Question

Evaluate the following :

${\displaystyle \underset{x\to \infty }{lim}\left[\sqrt{x}(\sqrt{x+1}-\sqrt{x})\right]}$

Answer 1

${\displaystyle \underset{x\to \infty }{lim}\left[\sqrt{x}(\sqrt{x+1}-\sqrt{x})\right]}$

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

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

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

= ${\displaystyle \underset{x\to \infty }{lim}\frac{1}{\frac{\sqrt{x+1}}{\sqrt{x}}+1} ...\left[\begin{array}{c}\text{Divide numerator and}\\ \text{denominator by} \sqrt{x}\end{array}\right]}$

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

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

= ${\displaystyle \frac{1}{\sqrt{1+0}+1} ...[\underset{x\to \infty }{lim}\frac{1}{x^{\text{k}}}=0,\text{k}>0]}$

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

= ${\displaystyle \frac{1}{2}}$