Proving there's a function $f$ such that $\lim_{x\to 0}f(x)=l \Rightarrow \lim_{x\to 0}f(\sin x)=l$ 
Prove that there's a function $f: (-a,a)\setminus\{ 0 \}$ such that if $\displaystyle\lim_{x\to 0}f(x)=l$ then $\displaystyle\lim_{x\to 0}f(\sin x)=l$

Well, the answer is simply the constant function $f(x)=c , \ \forall c\in\mathbb R$, do I prove it by contraposition: assume there's no such function then try the constant function thus getting a contradiction ?

Reading the question, the answer consists in any (explicit) function with the desired properties. You can answer $f(x)=x$ for $x \in (-a,a) \setminus \{0\}$ and prove that it has the desired property. This is very easy.
