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!
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!
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.
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 $\lim_{x\to +\infty}\left[\frac2{(1+ \frac1{x^m})^{{(n-1)}/n}+...+(1+ \frac1{x^m})^{{(n-1)}/n}}\right]=\frac2n$