Question

\( \lim_{x \to 0} \frac{\sqrt{1 - \cos(2x)}}{\sqrt{2} \cdot x} \) is

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

Hint:

$limx→0\sqrt1-cos 2x/\sqrt2· x}$ is

Answer 1

The Correct Option is D

Step 1: Concept
Use the trigonometric identity $1 - \cos 2x = 2 \sin^2 x$.
Step 2: Analysis
The limit becomes $lim_{x\rightarrow0}\frac{\sqrt{2 \sin^2 x}}{\sqrt{2} \cdot x} = lim_{x\rightarrow0}\frac{\sqrt{2} | \sin x |}{\sqrt{2} \cdot x} = lim_{x\rightarrow0}\frac{|\sin x|}{x}$.
Step 3: Evaluation
Right Hand Limit (RHL): $lim_{x\rightarrow0^+} \frac{\sin x}{x} = 1$. Left Hand Limit (LHL): $lim_{x\rightarrow0^-} \frac{-\sin x}{x} = -1$.
Step 4: Conclusion
Since RHL $\ne$ LHL, the limit does not exist.
Final Answer: (d)