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 &quot;un-linearize&quot; it?
It is always true if $f(n) \neq 0$ in a punctured neighborhood of $x=n$ otherwise we can't take the limit.
hint
Let $ \epsilon&gt;0$.
we know that
$$\lim_{X\to 0,\ne}\frac{\sin(X)}{X}=1$$
Thus
$\exists \eta&gt;0$ such that
$$\color{red}{0&lt;}|X|&lt;\eta \implies |\frac{\sin(X)}{X}-1|&lt;\epsilon$$
But
$$\lim_{x\to 0}f(x)=0$$
then
$\exists \alpha&gt;0 $ such that
$$|x|&lt;\alpha \implies |f(x)|&lt;\eta$$
The red condition $ \color{red}{0&lt;|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$$\frac{\sin(0)}{0}=1$$