Famous Theorems of Mathematics/L'Hôpital's rule

Consider some limit

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

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

$\underset{x\to a}{\lim }f(x)=\underset{x\to a}{\lim }g(x)=\mathrm{\infty }$

L'Hopital's rule says that

$\underset{x\to a}{\lim }\frac{f(x)}{g(x)}=\underset{x\to a}{\lim }\frac{f^{\prime }(x)}{g^{\prime }(x)}$

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

$f^{\prime }(a)=\underset{x\to a}{\lim }\frac{f(x)-f(a)}{x-a}$

Therefore,

$\frac{f^{\prime }(a)}{g^{\prime }(a)}=\underset{x\to a}{\lim }\frac{\frac{f(x)-f(a)}{x-a}}{\frac{g(x)-g(a)}{x-a}}=\underset{x\to a}{\lim }\frac{f(x)-f(a)}{g(x)-g(a)}$

$\frac{f^{\prime }(a)}{g^{\prime }(a)}=\frac{f(a)}{g(a)}=\underset{x\to a}{\lim }\frac{f^{\prime }(a)}{g^{\prime }(a)}$

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

$f^{\prime }(a)=\underset{x\to 0}{\lim }\frac{f(x+a)}{x}$

Then

$\frac{f^{\prime }(a)}{g^{\prime }(a)}=\underset{x\to 0}{\lim }\frac{xf(x+a)}{xg(x+a)}=\underset{x\to 0}{\lim }\frac{f(x+a)}{g(x+a)}=\frac{f(x+a)}{g(x+a)}$

QED