If $\lim_{x\to n}f(x) = 0$ and $f(n)=0$, does $\lim_{x\to n}\frac{\sin(f(x))}{f(x)} = 1$ always? If $\lim_{x\to n}f(x) = 0$ and $f(n)=0$, does $\lim_{x\to n}\frac{\sin(f(x))}{f(x)} = 1$ always?
I have been playing around with some graphs on desmos and there's always the indication that the limit equals $1$. I know that $\sin x$ becomes very linear, is there any function with the potential to "un-linearize" it?

hint
Let $ \epsilon>0$.
we know that
$$\lim_{X\to 0,\ne}\frac{\sin(X)}{X}=1$$
Thus
$\exists \eta>0$ such that
$$\color{red}{0<}|X|<\eta \implies |\frac{\sin(X)}{X}-1|<\epsilon$$
But
$$\lim_{x\to 0}f(x)=0$$
then
$\exists \alpha>0 $ such that
$$|x|<\alpha \implies |f(x)|<\eta$$
The red condition $ \color{red}{0<|f(x)|}$ is not always satisfied to insure the existence of $\frac{\sin(f(x))}{f(x)}$.
Your conclusion will be correct if you put$$\frac{\sin(0)}{0}=1$$
