What is the $\lim\limits_{x \rightarrow 0} \frac{f(x)}{g(x)}$ Assume that $f(x)$ and $g(x)$ are both differentiable functions such that 
$f(0) = g(0) = 0$ and $g'(0)$ does not equal $0$.
What is the $$\lim_{x\to0}\frac{f(x)}{g(x)}$$
in terms of $f(x)$, $g(x)$, $f'(x)$ and $g'(x)$.
Sidenote: Unable to use L'Hopital's rule
 A: Well, this problem is setup for L'Hopital, so I guess what you need to do is run through its proof:
$$\lim_{x\to 0}\frac{f(x)}{g(x)}=\lim_{x\to 0}\frac{f(x)-f(0)}{g(x)-g(0)}=\lim_{x\to 0}\frac{\frac{f(x)-f(0)}{x}}{\frac{g(x)-g(0)}{x}}=\frac{\lim_{x\to 0}\frac{f(x)-f(0)}{x}}{\lim_{x\to 0}\frac{g(x)-g(0)}{x}}=\frac{f'(0)}{g'(0)}$$
Make sure you can justify why each equality holds!
A: Use one of the forms of the definition of derivative:
$$
f(x)=f'(0)\,x+e_1(x),\quad g(x)=g'(0)\,x+e_2(x)
$$
where
$$
\lim_{x\to0}\frac{e_1(x)}{x}=\lim_{x\to0}\frac{e_2(x)}{x}=?
$$
A: Hint:
$$
\frac{f(x)}{g(x)}=\frac{\frac{f(x)-f(0)}{x-0}}{\frac{g(x)-g(0)}{x-0}}
$$
for $x\neq0$ sufficiently close to $0$.
A: Note that
$$
\frac{f(x)}{g(x)}=\frac{\frac{f(x)-f(0)}{x}}{\frac{g(x)-g(0)}{x}},
$$
and 
$$
\frac{f(x)-f(0)}{x}\to f'(0), \quad \frac{g(x)-g(0)}{x}\to g'(0).
$$
Hence
$$
\frac{f(x)}{g(x)}=\frac{\frac{f(x)-f(0)}{x}}{\frac{g(x)-g(0)}{x}}\to \frac{f'(0}{g'(0)}.
$$
A: The answer $\frac{f'(0)}{g'(0)}$ follows from the definition of the derivative. Rewrite $\frac{f(h)}{g(h)}$ as $\frac{\frac{f(h)-f(0)}{h}}{\frac{g(h)-g(0)}{h}}$.
That has been amply described in answers already. We look only at a a small technical issue. Everything is fine if $g(h)\ne 0$. But we need to make sure that there is an interval $(-\epsilon, \epsilon)$ such that if $h$ is in this interval and $h\ne 0$, then $g(h)\ne 0$. 
Suppose to the contrary that there is a sequence $h_1, h_2, \dots$ of non-zero numbers with limit $0$ such that $g(h_i)=0$ for all $i$. Then for these $h_i$ we have $\frac{g(h_i)-g(0)}{h_i}=0$, contradicting the fact that $g'(0)\ne 0$. 
