Question

Find the following limit.
$\underset{n\to \infty }{lim}\frac{\sqrt{2n^{2}1}-\sqrt{n^2+1}}{1+1}.$

Answer 1

$\underset{n\to \infty }{lim}\frac{\sqrt{2n^2-1}-\sqrt{n^2+1}}{n+1}$

here we take $n^2$ common from Numerator and $n$ from denominator.

So,

$=\underset{n\to \infty }{lim}\frac{\sqrt{n^2(2-\frac{1}{n^2})}-\sqrt{n-(1+\frac{1}{n^2})}}{n(1+\frac{1}{n})}$

$=\underset{n\to \infty }{lim}\frac{n(\sqrt{2-\frac{1}{n^2}}-\sqrt{1+\frac{1}{n^2}})}{n(1+\frac{1}{n})}$

$=\frac{\sqrt{1-\frac{1}{\infty }}-\sqrt{1+\frac{1}{\infty }}}{(1+\frac{1}{\infty })}=\frac{\sqrt{2-0}-\sqrt{1+0}}{1+0}$

$=\sqrt{2}-1$.