`lim_(xto0) (sqrt(1-cos 2x))/(sqrt2x)` is equal to-

Options

A. 1
B. `-1`
C. zero
D. Does not exist

Answer

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

### Step 1: Rewrite the limit
We start with the expression:
\[ \lim_{x \to 0} \frac{\sqrt{1 - \cos 2x}}{\sqrt{2x}} \]
We can rewrite this as:
\[ \lim_{x \to 0} \frac{\sqrt{1 - \cos 2x}}{\sqrt{2} \sqrt{x}} \]

### Step 2: Use the trigonometric identity
We know from trigonometric identities that:
\[ 1 - \cos \theta = 2 \sin^2\left(\frac{\theta}{2}\right) \]
Applying this to our expression where \( \theta = 2x \):
\[ 1 - \cos 2x = 2 \sin^2(x) \]
Thus, we can substitute this into our limit:
\[ \lim_{x \to 0} \frac{\sqrt{2 \sin^2 x}}{\sqrt{2} \sqrt{x}} \]

### Step 3: Simplify the expression
Now, we can simplify the limit:
\[ \lim_{x \to 0} \frac{\sqrt{2} \sin x}{\sqrt{2} \sqrt{x}} = \lim_{x \to 0} \frac{\sin x}{\sqrt{x}} \]

### Step 4: Analyze the limit
The limit can be rewritten as:
\[ \lim_{x \to 0} \frac{\sin x}{x} \cdot \frac{x}{\sqrt{x}} = \lim_{x \to 0} \frac{\sin x}{x} \cdot \sqrt{x} \]
As \( x \to 0 \), we know that \( \frac{\sin x}{x} \to 1 \) and \( \sqrt{x} \to 0 \).

### Step 5: Final evaluation
Thus, we have:
\[ \lim_{x \to 0} \frac{\sin x}{x} \cdot \sqrt{x} = 1 \cdot 0 = 0 \]

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