Question

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

Answer 1

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

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

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

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

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

Dividing numerator and denominator by $\sqrt{x}$, we get,
$=\underset{x\to \infty }{lim}\frac{\left(\frac{\sqrt{x}}{\sqrt{x}}\right)}{\frac{\{\sqrt{x+1}+\sqrt{x}\}}{\sqrt{x}}}$

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

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

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

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