If $\lim_{x\to 0}{f(x)} = \lim_{x\to 0}{g(x)}=0$, then is $\lim_{x\to 0}\frac{\sin f(x)}{g(x)}= \lim_{x\to 0}\frac {f(x)} {g(x)}$?

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.

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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 $$ \left|\frac{\sin f(x)}{f(x)}-1\right|<\varepsilon. $$