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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?

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    $\begingroup$ What are your thoughts on this? What have you tried? $\endgroup$
    – Michael
    Mar 8 '16 at 23:39
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    $\begingroup$ 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 $M = \dfrac{1}{\epsilon^{1/n}}$? $\endgroup$
    – Puzzled417
    Mar 8 '16 at 23:43
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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 $M?$

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