# Proving that $\lim\limits_{x\to 0} \sin(x)\cos(\tfrac1x) = 0$

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.

• For some basic information about writing maths at this site see e.g. here, here, here and here. Jun 27, 2013 at 19:19
• My own approach would be to observe that $\lim_{x\to 0}\sin x$ is zero and that $\cos(1/x)$ is bounded on $(0,\infty)$.
– MJD
Jun 27, 2013 at 19:21
• 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! Jun 27, 2013 at 19:21

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 $$\lim_{x\rightarrow 0}\sin (x)\cos\left(\frac{1}{x}\right)=0.$$

• For future reference, I suggest that you read this FAQ entry regarding answering homework questions. Jun 27, 2013 at 19:46
• To elaborate, for a small enough δ, we have $|\sin(u)|≤|u| –$ 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 $$\,0\,$$.