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?
 A: It is always true if $f(n) \neq 0$ in a punctured neighborhood of $x=n$ otherwise we can't take the limit.
A: Yes, assuming that:

$\exists \delta' > 0$ such that if $|x - n| < \delta'$, then $|f(x)| > 0$.


Since $f(n) = \lim_{x\to n}f(x) = 0$ and $\lim_{y \to 0} \frac{\sin y}{y} = 1$, we know that:

$\forall \varepsilon_1 > 0$, $\exists \delta_1 > 0$ such that if $|x - n| < \delta_1$, then $|f(x)| < \varepsilon_1$
$\forall \varepsilon_2 > 0$, $\exists \delta_2 > 0$ such that if $0 < |y| < \delta_2$, then $|\frac{\sin y}{y} - 1| < \varepsilon_2$

Now given any $\varepsilon > 0$, let $\delta > 0$ be $\min({\delta', \delta_1})$, where:

*

*$\delta_1$ corresponds to taking $\varepsilon_1 = \delta_2$

*$\delta_2$ corresponds to taking $\varepsilon_2 = \varepsilon$
Then observe that if $0 < |x - n| < \delta$, then:

*

*We know that $|x - n| < \delta \leq \delta'$, so $|f(x)| > 0$.

*We know that $|x - n| < \delta \leq \delta_1$, so $|f(x)| < \varepsilon_1 = \delta_2$.

*We know that $0 < |f(x)| < \delta_2$, so $|\frac{\sin f(x)}{f(x)} - 1| < \varepsilon_2 = \varepsilon$, as desired. $~~\blacksquare$
A: This is true, just by composition of limits. Indeed,
$$\lim_{x \rightarrow n} f(x)=0 \quad \text{and} \quad \lim_{y \rightarrow 0} \frac{\sin(y)}{y}=1$$
so composing the two limits, $$\lim_{x \rightarrow n} \frac{\sin(f(x))}{f(x)}=1$$
A: 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$$\frac{\sin(0)}{0}=1$$
