Showing that $\lim_{x \to 0}\frac{f(x)}{g(x)} = \frac{f'(0)}{g'(0)}$. If $f$ and $g$ are differentiable functions with $f(0) = g(0) = 0$ and $g'(0) \neq 0$, show that $\lim_{x \to 0}\frac{f(x)}{g(x)} = \frac{f'(0)}{g'(0)}$.
I consider that perhaps:
$$
\begin{align}
\\ \lim_{x \to 0}\frac{f(x)}{g(x)} &= \lim_{x \to 0}\frac{f(0+x) - f(0)}{x} \cdot \frac{1}{ \lim_{x \to 0}\frac{g(0+x) - g(0)}{x}} = f'(0) \cdot \frac{1}{g'(0)} = \frac{f'(0)}{g'(0)}
\end{align}
$$
But, it seems like that's maybe not quite right. I'm not certain.
Insight?
 A: \begin{align}
\lim_{x \to 0}\frac{f(x)}{g(x)} &= \lim_{x \to 0}\frac{f(0+x) - f(0)}{x} \cdot \frac{x}{g(0+x) - g(0)}
\end{align}
by simple algebra and substitution. 
Now the limit of a product is the product of the limits if both exist. So to perform the split you wanted, you'll need to show that. The first limit certainly exists, because $f$ is known to be differentiable at 0. What about the second? 
Well, if $\lim_{x \to 0} h(x) = m \ne 0$, then $\lim_{x \to 0} \frac{1}{h(x)} = \frac{1}{m}$. (This is Theorem 2 of chapter 5 of Spivak's Calculus, if you need a reference). We'd like to apply this to the function 
$$
h(x) = \frac{g(0+x) - g(0)}{x}
$$
but to do so, we need to know that $\lim_{x \to 0} h(x)$ exists and is nonzero. 
Fortunately for us, $g$ is differentiable at 0 (by assumption), and its derivative, $g'(0)$, is exactly defined to be $\lim_{x \to 0} h(x)$, which therefore exists. Furthermore, the hypothesis $g'(0) \ne 0$ tells us that it's nonzero. So we can conclude that 
\begin{align}
\lim_{x \to 0} \frac{1}{\frac{g(0+x) - g(0)}{x}} &= \frac{1}{\lim_{x \to 0} \frac{g(0+x) - g(0)}{x}}\\
&= \frac{1}{g'(0)}. 
\end{align}
Finally, the limit on the left is, through a little algebra, just 
\begin{align}
\lim_{x \to 0} \frac{1}{\frac{g(0+x) - g(0)}{x}} &= \lim_{x \to 0} \frac{x}{g(0+x) - g(0)}.
\end{align}
So both the limits in the product into which we want to split our original expression exist and hence we have
\begin{align}
\lim_{x \to 0}\frac{f(x)}{g(x)} &= \lim_{x \to 0}\frac{f(0+x) - f(0)}{x} \cdot \frac{x}{g(0+x) - g(0)} \\
 &= \lim_{x \to 0}\frac{f(0+x) - f(0)}{x} \cdot \frac{1}{ \lim_{x \to 0}\frac{g(0+x) - g(0)}{x}} \\
& = f'(0) \cdot \frac{1}{g'(0)} = \frac{f'(0)}{g'(0)}
\end{align}
