Question

Evaluate:

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

Answer 1

Now,

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

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

$=\underset{x\to \infty }{lim}\frac{\sqrt{x+1}}{\sqrt{x}\times \sqrt{x+1}}$

$=\underset{x\to \infty }{lim}\frac{1}{\sqrt{x}}$

$=0$.