The value of `lim_(xrarr0) (sqrt(1-cosx^(2)))/(1-cos x)` is

Options

A. `(1)/(2)`
B. `2`
C. `sqrt2`
D. none of these

Explanation

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

### Step 1: Rewrite the limit using trigonometric identities
We know that \( 1 - \cos \theta = 2 \sin^2\left(\frac{\theta}{2}\right) \). Therefore, we can rewrite both the numerator and the denominator:

\[ 1 - \cos(x^2) = 2 \sin^2\left(\frac{x^2}{2}\right) \]
\[ 1 - \cos x = 2 \sin^2\left(\frac{x}{2}\right) \]

Now substituting these into our limit gives:

\[ L = \lim_{x \to 0} \frac{\sqrt{2 \sin^2\left(\frac{x^2}{2}\right)}}{2 \sin^2\left(\frac{x}{2}\right)} \]

### Step 2: Simplify the expression
The square root in the numerator can be simplified:

\[ L = \lim_{x \to 0} \frac{\sqrt{2} \sin\left(\frac{x^2}{2}\right)}{2 \sin^2\left(\frac{x}{2}\right)} \]

### Step 3: Factor out constants
We can factor out the constant \( \frac{\sqrt{2}}{2} \):

\[ L = \frac{\sqrt{2}}{2} \lim_{x \to 0} \frac{\sin\left(\frac{x^2}{2}\right)}{\sin^2\left(\frac{x}{2}\right)} \]

### Step 4: Apply the limit
Using the limit property \( \lim_{y \to 0} \frac{\sin y}{y} = 1 \), we rewrite the limit:

\[ L = \frac{\sqrt{2}}{2} \lim_{x \to 0} \frac{\sin\left(\frac{x^2}{2}\right)}{\frac{x^2}{2}} \cdot \frac{\frac{x^2}{2}}{\sin^2\left(\frac{x}{2}\right)} \]

### Step 5: Rewrite the sine terms
Now, we can express \( \sin\left(\frac{x}{2}\right) \) in terms of \( x \):

\[ \sin\left(\frac{x}{2}\right) \approx \frac{x}{2} \text{ as } x \to 0 \]

Thus, \( \sin^2\left(\frac{x}{2}\right) \approx \left(\frac{x}{2}\right)^2 = \frac{x^2}{4} \).

### Step 6: Substitute and simplify
Now substituting back into our limit gives:

\[ L = \frac{\sqrt{2}}{2} \lim_{x \to 0} \frac{\sin\left(\frac{x^2}{2}\right)}{\frac{x^2}{2}} \cdot \frac{\frac{x^2}{2}}{\frac{x^2}{4}} = \frac{\sqrt{2}}{2} \lim_{x \to 0} 2 \cdot \frac{\sin\left(\frac{x^2}{2}\right)}{\frac{x^2}{2}} = \sqrt{2} \]

### Final Result
Thus, the value of the limit is:

\[ L = \sqrt{2} \]