# Complicated limit calculation

Find the limit of $$\left[(x^m + 1)^{1/n} - (x^m - 1)^{1/n}\right]x^{(mn-m)/n}$$ when $x\to\infty$ and $m,n$ are natural numbers.

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What have you tried so far? –  Christian Ivicevic Dec 21 '12 at 16:23

An even quicker way to the answer comes from observing that

$$\left ( x^m + 1 \right )^{\frac{1}{n}} - \left (x^m - 1 \right)^{\frac{1}{n}} = x^{\frac{m}{n}} \left [ \left ( 1 + \frac{1}{x^m} \right )^{\frac{1}{n}} - \left (1 - \frac{1}{x^m} \right)^{\frac{1}{n}} \right ]$$

$$\approx x^{\frac{m}{n}} \left [ \left ( 1 + \frac{1}{n x^m} \right ) - \left (1 - \frac{1}{n x^m} \right) \right ]$$

$$= \frac{2 x^{\frac{m}{n}}}{n x^m}$$

The value $2/n$ follows from the factor on the right of the original expression.

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Hi. Where did the second line come from? –  Smithnson Dec 21 '12 at 16:46
Taylor expansion of $(1+\epsilon)^{\alpha} \approx (1+ \alpha \epsilon)$ for | $\epsilon | \ll 1$. –  Ron Gordon Dec 21 '12 at 16:55

First observe that $$\lim_{x\to +\infty}\left[(x^m + 1)^{1/n} - (x^m - 1)^{1/n}\right]x^{(mn-m)/n} =\lim_{x\to +\infty}x^{\frac mn}\left[(1 + \frac1{x^m})^{1/n} - (1- \frac1{x^m})^{1/n}\right]x^{(mn-m)/n}= \lim_{x\to +\infty}\left[(1+ \frac1{x^m})^{1/n} - (1- \frac1{x^m})^{1/n}\right]x^{m}$$ Now, since $$\alpha^{1/n}-\beta^{1/n}=\frac{\alpha-\beta^n}{\alpha^{{(n-1)}/n}+ \alpha^{{(n-2)}/n}\beta+...+\alpha\beta^{{(n-2)}/n}+\beta^{{(n-1)}/n}}$$ the limit becomes $$\lim_{x\to +\infty}\left[\frac{1+ \frac1{x^m}- 1+ \frac1{x^m}}{(1+ \frac1{x^m})^{{(n-1)}/n}+...+(1+ \frac1{x^m})^{{(n-1)}/n}}\right]x^m= \lim_{x\to +\infty}\left[\frac{2}{(1+ \frac1{x^m})^{{(n-1)}/n}+...+(1+ \frac1{x^m})^{{(n-1)}/n}}\right]$$ The denominator tends to $1+1+...+1=n$ and so $$\lim_{x\to +\infty}\left[\frac2{(1+ \frac1{x^m})^{{(n-1)}/n}+...+(1+ \frac1{x^m})^{{(n-1)}/n}}\right]=\frac2n$$

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