

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

QED