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

Question

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. — Lord_Farin, Jun 27 '13 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 '13 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! — Nick Peterson, Jun 27 '13 at 19:21

score 4 · Answer 1 · answered Jun 27 '13 at 19:49

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.

score 1 · Answer 2 · edited Oct 23 '22 at 10:55

1

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

edited Oct 23 '22 at 10:55

Angelo

12,328

answered Jun 27 '13 at 19:41

teh

593

For future reference, I suggest that you read this FAQ entry regarding answering homework questions. – Lord_Farin Jun 27 '13 at 19:46
To elaborate, for a small enough δ, we have $|\sin(u)|≤|u| – $ – Lemon Jun 27 '13 at 19:51

score 0 · Answer 3 · edited Oct 23 '22 at 11:06

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 $\,0\,$.

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

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