I decided to try proof by contradiction since the negation of the definition of a limit is $\exists \epsilon > 0 \space \forall \delta > 0 \space , 0 < |x - c| < \delta \land |f(x) - L| \geq \epsilon$
Suppose $\lim_{x\to 0} {cos{1\over x}}$ exists, and let its limit be equal to L. Let $\epsilon = {L\over2}$. Let |x| = $1\over{n\pi\over2}$ for some odd n $\in N$, and let n be sufficently small such that ${1\over{n\pi\over2}} < \delta$. Then, $f(x) = cos({1\over{1\over{n\pi\over2}}}) = 0$. Therfore, |f(x) - L| = |0 - L| = |L| > $L\over2$ > $\epsilon$. Hence, the limit does not exist.
If $f(x)=\cos(1/x)$ and $x_n = \frac{1}{n \pi}$ for $n \in \mathbb N$ then $x_n \to 0$ but
$$f(x_n)=(-1)^n.$$
Conclusion ?