<s>[INST] Limit of $f(x)\cdot g(x)$ given that $\lim_{x\to 0}{\frac{f(x)}{\sin(2x)}}=2$ and $\lim_{x\to 0}{(\sqrt{x+4}-2)\cdot{g(x)}}=5$ Question:

If $$\lim_{x\to 0}{\left[\frac{f(x)}{\sin(2x)}\right]}=2$$ and $$\lim_{x\to 0}{\left[(\sqrt{x+4}-2)\cdot{g(x)}\right]}=5$$ then find $$\lim_{x\to 0}{[f(x)\cdot{g(x)}]}$$

But, from what I found, $\lim_{x\to 0}{[g(x)]}$ does not exist. So how can I find $\lim_{x\to 0}{[f(x)\cdot{g(x)}]}$?
 [/INST] Only one theorem applied.

Theorem. If $\lim\limits_{x\rightarrow c}f(x)=L$ and $\lim\limits_{x\rightarrow c}g(x)=M$, then $\lim\limits_{x\rightarrow c}[f(x)g(x)]=LM$ 

$$
\because \qquad \lim_{x\rightarrow0}\frac{f(x)}{\sin(2x)}=2\quad \text{and}\quad \lim_{x\rightarrow0}\left[(\sqrt{x+4}-2）\cdot g(x)\right]=5
$$
$$
\therefore\qquad \lim_{x\rightarrow0}\left\{\frac{f(x)}{\sin(2x)}\cdot \left[(\sqrt{x+4}-2）\cdot g(x)\right]\right\}=2\times5=10
$$
Rearranging:
\begin{align}
\frac{f(x)}{\sin(2x)}\cdot \left[(\sqrt{x+4}-2）\cdot g(x)\right]
&=\left[f(x)g(x)\right]\cdot\frac{\sqrt{x+4}-2}{\sin(2x)}\\
&=\left[f(x)g(x)\right]\cdot\frac{2x}{\sin(2x)}\cdot \frac{1}{2(\sqrt{x+4}+2)}
\end{align}
$$
\Rightarrow\qquad
\lim_{x\rightarrow0}\left\{\left[f(x)g(x)\right]\cdot\frac{2x}{\sin(2x)}\cdot \frac{1}{2(\sqrt{x+4}+2)}\right\}
=10
$$
$$
\because\qquad
\lim_{x\rightarrow0}\frac{\sin(2x)}{2x}=1\quad
\text{and}
\quad \lim_{x\rightarrow0}\frac{2(\sqrt{x+4}+2)}{1}=8
$$
$$
\therefore\qquad
\lim_{x\rightarrow0}
\left\{
\left[\left[f(x)g(x)\right]\cdot\frac{\sin(2x)}{2x}\cdot \frac{1}{2(\sqrt{x+4}+2)}\right]
\cdot
\left[
\frac{2x}{\sin(2x)}
\right]
\cdot
\left[
\frac{2(\sqrt{x+4}+2)}{1}
\right]
\right\}\\=10\cdot 1\cdot8=80
$$
$$
\Rightarrow\qquad
\lim_{x\rightarrow0}\left[f(x)g(x)\right]=80
$$
Tips. Treat the block of functions as one function.
</s>