If I know $\lim\_{x \to 0}g(x)=L$. For my particular problem $L=0$.
So I've been trying to prove: $\lim\_{h \to 0} g(f(x+h)-f(x)) = \lim\_{y \to 0}g(y)=0$ using $\epsilon-\delta$.  

(Also I'm assuming the $g(x)$ is continuous at $0$ and $f(x)$ is continuous at $x$)

So I know that $\forall\epsilon$, **there is** some $\delta$ such that
$0<|k|<\delta\Rightarrow|g(k)-L|<\epsilon$ **(1)**

I have to prove that $\forall\epsilon\_1$,$\exists\delta\_1$ such that
$0<|h|<\delta\_1\Rightarrow|g(f(x+h)-f(x))-L|<\epsilon\_1$

If $\exists\epsilon=\epsilon'$, and $\delta=\delta'$ satisfies **(1)**, such that
$0<|k|<\delta'\Rightarrow|g(k)-L|<\epsilon'$ **(2)**

If I to find value for $\delta\_1$, when $\epsilon\_1=\epsilon'$ such that
$0<|h|<\delta\_1\Rightarrow|g(f(x+h)-f(x))-L|<\epsilon'$ **(3)**  

I can prove $\forall\epsilon\_1,\exists\delta\_1$.

If I take $f(x+h)-f(x)=k$, then RHS of **(2)** and **(3)** are same.
$$\Rightarrow0<|f(x+h)-f(x)|<\delta'\Rightarrow|g(f(x+h)-f(x))-L|<\epsilon'$$
So I have to find $\delta\_1$ such that
$0<|h|<\delta\_1\Rightarrow0<|f(x+h)-f(x)|<\delta'$?

This seems similar to another $\epsilon-\delta$ type condition but...I'm not sure.  

I know $\lim\_{h\to0}f(x+h)=f(x)$, since I assumed $f(x)$ is continuous at $x$.  

Since $|h|>0\Rightarrow f(x+h)\neq f(x)\Rightarrow|f(x+h)-f(x)|>0$? If this is true, I think it has been proved.  

(Also please check if there are any other mistakes or misassumptions)