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.
Thanks in advance!
 A: 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.
A: 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 
$$\alpha^{1/n}-\beta^{1/n}=\frac{\alpha-\beta}{\alpha^{{(n-1)}/n}+
\alpha^{{(n-2)}/n}\beta+...+\alpha\beta^{{(n-2)}/n}+\beta^{{(n-1)}/n}}$$
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 $1+1+...+1=n$ therefore 
$$\lim_{x\to +\infty}\left[\frac2{(1+ \frac1{x^m})^{{(n-1)}/n}+...+(1+ \frac1{x^m})^{{(n-1)}/n}}\right]=\frac2n$$
A: Substitute $y=x^m$. Then after rework the limit is that of
$$\lim_{y\to\infty}y\,((1+y^{-1})^{1/n}-(1-y^{-1})^{1/n}).$$
Then


*

*by Taylor, $$\lim_{y\to\infty}y\,\left(1+\frac{y^{-1}}n-1+\frac{y^{-1}}n+o(y^{-1})\right)=\frac2n,$$

*or multiplying by the conjugate multinomial
$$\lim_{y\to\infty}y\frac{1+y^{-1}-1+y^{-1}}{\displaystyle\sum_{k=0}^{n-1}(1+y^{-1})^{k/n}}=\frac2n,$$


*

*or by L'Hospital with the variable $y^{-1}$,


$$\lim_{y\to\infty}\frac{(1+y^{-1})^{1/n}-(1-y^{-1})^{1/n}}{y^{-1}}=\lim_{y\to\infty}\frac{(1+y^{-1})^{1/n-1}+(1-y^{-1})^{1/n-1}}n=\frac2n.$$
