`lim_(xoto0) x log_(e) (sinx)` is equal to

Options

A. -1
B. `log_(e)1`
C. 1
D. none of these

Explanation

To solve the limit \( \lim_{x \to 0} x \log_e(\sin x) \), we can follow these steps:

### Step 1: Identify the form of the limit
First, we substitute \( x = 0 \) into the expression:
\[ \sin(0) = 0 \quad \text{and thus} \quad \log_e(\sin(0)) = \log_e(0). \]
Since \( \log_e(0) \) approaches \( -\infty \), we have:
\[ 0 \cdot (-\infty) \quad \text{which is an indeterminate form of type } 0 \cdot (-\infty). \]

### Step 2: Rewrite the limit
To resolve the indeterminate form, we can rewrite the expression:
\[ \lim_{x \to 0} x \log_e(\sin x) = \lim_{x \to 0} \frac{\log_e(\sin x)}{\frac{1}{x}}. \]
Now, this is in the form \( \frac{-\infty}{\infty} \), which is suitable for applying L'Hôpital's rule.

### Step 3: Apply L'Hôpital's Rule
According to L'Hôpital's rule, we differentiate the numerator and the denominator:
- The derivative of the numerator \( \log_e(\sin x) \) is:
\[ \frac{d}{dx} \log_e(\sin x) = \frac{1}{\sin x} \cdot \cos x = \cot x. \]
- The derivative of the denominator \( \frac{1}{x} \) is:
\[ \frac{d}{dx} \left(\frac{1}{x}\right) = -\frac{1}{x^2}. \]
Thus, we can rewrite the limit as:
\[ \lim_{x \to 0} \frac{\cot x}{-\frac{1}{x^2}} = -\lim_{x \to 0} x^2 \cot x. \]

### Step 4: Simplify the expression
We know that \( \cot x = \frac{\cos x}{\sin x} \), so:
\[ -\lim_{x \to 0} x^2 \cot x = -\lim_{x \to 0} x^2 \cdot \frac{\cos x}{\sin x} = -\lim_{x \to 0} \frac{x^2 \cos x}{\sin x}. \]
As \( x \to 0 \), \( \cos x \to 1 \) and \( \sin x \approx x \), so:
\[ -\lim_{x \to 0} \frac{x^2 \cdot 1}{x} = -\lim_{x \to 0} x = 0. \]

### Conclusion
Thus, we find that:
\[ \lim_{x \to 0} x \log_e(\sin x) = 0. \]

### Final Answer
The limit is equal to \( 0 \).

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