We will use an extension of the Mean Value Theorem:

Extended (Cauchy) Mean Value Theorem

Let $f$ and $g$ be differentiable on $(a,b)$ and continuous on $[a,b]$. Suppose that $g'(x)\neq 0$ in $(a,b)$. Then there is at least one point $c$ in $(a,b)$ such that \[\frac{f'(c)}{g'(c)}=\frac{f(b)-f(a)}{g(b)-g(a)}.\] The proof of this theorem is fairly simple and can be found in most calculus texts.

We will now sketch the proof of L'Hôpital's Rule for the $\frac{0}{0}$ case in the limit as $x\to c^+$, where $c$ is finite. The case $x\to c^-$ can be proven in a similar manner, and these two cases together can be used to prove L'Hôpital's Rule for a two-sided limit. This proof is taken from Salas and Hille's Calculus: One Variable.

Let $f$ and $g$ be defined on an interval $(c,b)$, where $f(x)\to 0$ and $g(x)\to 0$ as $x\to c^+$ but $\displaystyle \frac{f'(x)}{g'(x)}$ tends to a finite limit $L$. Then $f'$ and $g'$ exist on some set $(c, c+g]$ and $g'\neq 0$ on $(c, c+h]$. Also, $f$ and $g$ are continuous on $[c,c+h]$, where we define $f(c)=0$ and $g(c)=0$.

By the Extended Mean Value Theorem, there exists $c_h\in (c,c+h)$ such that \[\frac{f'(c_h)}{g'(c_h)}=\frac{f(c+h)-f(c)}{g(c+h)-g(c)}=\frac{f(c+h)} {g(c+h)}\] since $f(c)=g(c)=0$. Letting $h\to 0^+$, $\displaystyle \lim_{h\to 0^+}\, \frac{f'(c_h)}{g'(c_h)}=\lim_{x\to c^+}\, \frac{f'(x)}{g'(x)}$ while $\displaystyle \lim_{h\to 0^+}\, \frac{f(c+h)}{g(c+h)}=\lim_{x\to c^+}\, \frac{f(x)}{g(x)}$. Thus, \[\lim_{x\to c^+}\frac{f(x)}{g(x)}=\lim_{x\to c^+} \frac{f'(x)}{g'(x)}.\]