How would you prove that (1/x^n)-->0 as x-->infinity and (1/x^n)-->0 as x-->-infinity?
Show that for all $\displaystyle \varepsilon>0$ there exists a $\displaystyle X_{\varepsilon}$ such that:for all $\displaystyle x>X_{\varepsilon}$ we have $\displaystyle |x^{-n}-0|<\varepsilon$
In this case choose $\displaystyle X_{\varepsilon}=\max\left(1,\frac{1}{\varepsilon}\ right)$, then for $\displaystyle x>X_{\varepsilon}$:
$\displaystyle |x^{-n}-0|=|x^{-n}|<\frac{1}{x}<\frac{1}{X_{\varepsilon}}\le \varepsilon$
CB