Evaluate : `lim_(x to 0) (sqrt(5)-sqrt(4+cosx))/(3sin^(2)x)`.

Answer

To evaluate the limit

\[ \lim_{x \to 0} \frac{\sqrt{5} - \sqrt{4 + \cos x}}{3 \sin^2 x}, \]

we start by substituting \( x = 0 \):

1. **Substituting \( x = 0 \)**:
\[ \sqrt{5} - \sqrt{4 + \cos(0)} = \sqrt{5} - \sqrt{4 + 1} = \sqrt{5} - \sqrt{5} = 0. \]
The denominator:
\[ 3 \sin^2(0) = 3 \cdot 0^2 = 0. \]
This gives us the indeterminate form \( \frac{0}{0} \).

2. **Applying L'Hôpital's Rule**:
Since we have the indeterminate form \( \frac{0}{0} \), we can apply L'Hôpital's Rule, which states that we can take the derivative of the numerator and the denominator.

- **Numerator**:
\[ \frac{d}{dx}(\sqrt{5} - \sqrt{4 + \cos x}) = 0 - \frac{1}{2\sqrt{4 + \cos x}} \cdot (-\sin x) = \frac{\sin x}{2\sqrt{4 + \cos x}}. \]

- **Denominator**:
\[ \frac{d}{dx}(3 \sin^2 x) = 3 \cdot 2 \sin x \cos x = 6 \sin x \cos x. \]

3. **Re-evaluating the limit**:
Now, substituting back into the limit:
\[ \lim_{x \to 0} \frac{\frac{\sin x}{2\sqrt{4 + \cos x}}}{6 \sin x \cos x}. \]
We can simplify this:
\[ = \lim_{x \to 0} \frac{1}{2\sqrt{4 + \cos x} \cdot 6 \cos x} = \lim_{x \to 0} \frac{1}{12 \sqrt{4 + \cos x} \cos x}. \]

4. **Substituting \( x = 0 \) again**:
Now substituting \( x = 0 \):
\[ \cos(0) = 1 \quad \text{and} \quad \sqrt{4 + \cos(0)} = \sqrt{4 + 1} = \sqrt{5}. \]
Therefore, we have:
\[ = \frac{1}{12 \cdot \sqrt{5} \cdot 1} = \frac{1}{12\sqrt{5}}. \]

Thus, the final answer is:

\[ \lim_{x \to 0} \frac{\sqrt{5} - \sqrt{4 + \cos x}}{3 \sin^2 x} = \frac{1}{12\sqrt{5}}. \]