Prove $\lim _{x \to 0} \sin(\frac{1}{x}) \ne 0$ Prove $$\lim _{x \to 0} \sin\left(\frac{1}{x}\right) \ne 0.$$
I am unsure of how to prove this problem. I will ask questions if I have doubt on the proof. Thank you!
 A: HINT
Consider the sequences $$x_n = \dfrac1{2n \pi + \pi/2}$$ and $$y_n = \dfrac1{2n \pi + \pi/4}$$ and look at what happens to your function along these two sequences. Note that both sequences tend to $0$ as $n \to \infty$.
A: Here's an intuitive answer, just to shake things up.  The limit of a function as it approaches some point is the value you expect the function to take at that point$^1$ based on what it does nearby the point.  If you can't form any expectations about the function's value, the limit doesn't exist.  
Now take a look at the function in question:

Can anyone possibly form expectations about where that function should be at $0$?
Of course, once you have the intuition, you have to actually write down the proof.  For that, you need to remember the definition of the limit and formalize why you can't figure out where $\sin(x^{-1})$ "should" be at $0$.


*

*(Be careful, though!  The function might not even be defined at the point!  We're just forming an expectation of what we think the function should be, if it were to exist.  We don't know anything about what the function actually does.)

A: in order to prove the limit $\not=0$, only need to find a sequent $\{x_n\}\rightarrow0$,but $\sin(x_n) \not\rightarrow 0$. and the below is the construction.
$$x_n=\frac{1}{\frac{\pi}{2}+2n\pi}$$
$$\sin\left(\frac{1}{x_n}\right)=1$$
A: To show that, some limit of some function does not exist it is better often use Heine definition of limit, which states that a function $f(x)$ has a limit $L$ at $x = a$, if for every sequence $\{x_n\}$ , which has a limit at $a$, the sequence $f(\{x_n\})$ has a limit $L$.
Thus, by the above definition, you need to find(it is already done for you, in above answer) sequences $\{x_n\}$ and $\{y_n\}$ such that $\lim x_n=\lim y_n=0$, but $f(x_n)\neq f(y_n)$.
A: As a complement to Marvis' answer and to Neal's drawing, consider 
$$
x_n(\delta) = \frac 1{2n\pi + \delta}.
$$
Then 
$$
f(x_n) = \sin(2 n \pi + \delta) = \sin (\delta) \underset{n \to \infty}{\longrightarrow} \sin(\delta).
$$
Choosing $\delta$ such that $\sin(\delta) = y \in [-1,1]$, we see that using an appropriate sequence $x_n(\delta)$ converging to $0$, we can approach any value between $[-1,1]$. Since $|f(x_n)| \le -1$, this explains the ugly behavior of the function near $0$ as seen in Neal's graph.
Hope that helps,
