Proof: L’Hopital’s theorem (special case )

Statement:

$f(x)$ and $g(x)$ are differentiable on an open interval containing $a$
$f(c)=g(c)=0$
$g'(x)\neq 0$ for $x$ near (but not equal to) $c$

Then

$\lim_{x \to c}\frac{f(x)}{g(x)}=\lim_{x \to c}\frac{f'(x)}{g'(x)}$ provided that the limit on the right side of the equations exists.

Proof:

Assume the above conditions. Then,

$\lim_{x \to c}\frac{f(x)}{g(x)}=\lim_{x \to c}\frac{f(x)-0}{g(x)-0}=\lim_{x \to c}\frac{f(x)-f(c)}{g(x)-g(c)}$

Dividing the right-hand expression by $x-c$ , we get

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

Therefore,

$\lim_{x \to c}\frac{f(x)}{g(x)}=\lim_{x \to c}\frac{f'(x)}{g'(x)}$ , which is what we sought to prove.