Let $f:[-1,1]\rightarrow \Bbb R$ and $f(\sin(\frac{1}{n}))=\cos(\frac{1}{n})$ and $f'(0)$ exist. Prove that $f(0)=1$ 
Let $f:[-1,1]\rightarrow \mathbb{R}$ and $f(\sin(\frac{1}{n}))=\cos(\frac{1}{n})$
  and $f'(0)$ exist. Prove that $f(0)=1$.

Wwhat I did is because $f'(0)$ exist then $f$ is continuous at $0$ and 
$f(0)= \lim _{x\to 0}\left(f\left(x\right)\right) =\lim _{n\to \infty }\left(f\left(sin\left(\frac{1}{n}\right)\right)\right) =  \lim _{n\to \infty }\left(cos\left(\frac{1}{n}\right)\right) = 1$
Is this correct? if not how to prove it? 
thanks a lot 

Perfect answer! Your arguments are correct. 
