Let $f : \mathbb{R}\to \mathbb{R}$ be given.
Suppose that $\lim_{x\to 0} f(x) = L$ and $f(0) = L$ . Prove that $\lim_{x\to 0}(f \circ g)(x) = L$, where $g(x) = xD(x)$:
How do I approach this? Fairly new to Calc and I tried looking for examples for some time but I can't quite explain
If it is the case that for all $|x|<\epsilon$, for some $\epsilon>0$, that
$$|D(x)| Then by squeeze theorem you should be able to see that $$-M|x|\le xD(x)\le M|x|$$ $$\lim_{x\to0}xD(x)=0$$ And we know that if $f(x)$ is continuous around $x=0$, then $$\lim_{x\to0}f(g(x))=f(\lim_{x\to0}g(x))=f(0)=0$$