I already determined that $\lim_{x\to0}x\sin\frac{1}{x}=0$ via the Squeeze Theorem, but I'm interested solving the limit without it. This is the work I have thus far: $$\lim_{x\to0}x\sin{\frac{1}{x}}=\lim_{x\to0}x\times\lim_{x\to0}\sin\frac{1}{x}=\lim_{x\to0}x\times\sin(\lim_{x\to0}\frac{1}{x})=0\times\sin(\lim_{x\to0}\frac{1}{x})=0$$

My concerns with my work is that $\lim_{x\to0}\frac{1}{x}$ doesn't exist, which would technically make the entire limit undefined. I tried to use the fact $\lim_{x\to0}x=0$ to justify my answer, but it assumes $\times$ is evaluated before $\lim$. Therefore, my questions are as follows:

  1. Is $\times$ evaluated before $\lim$? If so, does that make my work correct?
  2. If not, how would I overcome the issue of $\lim_{x\to0}\frac{1}{x}$ to get a numerical answer to the limit without using the Squeeze Theorem?

EDIT: It appears I didn't clarify my question enough; I'm looking to solve the limit without the Squeeze Theorem

  • 3
    $\begingroup$ Note that $|x\sin\frac{1}{x}|\leq |x|$ then use the definition of limit. $\endgroup$
    – Robert Z
    Jul 15, 2017 at 17:14
  • 1
    $\begingroup$ Saying that $$\lim_{x\to0}x\sin{\frac{1}{x}}=\lim_{x\to0}x\times\lim_{x\to0}\sin\frac{1}{x}$$ is wrong! $\endgroup$
    – Juniven
    Jul 15, 2017 at 17:16
  • $\begingroup$ I suspected so, but where exactly is the mistake in that section? $\endgroup$
    – Brian Lee
    Jul 15, 2017 at 17:18
  • $\begingroup$ You can split if both limits exist. $\endgroup$
    – Juniven
    Jul 15, 2017 at 17:19
  • 2
    $\begingroup$ Do you mean solve the limit? I don't see any derivatives being taken here. I also don't see how the chain rule would get you the limit. $\endgroup$
    – user460426
    Jul 15, 2017 at 17:53

4 Answers 4



Since $|\sin y|\le 1, \forall y\in\mathbb{R},$ we have that

$$\left| x\sin\dfrac1x\right|\le \left| x\right|,$$ and, thus,

$$\left|\lim_{x\to 0}\left( x\sin\dfrac1x\right)\right|\le \left|\lim_{x\to 0} x\right|.$$


Your fundamental problem is the use of product rule of limits. The limit of $\sin(1/x)$ does not exist and unfortunately the limit of other factor is $0$. Note that the product rule can be applied if one of the factors has a non-zero limit.

So that reasoning does not work. Next you can notice that $|x\sin(1/x)|\leq |x|$ and hence you may guess that the desired limit is $0$ and formally prove this using definition of limit. More generally the definition of limit can be used to prove the following simple result:

Theorem: If $f(x) \to 0$ as $x\to a$ and $g(x) $ is bounded as $x\to a$ then $f(x) g(x) \to 0$ as $x\to a$.


this is not correct, use the squeeze Theorem: $$\left|x\sin\left(\frac{1}{x}\right)\right|\le |x|$$


Using squeeze theorem!

We know that The value of x can take a maximum of 1 and a minimum of negative one!

$$-1 \leq \sin x \leq 1$$

So the same applies to

$$-1 \leq \sin \frac{1}{x} \leq 1$$

Now let's multiply the inequality with $x$,

$$-x \leq x\sin \frac{1}{x} \leq x$$

Taking the limit for all three quantities!

$$\lim_{\text x \rightarrow 0}-x \leq \lim_{\text x \rightarrow 0} x\sin \frac{1}{x} \leq \lim_{\text x \rightarrow 0} x $$

Know that

$$\lim_{\text x \rightarrow 0}-x =0 $$

$$\lim_{\text x \rightarrow 0} x=0$$

So limit for $x \sin \frac{1}{x}$ is

$$\lim_{\text x \rightarrow 0} x\sin \frac{1}{x}=0$$

Another way of seeing this $$\lim_{x \rightarrow 0}x\sin\frac{1}{x}=\lim_{x \rightarrow 0}\frac{sin\frac{1}{x}}{\frac{1}{x}}$$

let $k=\frac{1}{x}$

$${x}\rightarrow0,k\rightarrow \infty$$

$$\lim_{k \rightarrow \infty}\frac{\sin k}{k}=0$$


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