Question

Evaluate $\underset{x\to 0}{\lim }$ $\frac{\sqrt{1-\sqrt{cos2x}}}{x}$ is

• A. $0$
• B. $1$
• C. $-1$
• D. Does not exist

Answer 1

Answer: (D)

$\underset{x\to 0}{\lim }$ $\frac{\sqrt{1-\sqrt{cos2x}}}{x}$
$=\underset{x\to 0}{\lim }$ $\frac{\sqrt{1-cos2x}}{x}\times \frac{1}{\sqrt{1+\sqrt{cos2x}}}$
$=\underset{x\to 0}{\lim }$ $\frac{\sqrt{2}|sinx|}{x}\cdot \left(\frac{1}{\sqrt{2}}\right)$
$=\underset{x\to 0}{\lim }$ $\frac{|sinx|}{x}$
$\therefore L.H.L.$ $=\underset{x\to 0^{-}}{\lim }\frac{\left|sinx\right|}{x}$
$=\underset{x\to 0}{\lim }$ $\frac{-sinx}{x}=-1$
$\therefore R.H.L.=$ $\underset{x\to 0^{+}}{\lim }\frac{\left|sinx\right|}{x}$
$=\underset{x\to 0}{\lim }$$\frac{sinx}{x}=1$
$\because L.H.L.\ne R.H.L.$
$\therefore$ Limit does not exist.