$\lim\limits_{x\to0}\sin1/x$

For every $a\in[-1,1]$, there is some sequence $(x_{n})$ such that $x_{n}\rightarrow 0$ and $\sin(1/x_{n})\rightarrow a$.

$$\Box \ \nexists \lim_{x\to 0}\bigg(\sin\frac1x\bigg).$$ Proof: Let $u = \dfrac{1}{x}$ then $$\lim_{x\to 0}\frac{1}{x} = \infty$$ since $$\lim_{x\to\infty}\frac{1}{x} = 0.$$ Therefore, we get $$\lim_{x\to 0}\bigg(\sin\frac 1x\bigg) = \lim_{u\to\infty}(\sin u)$$ but this cannot exist because sine is a periodic fluctuating function. $\qquad \qquad\qquad\qquad\quad\,\,\,\,$

The limit $$\lim_{x\rightarrow0}{\left( \sin{\frac{1}{x}}\right)}$$

does not exist.

Note that as x approaches $0$, $\sin( 1/x )$ covers the closed interval $[-1, 1]$ infinitely many times.

For example at $ x= \frac {2}{(2n+1)\pi } $ we have $\sin(1/x)=\pm 1.$

Thus there is no limit for $\sin(1/x)$ as $x$ approaches $0$.