To show $\lim_{x \to 0} \sin(1/x)$ does not exist: the sequences $x_n = \frac{1}{2n\pi} \to 0$ and $y_n = \frac{2}{(4n+1)\pi} \to 0$ satisfy $\sin(1/x_n) = 0$ and $\sin(1/y_n) = 1$. Since two sequences give different values, the limit fails to exist.

To prove $\lim_{x \to c} f(x) = L$, it suffices to exhibit ONE sequence $x_n \to c$ (with $x_n \neq c$) such that $f(x_n) \to L$.

If $f(x_n) \to L$ for every sequence $x_n \to c$ with $x_n \neq c$, then $\lim_{x \to c} f(x) = L$.

The value $f(c)$ plays no role in $\lim_{x \to c} f(x)$, so $\lim_{x \to c} f(x) = L$ is compatible with $f(c) \neq L$.