Prove that: $$\lim_{x\to 0} \sin(x)\cos(\tfrac1x) = 0$$

I am completely confused in knowing where to begin problems like this.

I have the beginning:
For all $\epsilon > 0$ choose $\delta = ?$...
When $|x-0|<\delta$ then $|f(x)-0| = |\sin(x)\cos(1/x)-0| \quad \ldots \quad < \epsilon$.

I need step by step help because I am completely lost in understanding how to complete this type of problem.

  • $\begingroup$ For some basic information about writing maths at this site see e.g. here, here, here and here. $\endgroup$
    – Lord_Farin
    Jun 27, 2013 at 19:19
  • 4
    $\begingroup$ My own approach would be to observe that $\lim_{x\to 0}\sin x$ is zero and that $\cos(1/x)$ is bounded on $(0,\infty)$. $\endgroup$
    – MJD
    Jun 27, 2013 at 19:21
  • 4
    $\begingroup$ This proof is almost identical to the one I just helped you through here. Please, consider going to your instructor for help understanding the definition! $\endgroup$ Jun 27, 2013 at 19:21

3 Answers 3


This is quite straightforward:

  • $|\sin(x) \cdot \cos(1/x)| = |\sin(x)|\cdot |\cos(1/x)|\le |\sin(x)|$ for all $x \ne 0$.
  • $|\sin(x)|$ tends to zero as $x$ tends to zero.
  • By the Sandwich Theorem, $|\sin(x) \cdot \cos(1/x)|$ tends to zero as $x$ tends to zero.

This is indeed an easy question. But for beginners, it may not be so, especially we require to show it in $\epsilon-\delta$ language. Anyway, here it is:

First note that $|\sin (x)|\leq |x|$, this plainly gives $\lim_\limits{x\rightarrow 0}\sin (x)=0$.

Now, for any given $\epsilon>0$, choose $\delta=\epsilon$. Then note $|\cos(1/x)|\leq 1$, one has $$ |\sin(x)\cos(1/x)|\leq |x|<\delta=\epsilon, \quad \text{whenever} \quad 0<|x|<\delta, $$ showing $$\lim_{x\rightarrow 0}\sin (x)\cos\left(\frac{1}{x}\right)=0.$$

  • $\begingroup$ For future reference, I suggest that you read this FAQ entry regarding answering homework questions. $\endgroup$
    – Lord_Farin
    Jun 27, 2013 at 19:46
  • $\begingroup$ To elaborate, for a small enough δ, we have $|\sin(u)|≤|u| – $ $\endgroup$
    – Lemon
    Jun 27, 2013 at 19:51

It’s quite easy, first of all,

$\sin(x)\cos\left(\dfrac1x\right)=\dfrac12\left[\sin\left(x +\dfrac1x\right) +\sin\left(x -\dfrac1x\right)\right]$

expand the above using taylor's series $\sin(x)=x-\dfrac{x^3}{3!} +\dfrac{x^5}{5!}+\ldots$

on simplifying for some degrees we see that all $\;\dfrac1x ,\dfrac1{x^2},\ldots,\;$ etc, cancels out and the answer appears.

Note: - the cancellation takes place due to the odd degrees in the taylor's series.

Finally, we get the answer as $\,0\,$.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .