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.
 A: 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.

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