Why are the two limits equal? I want to show that if $g$ is continuous at $a$ and $f$ at $g(a)$, then
$$\lim_{x \to a}{\frac{f(g(x))-f(g(a))}{g(x)-g(a)}} = \lim_{x \to g(a)}{\frac{f(x)-f(g(a))}{x-g(a)}}$$
Now I know that continuity implies that 
$$\lim_{x \to a}{f(g(x))} = \lim_{x \to g(a)}{f(x)} = f(g(a))$$
and
$$\lim_{x \to a}{g(x)} = \lim_{x \to g(a)}{x} = g(a)$$
so it is quite easy to see that the two original limits are equal. How do I prove this?
I cannot repeatedly use the limit laws since I get a limit that is zero ($\lim_{x \to a}{(g(x)-g(a))}$) and so cannot apply the quotient rule.
 A: Right, you need to show that:
$$\lim_{x\to a} h(g(x))=\lim_{y\to g(a)} h(y)$$
when $g$ is continuous at $a$. This is true if the right hand side exists, and if $g(x)\neq g(a)$ in some neighborhood of $a$ except when $x=a$ (with this condition, you don't need $h(g(a))$ defined, which is good for your above case.)
To show that the left side can exist, but the right side not, just take;
$$g(x)=\begin{cases}0&x=0\\
\frac{1}{\lceil 1/|x|\rceil}&x\neq 0
\end{cases}$$
The $\lim_{x\to 0} g(x)=0$. 
Now define $h(x)=\sin(\pi/x)$.
Then $\lim_{y\to 0} h(y)$ does not exist, but $\lim_{x\to 0} h(g(x))$ does.
So, let's assume $\lim_{y\to g(a)} h(y)$ exists and equals $L$, and that $g(x)\neq g(a)$ for $x\neq a$ and $|x-a|<\alpha$ for some $\alpha>0$, and that $g$ is continuous at $a$.
Given $\epsilon>0$ this means there is a $\delta>0$ so that if $|y-g(a)|<\delta$ and $y\neq g(a)$ then $|h(y)-L|<\epsilon$.
Since $g$ is continuous at $a$, this means that there is a $\delta_2$ such that if $|x-a|<\delta_2$ and $x\neq a$, we have $|g(x)-g(a)|<\delta$ and if $g(x)\neq g(a)$, we can conclude that $|h(g(x))-L|<\epsilon.$ So if $|x-a|<\min(\delta_2,\alpha)=\delta_3$ we know that $h(g(x))-L|<\epsilon$. 
