$f(\sin(1/n))=\cos(1/n)$ Let $f:\mathbb R\to\mathbb R$ with $f(\sin(1/n))=\cos(1/n)$ for $n\in\mathbb N$.
In addition, the derivative $f'$ exists at $0$.
I was asked to find $f'(0)$ = $?$
I found that $f(0) = 1$ and started solving using the definition of $f'(x)$.
So I got to:
$\lim_{h\to0} \frac{f(0+h)-f(0)}{h} = \lim_{n\to\infty} \frac{f(0+\sin(1/n))-f(0)}{\sin(1/n)}= \lim_{n\to\infty} \frac{f(\sin(1/n))-1}{\sin(1/n)} $
I get a $`` \frac{0}{0} " $
I do not know how to continue. 
Thank you!
 A: Final answer:
$g(x) = sin(x)$ is a continuous function in $\mathbb R$.
Let {$a_n$} be a sequence that $ \lim_{n\to\infty} a_n = 0$. Then, according to Heine theorem:
 $ \lim_{n\to\infty} sin(a_n) = sin(0)=0$.
According to the definition of $f'(x)$
\begin{align}
f'(0) 
&= \lim_{h\to 0} \frac{f(0+h)-f(0)}{h}\\
&= \lim_{n\to\infty} \frac{f(0+\sin(1/n))-f(0)}{\sin(1/n)}\\
&= \lim_{n\to\infty} \frac{f(\sin(1/n))-1}{\sin(1/n)}.
\end{align}
Let $x=1/n$:
\begin{align}
\lim_{x\to 0} \frac{f(0+\sin(x))-f(0)}{\sin(x)}
&= \lim_{x\to 0} \frac{f(\sin(x))-1}{\sin(x)}\\
&= \lim_{x \to 0} \frac{\cos(x) - 1}{\sin(x)}.
\end{align}
Using L'Hospital's Rule we get that:
$$
\lim_{x \to 0} \frac{\cos(x) - 1}{\sin(x)}=\lim_{x \to 0} \frac{-\sin(x)}{\cos(x)}=0=f'(0)
$$
Thank you all for your helpful and creative answers!
A: Note that, since $f\left(\sin\left(\frac1n\right)\right)=\cos\left(\frac1n\right)$, you have$$f\left(\sin\left(\frac1n\right)\right)-1=\cos\left(\frac1n\right)-1\approx-\frac1{2n^2}.$$But $\sin\left(\frac1n\right)\approx\frac1n$. Since$$\lim_{n\to\infty}\frac{-\frac1{2n^2}}{\frac1n}=0,$$$f'(0)=0$.
A: Hint: $$f(x)=\sqrt{1-x^2}$$ for every $x$ of the form $\sin\left(\frac 1n\right)$.
A: Hint: $\cos \frac{1}{n}=1-2\sin^2(\frac{1}{2n})$
