When $x$ goes to $0$ , what happens to $\sin\left(\frac{1}{x}\right) $ and $\cos\left(\frac{1}{x}\right)$? When $x$ approaches $0$, do $\sin\frac{1}{x}$ and $\cos\frac{1}{x}$ converge or diverge?
How do you show this?
 A: You can construct sequences $a_i,b_i$ such that $a_i,b_i\rightarrow0$ and $\sin(1/a_i)=0,\sin(1/b_i)=1$ for all $a_i,b_i$.
The some goes for $\cos(\frac{1}{x})$.
Use the fact that $\lim_{x\rightarrow a} f(x)=c$ if and only if for all sequences $w_i$ with $w_i \rightarrow a$ you have $f(w_i) \rightarrow c$.
A: When $x\to 0$ then $$\lim ~\sin(1/x)$$ does not exist. Indeed, when $x$ approaches to the origin, this function is being faced too much oscillations between $y=-1$ and $y=+1$. I think, one way to see that this limit is not exist is to assume that the limit exist and then we face to a contradiction!. Assume that the limit is $a$, then by choosing $\epsilon=1/2$ and use the definition of the limit, we can find $\delta>0$ such that $0<|x|<\delta$ leads us to $$|\sin(1/x)-a|<1/2$$. Now choose two points in reals for an sufficiently large integer number $n$ such that: $$x_1=1/(2n+1/2)\pi,~~~x_2=1/(2n-1/2)\pi$$ and try to find the contradiction. :)
A: We say the sin(1/x) and cos(1/x) is divergent because the limit does not exist. We know this because the sin(1/x) and cos(1/x) fluctuates infinitely between zero and 1 at 0. 
the limit exists (and is a number), in this case we say the function is convergent
the limit does not exist or it is infinite, then we say that function is divergent
A: Try by contradiction, suppose $\lim_{x\rightarrow 0}\sin{(\frac{1}{x})}=l, \:l\in \mathbb{R}$
therefore, $\forall \epsilon >0, \exists \delta>0$ such that if $| x - 0|<\delta$ the $| f(x)-l | < \epsilon$.
By archimedian property there exists $a_{1},a_{2}$ such that $0<a_{1},a_{2}< \delta$ and $a_{1}=\frac{1}{2k\pi\ + \frac{\pi}{2}}$ and $a_{2}=\frac{1}{2k\pi}$, $k\in \mathbb{N}$
$\Rightarrow | 1-l|<\epsilon$ and $ |l| <\epsilon$
Which concludes the proof.
A: The idea is that for both $\cos(\frac{1}{x})$ and $\sin(\frac{1}{x})$ when $x\to0$, they fluctuate indefinitely between 1 and 0 (you can graph it and zoom in closer to zero, the graph will look the same no matter how much you close in). 
So another way to prove this is to use the $\delta, \epsilon$ definition of a limit and show that no matter what $\delta$ you choose, there will always be a value (for any limit $L$ you may choose) $x$ such that $|\cos(\frac{1}{x}) - L|$ is not less than $\epsilon$.
A: Substitute $y=1/x$, recognize that $\lim_{z\to 0^+}1/z=\infty$ and ask yourself if 
$$
\lim_{x\to 0^+}\sin(1/x)= \lim_{y\to \infty}\sin(y)
$$
diverges or converges?
