Calculating $ \lim_{x \to 0} (\frac{x\cdot\sin{x}}{|x|}) $ To solve $ \lim_{x \to 0} (\frac{x\cdot\sin{x}}{|x|}) $ I have separated the function in two limits:
\begin{align}
\lim_{x \to 0} (\frac{x\cdot\sin{x}}{|x|}) =
\lim_{x \to 0} \frac{x}{|x|} \cdot \lim_{x \to 0} \sin{x}
\end{align}
Now solving independently:
\begin{align}
\lim_{x \to 0} \frac{x}{|x|} =  \lim_{x \to 0} \text{sign}(x) \, \, \, \text{does not exist}
\end{align}
And:
\begin{align}
\lim_{x \to 0} \sin{x} = 0
\end{align}
I've checked graphically and I'm aware that the limit is $ 0 $, however, I'm not sure that multiplying "undefined" by $ 0 $ is something allowed.
Can I do this? If not, what could be another way of solving the limit?
Edit:
Just realized I can split the function in a different way, knowing that $ \lim_{x \to 0} \frac{\sin{x}}{|x|} = 1 $
\begin{align}
\lim_{x \to 0} (\frac{x\cdot\sin{x}}{|x|}) =
\lim_{x \to 0} (\frac{|x|\cdot\sin{x}}{x}) =
\lim_{x \to 0} |x| \cdot \lim_{x \to 0} \frac{\sin{x}}{x} = 0 \cdot 1 = 0
\end{align}
How about this second approach?
 A: You can separate a limit into two only if both exist. Here as you point out, one of them does not, so this step is wrong.
You can say that $\operatorname{sgn}(x)$ is bounded, $\sin(x)$ goes to $0$, so their multiplication goes to $0$ as well.
Alternatively, calculate $x\to 0^+$ and $x\to 0^-$ and see they are equal.
A: Approach $0$ from the right:
$$\lim_{x \to 0^+} \frac{x\sin{x}}{|x|}=\lim_{x \to 0^+}\frac{x\sin{x}}{x}=\lim_{x \to 0^+}\sin x=0.$$
Then approach $0$ from the left:
$$\lim_{x \to 0^-} \frac{x\sin{x}}{|x|}=\lim_{x \to 0^-}\frac{x\sin{x}}{-x}=\lim_{x \to 0^-}-\sin x=0.$$
Since $\lim_{x \to 0^+}\frac{x\sin{x}}{|x|}=0=\lim_{x \to 0^-}\frac{x\sin{x}}{|x|}$, the limit is zero.
A: The important fact is that $\frac{x}{|x|}$ is bounded that is
$$-1\le\frac{x}{|x|}\le 1$$
therefore
$$0\le \left|\frac{x\cdot\sin{x}}{|x|}\right|\le|\sin x| \to 0$$
and we can conclude by squeeze theorem.
Alternatively we can use that
$$\frac{|\sin{x}|}{|x|} \to 1$$
and therefore
$$0\le \left|\frac{x\cdot\sin{x}}{|x|}\right|=|x|\frac{|\sin{x}|}{|x|} \to 0\cdot 1=0$$
A: $$\lim_{x\to 0} \frac{ x\cdot \sin x}{|x|} = \lim_{x\to 0} \frac{ |x| \cdot \sin x}{x} = \lim_{x\to0} |x| \cdot \lim_{x\to0}\frac{\sin x}{x} = 0\cdot 1 = 0.$$
A: Let $f(x)=\frac{x\sin(x)}{\left| x\right|}$. If $x$ approaches $0$ from the right, $x>0$, so $\left| x\right| = x$. Hence, $$\lim_{x\rightarrow 0^+} f(x)$$$$=\lim_{x\rightarrow 0^+}\frac{x\sin(x)}{x}$$$$=\lim_{x\rightarrow 0^+}\sin(x)$$$$=0.$$ Because $f(x)=f(-x)$, $$\lim_{x\rightarrow 0^-}f(x)=\lim_{x\rightarrow 0^+}f(x)=0.$$ Therefore, $\lim_{x\rightarrow 0}f(x)=0.$
