Limit of strange function composition Known facts. 
$f$ and $g$ are continious and exist in the neighbourhood of $x=0$. 
$\lim_{x\to 0}a(x)=\lim_{x\to 0}b(x)=0$.
$\lim_{x\to 0} \frac{f(x)}{g(x)}$ exists.
$\lim_{x\to 0} \frac{a(x)}{b(x)} = 1$
Problem. 
I want to investigate and possibly also prove the statement
$$\lim_{x\to 0}\frac{f\left(a(x)\right)}{g\left(b(x)\right)}=\lim_{x\to 0}\frac{f(x)}{g(x)}$$
My thoughts.
I know that $$\lim_{x\to 0} f\left(a(x)\right)=\lim_{x\to 0} f(x)=f(0)$$ and $$ \lim_{x\to 0} g(b(x))=\lim_{x\to 0} g(x)=g(0)$$
In the case of $g(0)\neq 0$, we get
$$\lim_{x\to 0} \frac{f(a(x))}{g(b(x))}=\frac{\lim_{x\to 0} f(a(x))}{\lim_{x\to 0} g(b(x))}=\frac{\lim_{x\to 0} f(x)}{\lim_{x\to 0} g(x)}=\lim_{x\to 0}\frac{f(x)}{g(x)}$$
but this is not valid when $g(0)=0$ (since it would give division by zero) and that may be the case here.
Now I really don't know how to proceed to prove t. I'm thinking that the $\epsilon-\delta$ defintions of the limits may be useful somehow, but I can't see how. This problem doesn't really seem to resemble anything I have seen before...

Under the assumptions given, the desired conclusion is not true in general.  Just let $f(x)=g(x)=a(x)=x$ and let $b(x)=2x$.  Clearly $$\lim_{x\to0}a(x)=\lim_{x\to0}b(x)=0$$
but
$$\lim_{x\to0}{f(x)\over g(x)}=1$$
while
$$\lim_{x\to0}{f(a(x))\over g(b(x))}={1\over2}$$
A suggestion:  See what you can prove if you add the assumption
$$\lim_{x\to0}{a(x)\over b(x)}=1$$
