If $\lim_{x \to 0}g(x)=L$. How do I prove $\lim_{h \to 0} g(f(x+h)-f(x)) = \lim_{y \to 0}g(y)$ using $\epsilon-\delta$

Question

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)

Answer 1

For given $\epsilon >0 ~~~\exists~ \delta'>0$ such that $|g(x)-L|<\epsilon$ whenever $|x|<\delta'$ $\hspace{1cm}$ $\cdots$ (1).

Now since $f$ is continuous at $x$, for the above $\delta'>0~~~\exists~\delta>0$ such that $|f(x+h)-f(x)|<\delta'$ whenever $|h|<\delta$. From (1) we get $|g(f(x+h)-f(x))-L|<\epsilon$ whenever $|h|< \delta$. This proves it.

Answer 2

Your statement $(1)$ is correct but incomplete as the way it stands it doesn't use continuity of $g$. Note that since $g$ is continuous, it's safe to write $|k|\lt \delta$ rather than restricting to $0\lt |k|\lt \delta$.

Statement $(2)$ is not required as it is same as $(1)$

Then "if I take $f(x+h)-f(x)=k$, then RHS of $(2)$ and $(3)$ are same" Why? That's not true $f(x+h)-f(x)$ may not be a constant. You think that it is always true that $f(x+10^{-Googol})=f(x+10^{-100})?$

You may proceed as below also:

$g$ is continuous at $0$ and $\lim_{x\to 0}g(x)=L$

$\begin{align} &\implies \forall\epsilon\gt 0, \exists \delta\gt 0: |x|\lt \delta\implies |g(x)-L|\lt\epsilon \end{align}$$\tag 1$
$f$ is continuous at $x$ and therefore, $\lim_{h\to 0} f(x+h)-f(x)=0$. It follows that $\exists \delta'\gt 0: |h|\lt \delta'\implies |f(x+h)-f(x)|\lt\delta \tag 2 $
From (1) and (2) above, it follows that $\forall\epsilon \gt 0,\exists \delta': |h|\lt \delta'\implies |g(f(x+h)-f(x))-L|\lt \epsilon $. This proves that $\lim_{h\to 0}g(f(x+h)-f(x))=L$.