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}


\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?


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?

  • $\begingroup$ Note that $\lim_{x \to a}(f \cdot g)=(\lim_{x \to a} f) \cdot (\lim_{x \to a}g)$ requires the existence of both limits on the right side which is not the case here. Instead you can say as $x \to 0^{+}$, then the function is $\sin x$ and as $x \to 0^{-}$, the function is $-\sin x$. $\endgroup$ – Anurag A Aug 7 '20 at 19:00
  • $\begingroup$ Why on the right side specifically? Shouldn't be, in general, the existence of both limits? $\endgroup$ – Jon Aug 7 '20 at 19:01
  • $\begingroup$ $$\lim_{x \to 0} f(x) = 0 \iff \lim_{x \to 0} \lvert f(x)\rvert = 0$$ $\endgroup$ – Daniel Fischer Aug 7 '20 at 19:02
  • $\begingroup$ We have that $\lim_{x \to 0} \frac{\sin{x}}{|x|} = \pm 1$. $\endgroup$ – user Aug 7 '20 at 19:11
  • $\begingroup$ Thanks, @user. Made a typo with the $ |x| $ in the denominator, should be just $ x $ $\endgroup$ – Jon Aug 7 '20 at 19:13

$$\lim_{x\to 0} \frac{ x\cdot \sin x}{|x|} = \lim_{x\to 0} \frac{ |x| \cdot \sin x}{x} = \lim_{x\to0} |x| \cdot \lim_{x\to0}\frac{\sin x}{x} = 0\cdot 1 = 0.$$

  • $\begingroup$ Very clever method! $\endgroup$ – user Aug 7 '20 at 19:31

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.


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.


The important fact is that $\frac{x}{|x|}$ is bounded that is

$$-1\le\frac{x}{|x|}\le 1$$


$$0\le \left|\frac{x\cdot\sin{x}}{|x|}\right|\le|\sin x| \to 0$$

and we can conclude by squeeze theorem.

Alternatively we can use that

$$\frac{|\sin{x}|}{|x|} \to 1$$

and therefore

$$0\le \left|\frac{x\cdot\sin{x}}{|x|}\right|=|x|\frac{|\sin{x}|}{|x|} \to 0\cdot 1=0$$

  • $\begingroup$ Thanks for your answer. I knew about the approach using the squeeze theorem but I'm afraid I don't understand how to use in this context. I'm aware that the signum function is bounded between -1 and 1, and sin(x) is also bounded in the same way. With that information, how can I discover that my original function is between them? $\endgroup$ – Jon Aug 7 '20 at 19:23
  • $\begingroup$ @Jon We know that $x/|x|$ is bounded whereas $\sin x \to 0$, for that reason the product goes to zero. To formalize that we invoque squeeze theorem but it is a very intuitive fact. $\endgroup$ – user Aug 7 '20 at 19:26

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, $\lim_{x\rightarrow 0}f(x)=0.$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.