Suppose that $f(a)=g(a)=0$ and $g'(a) \ne 0$. Then, for any $x$, $f(x)=f(x)-f(a)$ and $g(x)=g(x)-g(a)$. But then,\begin{eqnarray*} \displaystyle{\lim_{x \to a} \frac{f(x)}{g(x)}} &=& \displaystyle{\lim_{x \to a} \,\frac{f(x)-f(a)}{g(x)-g(a)}}\\\\ &=& \displaystyle{ \lim_{x\to a}\, \frac{\displaystyle\quad\frac{f(x)-f(a)}{x-a}\quad}{\displaystyle\quad\frac{g(x)-g(a)}{x-a}\quad}} \\\\ &=& \displaystyle\frac{\displaystyle\lim_{x\to a} \,\,\Big(\frac{f(x)-f(a)}{x-a}\Big)\,\,}{\displaystyle\lim_{x\to a} \,\,\Big(\frac{g(x)-g(a)}{x-a}\Big)\,\,} \\\\ &=& \displaystyle{\frac{f'(a)}{g'(a)}}. \end{eqnarray*} By definition, $f'(a) = \displaystyle{\lim_{x\to a} \frac{f(x)-f(a)}{x-a}}$ and $g'(a) = \displaystyle{\lim_{x\to a}\frac{g(x)-g(a)}{x-a}}$. Since $f'$ and $g'$ are assumed to be continuous, $f'(a)/g'(a)$ is also $$\frac{\lim_{x \to a} f'(x)}{\lim_{x\to a}g'(x)} = \lim_{x\to a}\frac{f'(x)}{g'(x)}.\qquad \hbox{QED}$$

Souped Up Mean Value Theorem: If $f(x)$ and $g(x)$ are continuous on a closed interval $[a,b]$ and differentiable on the open interval $(a,b)$, then there is a point $c$, between $a$ and $b$, where $$\big(f(b)-f(a)\big)g'(c) = \big(g(b)-g(a)\big)f'(c). $$ (When $g(x)=x$, this is the same as the usual MVT.)

Proof: Consider the function $$ h(x) = \big(f(x)-f(a)\big)\big(g(b)-g(a)\big) - \big(f(b)-f(a)\big)\big(g(x)-g(a)\big).$$ This is continuous on $[a,b]$ and differentiable on $(a,b)$, with $$h'(x) = f'(x)\big(g(b)-g(a)\big) - g'(x) \big(f(b) - f(a)\big).$$ Note that $h(a)=0=h(b)$. By Rolle's Theorem, there a spot $c$ where $h'(c)=0$. But $h'(c)= \big(f(b)-f(a)\big)g'(c) - \big(g(b)-g(a)\big)f'(c).$ Since this is zero, $\big(f(b)-f(a)\big)g'(c) = \big(g(b)-g(a)\big)f'(c)$. $\qquad \hbox{QED}$

By assumption, $f$ and $g$ are differentiable to the right of $a$, and the limits of $f$ and $g$ as $x \to a^+$ are zero. Define $f(a)$ to be zero, and likewise define $g(a)=0$. Since these values agree with the limits, $f$ and $g$ are continuous on some half-open interval $[a,b)$ and differentiable on$(a,b)$.

For any $x \in (a,b)$, we have that $f$ and $g$ are differentiable on $(a,x)$ and continuous on $[a,x]$. By the Souped up MVT, there is apoint $c$ between $a$ and $x$ such that $f'(c) g(x) = f'(x) g(c)$. In other words, $f'(c)/g'(c) = f(x)/g(x)$. Also, as $x$ approaches $a$, $c$ also approaches $a$, since $c$ is somewhere between $x$ and $a$. But then $$\lim_{x \to a^+}\, \frac{f(x)}{g(x)} = \lim_{x \to a^+} \,\frac{f'(c)}{g'(c)} = \lim_{c \to a^+}\,\frac{f'(c)}{g'(c)}.$$ That last expression is the same as $\lim_{x \to a^+} f'(x)/g'(x)$. $\qquad \hbox{QED}$

We're going to use a single trick, over and over again. Namely, we can always rewrite $x$ as $\frac{1}{1/x}$, $f(x)$ as $\frac{1}{1/f(x)}$ and $g(x)$ as $\frac{1}{1/g(x)}$.
Suppose $L= \lim_{x \to a} \frac{f(x)}{g(x)}$, where both $f$ and $g$ go to $\infty$ (or $-\infty$) as $x \to a$. Also suppose that $L$ is neither 0 nor infinite. Then $$ L = \lim_{x \to a}\, \frac{f(x)}{g(x)} = \lim_{x \to a}\, \frac{1/g(x)}{1/f(x)}.$$ Since $1/g(x)$ and $1/f(x)$ go to zero as $x \to a$, we can apply the (baby or macho) L'Hospital's rule to this limit: \begin{eqnarray*}L &=& \lim_{x \to a}\, \frac{ (1/g)'}{(1/f)'} \\\\&=& \lim_{x \to a} \,\frac{-g'(x)/g(x)^2}{-f'(x)/f(x)^2} \cr\cr\cr&=& \lim_{x\to a}\, \frac{f(x)^2 \cdot g'(x)}{g(x)^2\cdot f'(x)} \cr\cr\cr&=& \lim_{x \to a}\,\frac{f(x)^2}{g(x)^2}\,\cdot\, \lim_{x \to a}\, \frac{g'(x)}{f'(x)} \\\\& = & \frac{L^2}{\displaystyle\lim_{x \to a}\, \Big(f'(x)/g'(x)\Big)}.\end{eqnarray*} Since $L = \displaystyle\frac{L^2}{\lim_{x \to a} \big(f'(x)/g'(x)\big)}$, $L$ must equal $\displaystyle \lim_{x \to a}\Big(f'(x)/g'(x)\Big)$, which is what we wanted to prove.

This argument only works for finite and nonzero values of $L$. However, if $L=0$, we can apply the same argument to the limit of $\big(f(x)+g(x)\big)/g(x)$, which then does not equal zero. The upshot is that $$ 1 + \lim_{x \to a}\,\frac{f(x)}{g(x)} = \lim_{x\to a}\, \frac{f(x)+g(x)}{g(x)} = \lim_{x \to a}\, \frac{f'(x) + g'(x)}{g'(x)} = 1 + \lim_{x \to a}\,\frac{f'(x)}{g'(x)},$$ hence that $\lim\, (f/g) = \lim \,(f'/g')$. Finally, if $\lim\,(f/g)=\pm \infty$, look instead at $\lim\,(g/f)$, which is then zero, so the previous reasoning applies. Since $0=\lim\,(g/f)=\lim\,(g'/f')$, $\lim\,(f'/g')$ must be infinite. By the Souped up MVT, $f/g$ has the same sign as $f'/g'$, so we must have $\lim\,(f/g)=\lim\,(f'/g')$.

Now that we have L'Hopital's Rule for limits as $x \to a$ (or $x \to a^+$ or $x \to a^-$), we consider what happens as $x \to \infty$. Define a new variable $t = 1/x$, so that $x \to \infty$ is the same as $t \to 0^+$.Then $$ \lim_{x \to \infty}\, \frac{f(x)}{g(x)} = \lim_{t \to 0^+}\, \frac{f(1/t)}{g(1/t)}.$$ But we know how to apply L'Hospital's Rule to limits as $t \to 0$, so this turns into $$\lim_{t \to 0^+}\, \frac{\frac{d}{dt} f(1/t)}{\frac{d}{dt}g(1/t)} = \lim_{t \to 0^+} \frac{-f'(1/t)\cdot \frac{1}{t^2}}{-g'(1/t)\cdot \frac{1}{t^2}}= \lim_{t \to 0^+} \frac{f'(1/t)}{g'(1/t)}.$$ Converting back to $x=1/t$, we get $$ \lim_{x \to \infty} \frac{f'(x)}{g'(x)},$$ which is what we wanted.

Computing a limit as $x \to -\infty$ is similar, only with $t \to 0^-$ instead of $t \to 0^+$. That completes the proof of L'Hospital's Rule. $\qquad \hbox{QED}$