calculus - Given $lim_{xto0}frac{sin x}{x}=1$, why is it true $lim_{xto 0}frac{sin(tan x)}{tan x} = lim_{tan xto 0}frac{sin(tan x)}{tan x}=1$?

Given $\lim_{x\to0}\frac{\sin(x)}{x}=1$, why is it true $\lim_{x\to > 0}\frac{\sin(\tan(x))}{\tan(x)} = \lim_{\tan(x)\to > 0}\frac{\sin(\tan(x))}{\tan(x)}=1$?

I know the theorem that sates that $\lim_{x\to0}\frac{\sin(x)}{x}=1$; nonetheless, I have seen numerous times that, given a function $f$ for which this works, they do something like $\lim_{x\to0}\frac{\sin(f(x))}{f(x)}=1$, the last time I saw something like this is $\lim_{x\to 0}\frac{\sin(\tan(x))}{\tan(x)} = \lim_{\tan(x)\to 0}\frac{\sin(\tan(x))}{\tan(x)}=1$.

I want to know how can I apply this in general, correctly, since up until now, I've applying it informally kind of $\lim_{x\to0}\frac{\sin(\text{something in terms of $x$})}{\text{same something in terms of $x$}}=1$.

Could you help me with this?

thanks in advance.

Answer

It is true in general that if $\lim_{x\to a}f(x)=0$, and $f(x)\neq0$ close to $a$, then
$$\lim_{x\to a}\frac{\sin(f(x))}{f(x)}=1$$
This is rather straightforward to prove from the $\epsilon$-$\delta$ definition of limits.

For your special case, we have $f(x)=\tan(x)$ and $a=0$.