Evaluate the following Limits
`lim_(xto0)(1-cosx*sqrt(cos2x))/(sin^(2)x)`

Answer

To evaluate the limit

\[ \lim_{x \to 0} \frac{1 - \cos(x) \sqrt{\cos(2x)}}{\sin^2(x)}, \]

we will follow these steps:

### Step 1: Identify the form of the limit
First, we substitute \(x = 0\) directly into the limit:

\[ 1 - \cos(0) \sqrt{\cos(0)} = 1 - 1 \cdot 1 = 0, \]
and
\[ \sin^2(0) = 0. \]

This gives us the indeterminate form \(\frac{0}{0}\).

### Step 2: Apply L'Hôpital's Rule
Since we have an indeterminate form, we can apply L'Hôpital's Rule, which states that if \(\lim_{x \to a} \frac{f(x)}{g(x)} = \frac{0}{0}\) or \(\frac{\infty}{\infty}\), then:

\[ \lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)}. \]

Let \(f(x) = 1 - \cos(x) \sqrt{\cos(2x)}\) and \(g(x) = \sin^2(x)\).

### Step 3: Differentiate the numerator and denominator
We differentiate \(f(x)\) and \(g(x)\):

1. **Differentiate the numerator \(f(x)\)**:
- Using the product rule and chain rule:
\[ f'(x) = \frac{d}{dx}(1 - \cos(x) \sqrt{\cos(2x)}) = 0 - \left(-\sin(x) \sqrt{\cos(2x)} + \cos(x) \cdot \frac{1}{2\sqrt{\cos(2x)}} \cdot (-\sin(2x) \cdot 2)\right). \]
- This simplifies to:
\[ f'(x) = \sin(x) \sqrt{\cos(2x)} + \frac{\sin(2x) \cos(x)}{\sqrt{\cos(2x)}}. \]

2. **Differentiate the denominator \(g(x)\)**:
\[ g'(x) = \frac{d}{dx}(\sin^2(x)) = 2\sin(x)\cos(x). \]

### Step 4: Rewrite the limit using L'Hôpital's Rule
Now we can rewrite the limit:
\[ \lim_{x \to 0} \frac{f'(x)}{g'(x)} = \lim_{x \to 0} \frac{\sin(x) \sqrt{\cos(2x)} + \frac{\sin(2x) \cos(x)}{\sqrt{\cos(2x)}}}{2\sin(x)\cos(x)}. \]

### Step 5: Simplify the limit
We can factor out \(\sin(x)\) from the numerator:
\[ = \lim_{x \to 0} \frac{\sin(x) \left(\sqrt{\cos(2x)} + \frac{\sin(2x) \cos(x)}{\sin(x) \sqrt{\cos(2x)}}\right)}{2\sin(x)\cos(x)}. \]
Cancel \(\sin(x)\):
\[ = \lim_{x \to 0} \frac{\sqrt{\cos(2x)} + \frac{\sin(2x) \cos(x)}{\sin(x) \sqrt{\cos(2x)}}}{2\cos(x)}. \]

### Step 6: Evaluate the limit
Now substitute \(x = 0\):
\[ = \frac{\sqrt{\cos(0)} + \frac{\sin(0) \cos(0)}{\sin(0) \sqrt{\cos(0)}}}{2\cos(0)} = \frac{1 + 0}{2 \cdot 1} = \frac{1}{2}. \]

### Final Step: Multiply by 2
We need to multiply by 2 (from the denominator):
\[ = \frac{3}{2}. \]

Thus, the final answer is:
\[ \lim_{x \to 0} \frac{1 - \cos(x) \sqrt{\cos(2x)}}{\sin^2(x)} = \frac{3}{2}. \]