Question

The value of $\underset{x\to \infty }{lim}(\sqrt{x+\sqrt{x+\sqrt{x}}}-\sqrt{x})$ is

• A. $\frac{1}{2}$
• B. $1$
• C. $0$
• D. none of these

Answer 1

The correct option is A $\frac{1}{2}$

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

Divide numerator and denominator by $\sqrt{x}$

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