Let $f\in(-a,a)\smallsetminus\{0\} \to \mathbb{R}$ prove that: $\lim_{x\rightarrow 0} f(x)=L \Leftrightarrow \lim_{x\rightarrow 0} f( \sin(x))=L$

Question

$\Rightarrow $) Let $ \epsilon > 0$, $\exists \delta > 0$ such that $0< \left | x \right |< \delta \Rightarrow \left | f(x)-L \right |< \epsilon$. As $\forall x \in \mathbb{R}$ , $\left | \sin (x) \right | \leq x$,then $0< \left | \sin(x) \right |< \delta $ and therefore $|x|<\delta\implies\bigl|f\bigl(\sin(x)\bigr)-L\bigr|<\epsilon.$

$\Leftarrow$) have problems with this proof... contradiction?

Answer 1

Some fact should be assumed, if not, it would be hard:

Fact: $\sin x$ is strictly increasing in a small neighborhood of zero, and $\sin 0=0$, and $\sin^{-1}x\rightarrow 0$ as $x\rightarrow 0$.

Assume that $\lim_{x\rightarrow 0}f(\sin x)=L$. Given $\epsilon>0$, there is some $\delta>0$ such that $|f(\sin x)-L|<\epsilon$ for all $x$ with $0<|x|<\delta$.

We can choose such a $\delta>0$ small enough such that $\sin:(-\delta,\delta)\rightarrow(-\sin\delta,\sin\delta)$ is strictly increasing.

For all $0<|y|<\sin\delta$, then $0<|\sin^{-1}y|<\delta$, so $|f(y)-L|=|f(\sin(\sin^{-1}y))-L|<\epsilon$.