Question

Evaluate the limit:
$\underset{x\to \infty }{lim}\sqrt{x+1}-\sqrt{x}$

Answer 1

We have,
$\underset{x\to \infty }{lim}\sqrt{x+1}-\sqrt{x}$

It is of $(\infty -\infty )$ form.

On rationalising the numerator, we get
$=\underset{x\to \infty }{lim}\frac{\left(\right(\sqrt{x+1}-\sqrt{x})\times (\sqrt{x+1}+\sqrt{x}\left)\right)}{(\sqrt{x+1}+\sqrt{x})}$

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

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

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

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

$=\frac{1}{\infty }=0$

Therefore,
$\underset{x\to \infty }{lim}\sqrt{x+1}-\sqrt{x}=0$