If $\lim_{x\to 0}{f(x)} = \lim_{x\to 0}{g(x)}=0$, then is $\lim_{x\to 0}\frac{\sin f(x)}{g(x)}= \lim_{x\to 0}\frac {f(x)} {g(x)}$? I have been playing around with this identity for a while:
$$\lim_{x\to 0}\frac{\sin ax}{bx} = \frac {a} {b}$$
What I realized is that I can generalize this as such:
$$\lim_{x\to 0}\frac{\sin f(x)}{g(x)}= \lim_{x\to 0}\frac {f(x)} {g(x)}$$
where
$$\lim_{x\to 0}{f(x)} = \lim_{x\to 0}{g(x)}=0$$
Can you help me figure out whether this is true or not?
I have literally tried all functions I could think of and couldn't find any counterexamples.
I tried using L'Hôpital's rule but I couldn't prove it that way.
I think an epsilon-delta sytle proof could be nice but I don't know how to do it.
I even asked a few math teachers about this in my high school but none could make sense of it. 
I would really appreciate some help here.
 A: If $\lim_{x\to0}f(x)=0$, then 
$$\tag1
\lim_{x\to0}\frac{\sin f(x)}{f(x)}=1. 
$$
Then, if $\lim_{x\to0}\frac{f(x)}{g(x)}=L$
$$
\lim_{x\to0}\frac{\sin f(x)}{g(x)}=\lim_{x\to0}\frac{\sin f(x)}{f(x)}\,\lim_{x\to0}\frac{f(x)}{g(x)}=1\times L=L,
$$
using the fact that if both limits exist, then the product of the limit is the limit of the product. 
The case where $f$ is identically $0$ requires a different (but trivial!) argument and the result still holds. 

To justify $(1)$ formally, we the fact that $\lim_{x\to0}\frac{\sin x}x=1$ means that given any $\varepsilon>0$ there exists $\delta>0$ such that $|x|<\delta$ implies $\left|\frac{\sin x}x-1\right|<\varepsilon$. Now, given $\varepsilon_1>0$, from $\lim_{x\to0}f(x)=0$ there exists $\delta_1$ such that $|x|<\delta_1$ implies $|f(x)|<\varepsilon_1$. 
So, given $\varepsilon>0$, use the $\delta$ from above as the $\varepsilon_1$ for $f$, and so we get a $\delta_1$ such that $|x|<\delta_1$ implies $|f(x)|<\delta$, which in turn implies 
$$
\left|\frac{\sin f(x)}{f(x)}-1\right|<\varepsilon. 
$$
A: Assuming that $f(x)\neq0$ for $x\sim0$,$$\lim\limits_{x\to0}\frac{\sin(f(x))}{g(x)}=\lim\limits_{x\to0}\frac{\sin(f(x))}{f(x)}\cdot\frac{f(x)}{g(x)}$$It's not hard to prove that $\lim\limits_{x\to0}\frac{\sin(f(x))}{f(x)}=1$ if $\lim\limits_{x\to0}f(x)=0$ by a standard $\epsilon-\delta$ proof, so this ensures that $$\lim\limits_{x\to0}\frac{\sin(f(x))}{g(x)}=\lim\limits_{x\to0}\frac{\sin(f(x))}{f(x)}\cdot\lim\limits_{x\to0}\frac{f(x)}{g(x)}=\lim\limits_{x\to0}\frac{f(x)}{g(x)}$$
A: Assume that for some $\delta_{0}>0$, $f(x)/g(x)$ is not of the form $0/0$ for all $x\in(-\delta_{0},\delta_{0})$ and that $\lim_{x\rightarrow 0}f(x)/g(x)=L$.
We have to consider two cases. First, since $\lim_{x\rightarrow 0}f(x)=0$, if there were $x_{n}\downarrow 0$ such that $f(x_{n})\rightarrow 0$, then for large $n$, $x_{n}\in(-\delta_{0},\delta_{0})$, we deduce that $f(x_{n})/g(x_{n})\rightarrow 0$, this means that $L=0$.
The following $\epsilon$-argument will treat the case that $L\ne 0$, for then we can find some $\delta_{0}'>0$ such that $|f(x)|>0$ for all $x\in(-\delta_{0}',\delta_{0}')$. Anyway, the $\epsilon$-argument for the case that $\lim_{x\rightarrow 0}\dfrac{\sin f(x)}{g(x)}=L=0$ is similar and will be omitted.
We know that $\lim_{u\rightarrow 0}\dfrac{\sin u}{u}=1$, so given $\epsilon>0$, there exists some $\eta\in(0,\epsilon)$ such that $0<|u|<\eta$ implies that $|\varphi(u)-1|<\epsilon$.
Now choose a $\delta\in(0,\delta_{0}')$ such that for all $0<|x|<\delta$, $\left|\dfrac{f(x)}{g(x)}-L\right|<\epsilon$ and that $|f(x)|<\eta$. For the chosen $\delta$, we also have $|f(x)|>0$, so $0<|f(x)|<\eta$, then 
\begin{align*}
\left|\dfrac{\sin f(x)}{g(x)}-L\right|&=\left|\dfrac{\sin f(x)}{f(x)}\dfrac{f(x)}{g(x)}-L\right|\\
&=\left|\left(\dfrac{\sin f(x)}{f(x)}-1\right)\dfrac{f(x)}{g(x)}+\dfrac{f(x)}{g(x)}-L\right|\\
&\leq\left|\dfrac{\sin f(x)}{f(x)}-1\right|\left|\dfrac{f(x)}{g(x)}\right|+\left|\dfrac{f(x)}{g(x)}-L\right|\\
&=\left|\varphi(f(x))-1\right|\left|\dfrac{f(x)}{g(x)}\right|+\left|\dfrac{f(x)}{g(x)}-L\right|\\
&\leq\epsilon(|L|+\eta)+\epsilon\\
&\leq\epsilon(|L|+\epsilon)+\epsilon,
\end{align*}
we are done.
A: More generally, for any functions $\phi,\psi$, and functions $f,g$ as in the OP, if $$\lim_{x\to0}\frac{\phi(x)}x=L$$ and $$\lim_{x\to0}\frac{\psi(x)}x=M$$ then
$$\lim_{x\to 0}\frac{\phi(f(x))}{\psi(g(x))}=\frac LM\lim_{x\to 0}\frac{f(x)}{g(x)}.$$
This is because$$\lim_{x\to 0}\frac{\phi(f(x))}{\psi(f(x))}=\lim_{x\to 0}\frac{\phi(f(x))}{f(x)}\frac{g(x)}{\psi(f(x))}\frac{f(x)}{g(x)}=\lim_{x\to 0}\frac{\phi(f(x))}{f(x)}\lim_{x\to 0}\frac{g(x)}{\psi(f(x))}\lim_{x\to 0}\frac{f(x)}{g(x)}.$$

E.g.
$$\lim_{x\to0}\frac{\sin(2f(x))}{e^{g(x)}-1}=2\lim_{x\to0}\frac{f(x)}{g(x)}.$$
