Question

Find the value of the following limit:
$\underset{x\to \infty }{lim}\frac{\sqrt{x^2+1}-x}{x+1}$

Answer 1

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

taking $x$ common

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

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

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

$=0$.