`Lt_(x to 0)sqrt(1- cos 2x)/x`=

Options

A. 0
B. 1
C. `sqrt(2)`
D. doest not exist

Answer

To solve the limit \( \lim_{x \to 0} \frac{\sqrt{1 - \cos 2x}}{x} \), we can follow these steps:

### Step 1: Use the identity for \(1 - \cos 2x\)
We know that:
\[ 1 - \cos 2x = 2 \sin^2 x \]
Thus, we can rewrite the limit as:
\[ \lim_{x \to 0} \frac{\sqrt{1 - \cos 2x}}{x} = \lim_{x \to 0} \frac{\sqrt{2 \sin^2 x}}{x} \]

### Step 2: Simplify the expression
We can simplify the square root:
\[ \sqrt{2 \sin^2 x} = \sqrt{2} \cdot |\sin x| \]
Since \(x\) approaches \(0\), \(\sin x\) is positive in the vicinity of \(0\):
\[ \lim_{x \to 0} \frac{\sqrt{2} |\sin x|}{x} = \sqrt{2} \lim_{x \to 0} \frac{\sin x}{x} \]

### Step 3: Evaluate the limit
We know from the standard limit that:
\[ \lim_{x \to 0} \frac{\sin x}{x} = 1 \]
Therefore, we can substitute this into our limit:
\[ \sqrt{2} \cdot 1 = \sqrt{2} \]

### Final Result
Thus, the limit is:
\[ \lim_{x \to 0} \frac{\sqrt{1 - \cos 2x}}{x} = \sqrt{2} \]

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