Question

The value of \(\lim_{x \to 0} \frac{\sqrt{1 - \cos 2x}}{|x|}\) is equal to

• A. $-2$
• B. $-\sqrt{2}$
• C. $\sqrt{2}$
• D. $1$
• E. $2$

Hint:

Always convert $\cos 2x$ into $\sin^2 x$ when square root is involved.

Answer 1

The Correct Option is C

Concept: Use identity: \[ 1 - \cos 2x = 2\sin^2 x \]

Step 1: Apply identity
\[ \sqrt{1 - \cos 2x} = \sqrt{2\sin^2 x} = \sqrt{2}|\sin x| \]

Step 2: Substitute in limit
\[ \lim_{x \to 0} \frac{\sqrt{2}|\sin x|}{|x|} = \sqrt{2} \lim_{x \to 0} \frac{|\sin x|}{|x|} \]

Step 3: Use standard limit
\[ \lim_{x \to 0} \frac{\sin x}{x} = 1 \Rightarrow \frac{|\sin x|}{|x|} \to 1 \] \[ \Rightarrow \sqrt{2} \] Final Conclusion:
Option (C)