sin(1/x) has no limit at 0

The method. To show that a limit does not exist, it is enough to find two sequences tending to $0$ along which $f$ has different limits.

First sequence. Let $a_n=\dfrac1{n\pi}$. Then $a_n\to0$ and $f(a_n)=\sin(n\pi)=0$ for all $n$. So if the limit exists, it equals $0$.

Second sequence. Let $b_n=\dfrac1{\frac\pi2+2n\pi}$. Then $b_n\to0$ and $f(b_n)=\sin\left(\frac\pi2+2n\pi\right)=1$ for all $n$. So if the limit exists, it equals $1$.

Conclusion. The limit would have to equal both $0$ and $1$, which is impossible. Hence $\sin\frac1x$ has no limit at $0$.

Remark. On every interval $(0,\varepsilon)$ the function takes every value in $[-1,1]$ infinitely often, which is why it cannot settle down.