`lim_(x to 0) x log sin x=`

Options

A. 1
B. `-1`
C. 0
D. none

Answer

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

### Step 1: Identify the form of the limit
As \( x \to 0 \), we have:
- \( \sin x \to 0 \)
- Therefore, \( \log(\sin x) \to \log(0) \to -\infty \)

Thus, the expression \( x \log(\sin x) \) approaches the form \( 0 \cdot (-\infty) \), which is indeterminate.

### Step 2: Rewrite the limit
To resolve this indeterminate form, we can rewrite the limit:
\[ \lim_{x \to 0} x \log(\sin x) = \lim_{x \to 0} \frac{\log(\sin x)}{1/x} \]
This transforms our limit into a \( \frac{-\infty}{\infty} \) form, which is suitable for applying L'Hôpital's Rule.

### Step 3: Apply L'Hôpital's Rule
According to L'Hôpital's Rule, we take the derivative of the numerator and the derivative of the denominator:
- The derivative of \( \log(\sin x) \) is \( \frac{1}{\sin x} \cdot \cos x = \frac{\cos x}{\sin x} \).
- The derivative of \( 1/x \) is \( -\frac{1}{x^2} \).

Thus, we have:
\[ \lim_{x \to 0} \frac{\log(\sin x)}{1/x} = \lim_{x \to 0} \frac{\frac{\cos x}{\sin x}}{-\frac{1}{x^2}} = \lim_{x \to 0} -x^2 \frac{\cos x}{\sin x} \]

### Step 4: Simplify the limit
Now we can express this limit as:
\[ \lim_{x \to 0} -x^2 \frac{\cos x}{\sin x} \]
We know that \( \frac{x}{\sin x} \to 1 \) as \( x \to 0 \), hence \( \frac{\sin x}{x} \to 1 \) implies \( \frac{\cos x}{\sin x} = \frac{\cos x}{x} \cdot \frac{x}{\sin x} \).

Thus:
\[ \lim_{x \to 0} -x^2 \cdot \frac{\cos x}{\sin x} = \lim_{x \to 0} -x^2 \cdot \frac{\cos x}{x} \cdot \frac{x}{\sin x} \]
As \( x \to 0 \):
- \( \cos x \to 1 \)
- \( \frac{x}{\sin x} \to 1 \)

So we get:
\[ \lim_{x \to 0} -x^2 \cdot 1 \cdot 1 = 0 \]

### Final Result
Thus, the limit is:
\[ \lim_{x \to 0} x \log(\sin x) = 0 \]