Question

Find the value of the following limit:
$\underset{x\to \infty }{lim}\frac{\sqrt{5+x}-1}{x^2+4x}$

Answer 1

$\underset{x\to \infty }{lim}\frac{\sqrt{5+x}-1}{x^2+4x}$

Using Rationalization here

$=\underset{x\to \infty }{lim}\frac{\sqrt{5+x}-1}{x^2+4x}\times \frac{\sqrt{5+x}+1}{\sqrt{5+x}+1}$

$=\underset{x\to \infty }{lim}\frac{(5+x-1)}{x(x+4)(1+\sqrt{x+5})}$

$=\underset{x\to \infty }{lim}\frac{(x+4)}{x(x+4)(1+\sqrt{x+5})}$

$=\underset{x\to \infty }{lim}\frac{1}{x(1+\sqrt{x+5})}$

$=\frac{1}{\infty }$

$=0$.