What will be $\lim_{x \to 0}\sin(1/x)$? What will be the following limits?

$\lim\limits_{x \to 0}\sin(\frac{1}{x})$

for $x \to 0 ,$ $\frac{1}{x} \to \infty$

Now what will be the $\sin$ value on $\infty$ ?
Let $a_n=\dfrac{1}{n\pi}$ and $b_n=\dfrac{1}{\dfrac{\pi}{2}-n\pi}$. We have
$$\underset{n\to \infty }{\mathop{\lim }}\,f({{a}_{n}})=0\,\,\,\,\,,\,\,\,\,\underset{n\to \infty }{\mathop{\lim }}\,f({{b}_{n}})=\underset{n\to \infty }{\mathop{\lim }}{{(-1)}^{n}}$$
There is no definite value of $\sin$ at infinity. As you might know that infinity is not a number . It is a notion that shows that whatever the value we are considering it is big. So the limit does not exist.

One thing to note here is that the reason of the non-existing limit is not that the number in consideration is going to infinity. The reason is that we don't really know that value of $\sin$ at that angle. The value of sin can be anywhere between $-1$ and $1$.