I am reading Calculus, Purcell and there is a solved example
Show that if $k$ a positive integer, then
$$\lim_{x\rightarrow\infty}\frac{1}{x^k}=0\ \text{ and} \ \lim_{x\rightarrow-\infty}\frac{1}{x^k}=0$$ The first part was proven in the book by taking $M=\sqrt[k]{\frac{1}{\epsilon}}$, then $x>M$ implies that $$|\frac{1}{x^k}-0|=\frac{1}{x^k}<\frac{1}{M^k}=\epsilon.$$
For the second statement, the book states that the proof is similar. However, we assume that $x<M$. Therefore, $$|\frac{1}{x^k}-0|=\frac{1}{x^k}>\frac{1}{M^k}=\epsilon.$$
which is not what we want. Am I missing something?