Question

The value of \(\displaystyle \lim_{x \to0} \frac{\sqrt{\frac{1}{2} \left(1-\cos2x\right)}}{x}\)

• A. 1
• B. -1
• C. 0
• D. none of these

Answer 1

The Correct Option is D

Answer (d) none of these

Answer 2

To evaluate the limit 
\[\lim_{x \to 0} \frac{\sqrt{1 - \cos 2x}}{\sqrt{2}x},\]
we need to proceed carefully with the trigonometric identities and the properties of limits.

First, we use the double-angle identity for cosine:
\[\cos 2x = 1 - 2 \sin^2 x.\]

Thus,
\[1 - \cos 2x = 1 - (1 - 2 \sin^2 x) = 2 \sin^2 x.\]

Substituting this into the expression, we get:
\[\frac{\sqrt{1 - \cos 2x}}{\sqrt{2}x} = \frac{\sqrt{2 \sin^2 x}}{\sqrt{2}x}.\]

Since \(\sqrt{2 \sin^2 x} = \sqrt{2} |\sin x|\), the limit becomes:
\[\frac{\sqrt{2} |\sin x|}{\sqrt{2} x} = \frac{|\sin x|}{x}.\]

We now consider the limit:
\[\lim_{x \to 0} \frac{|\sin x|}{x}.\]

For \(x\) close to 0, \(\sin x \approx x\), so:
\[\frac{|\sin x|}{x} \approx \frac{|x|}{x}.\]

We analyze the behavior of \(\frac{|x|}{x}\) as \(x\) approaches 0 from the left and the right:
- As \(x \to 0^+\), \(\frac{|x|}{x} = \frac{x}{x} = 1\).
- As \(x \to 0^-\), \(\frac{|x|}{x} = \frac{-x}{x} = -1\).

Since the left-hand limit and the right-hand limit are not equal, the overall limit does not exist. Therefore, the limit
\[\lim_{x \to 0} \frac{\sqrt{1 - \cos 2x}}{\sqrt{2} x}\]
does not exist.
So The Correct Answer is Option (D): None of these