Question

Evaluate the limit:
$\underset{x\to \infty }{lim}\{\sqrt{x+1}-\sqrt{x}\}\sqrt{x+2}$

Answer 1

We have,
$\underset{x\to \infty }{lim}\{\sqrt{x+1}-\sqrt{x}\}\sqrt{x+2}$

On rationalising numerator, we get

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

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

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

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

Dividing numerator and denominator by $\sqrt{x}$, we get

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

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

When $x\to \infty$, then $\frac{1}{x}\to 0$

$=\frac{\sqrt{1+2\left(0\right)}}{\{\sqrt{1+\left(0\right)}+\sqrt{1}\}}$

$=\frac{\sqrt{1}}{\{\sqrt{1}+\sqrt{1}\}}$

$=\frac{1}{2}$

Therefore,
$\underset{x\to \infty }{lim}\{\sqrt{x+1}-\sqrt{x}\}\sqrt{x+2}=\frac{1}{2}$