Evaluate `lim_(xto pi//2) tanxlogsinx`

Options

A. -2
B. -1
C. `-(2)/(7)`
D. 0

Explanation

To evaluate the limit \( \lim_{x \to \frac{\pi}{2}} \tan x \log(\sin x) \), we can follow these steps:

### Step 1: Rewrite the limit
We start by rewriting the limit in a more manageable form. We know that:
\[ \tan x = \frac{\sin x}{\cos x} \]
Thus, we can express the limit as:
\[ \lim_{x \to \frac{\pi}{2}} \tan x \log(\sin x) = \lim_{x \to \frac{\pi}{2}} \frac{\log(\sin x)}{\cot x} \]
where \( \cot x = \frac{\cos x}{\sin x} \).

### Step 2: Analyze the limit
As \( x \to \frac{\pi}{2} \):
- \( \sin x \to 1 \) and thus \( \log(\sin x) \to \log(1) = 0 \).
- \( \cot x \to 0 \) because \( \cos x \to 0 \) and \( \sin x \to 1 \).

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

### Step 3: Apply L'Hôpital's Rule
Since we have an indeterminate form, we can apply L'Hôpital's Rule, which states that:
\[ \lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)} \]
if the limit exists. Here, let:
- \( f(x) = \log(\sin x) \)
- \( g(x) = \cot x \)

We differentiate both the numerator and the denominator.

### Step 4: Differentiate the numerator and denominator
The derivative of \( f(x) = \log(\sin x) \) is:
\[ f'(x) = \frac{1}{\sin x} \cdot \cos x = \frac{\cos x}{\sin x} = \cot x \]
The derivative of \( g(x) = \cot x \) is:
\[ g'(x) = -\csc^2 x \]

### Step 5: Substitute back into the limit
Now we substitute back into the limit:
\[ \lim_{x \to \frac{\pi}{2}} \frac{\cot x}{-\csc^2 x} = \lim_{x \to \frac{\pi}{2}} \frac{\cot x \cdot \sin^2 x}{-1} \]
Since \( \cot x = \frac{\cos x}{\sin x} \), we have:
\[ \lim_{x \to \frac{\pi}{2}} \frac{\cos x \cdot \sin^2 x}{-1} \]

### Step 6: Evaluate the limit
As \( x \to \frac{\pi}{2} \):
- \( \cos x \to 0 \)
- \( \sin^2 x \to 1 \)

Thus, the limit evaluates to:
\[ \lim_{x \to \frac{\pi}{2}} \frac{0 \cdot 1}{-1} = 0 \]

### Final Answer
Therefore, the limit is:
\[ \lim_{x \to \frac{\pi}{2}} \tan x \log(\sin x) = 0 \]

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