Question

Evaluate $\displaystyle \lim_{x \to 0}$ $\frac{\sqrt{1-\sqrt{cos\,2x}}}{x}$ is

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

Answer 1

Answer: (D)

$\displaystyle \lim_{x \to 0}$ $\frac{\sqrt{1-\sqrt{cos\,2x}}}{x}$
$=\displaystyle \lim_{x \to 0}$ $\frac{\sqrt{1-cos\,2x}}{x} \times \frac{1}{\sqrt{1+\sqrt{cos\,2x}}}$
$=\displaystyle \lim_{x \to 0}$ $\frac{\sqrt{2}\left|sin\,x\right|}{x}\cdot\left(\frac{1}{\sqrt{2}}\right)$
$=\displaystyle \lim_{x \to 0}$ $\frac{\left|sin\,x\right|}{x}$
$\therefore L.H.L.$ $=\displaystyle \lim_{x \to 0^-}\frac{\left|sin\,x\right|}{x}$
$=\displaystyle \lim_{x \to 0}$ $\frac{-sin\,x}{x}=-1$
$\therefore R.H.L.=$ $\displaystyle \lim_{x \to 0^+} \frac{\left|sin\,x\right|}{x}$
$=\displaystyle \lim_{x \to 0}$$\frac{sin\,x}{x}=1$
$\because L.H.L. \ne R.H.L.$
$\therefore $ Limit does not exist.