Question

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

• A. 1
• B. -1
• C. 0
• D. does not exist

Answer 1

Answer: (D)

does not exist

Answer 2

$\lim\frac{\sqrt{1-\cos 2x}}{\sqrt{2} x} \Rightarrow \lim \frac{\sqrt{1-\left(1-2 \sin^{2} x\right)}}{\sqrt{2} x}$ $\displaystyle\lim_{x \to0} \frac{\sqrt{2 \sin^{2} x}}{\sqrt{2} x} \Rightarrow \lim_{x \to0} \frac{\left|\sin x\right|}{x}$ The limit of above does not exist as LHS = -1 $\neq$ RHL = 1