# Prove that $\lim_{x \to \infty} \frac{a}{x^n} = 0$

Prove using the $\delta-\epsilon$ definition of the limit that for positive integer $n$ and constant $a$, $\displaystyle \lim_{x \to \infty} \frac{a}{x^n} = 0$

Attempt

We have to show that $$\forall \epsilon, \exists M > 0 \quad x > M \quad \implies \quad \left|\frac{1}{x^n}\right| < \epsilon.$$

How does I show this?

• What are your thoughts on this? What have you tried? Mar 8 '16 at 23:39
• I am not used to finding $M$s. We can try $|x^n| > \dfrac{1}{\epsilon}$ and so $|x| > \dfrac{1}{\epsilon^{1/n}}$ so let $M = \dfrac{1}{\epsilon^{1/n}}$? Mar 8 '16 at 23:43

Hint Assuming $x$ positive:
$$\left|\dfrac{a}{x^n}\right|<\epsilon \iff \left|\dfrac{a}{\epsilon}\right|<x^n \iff \sqrt[n]{\dfrac{|a|}{\epsilon}}<x.$$ What should be $M?$