`lim_(x to pi//2) tan x log_(e) sin x=`

Options

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

Answer

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

### Step 1: Rewrite the limit
We start with the limit:
\[ \lim_{x \to \frac{\pi}{2}} \tan x \log_e \sin x \]
We know that \( \tan x = \frac{\sin x}{\cos x} \), so we can rewrite the limit as:
\[ \lim_{x \to \frac{\pi}{2}} \frac{\log_e \sin x}{\cot x} \]

### Step 2: Evaluate the limit
Now, we evaluate the limit by substituting \( x = \frac{\pi}{2} \):
- As \( x \to \frac{\pi}{2} \), \( \sin x \to 1 \) and thus \( \log_e \sin x \to \log_e 1 = 0 \).
- Also, \( \cot x = \frac{\cos x}{\sin x} \) approaches \( \cot \frac{\pi}{2} = 0 \).

This gives us the indeterminate form \( \frac{0}{0} \), so we can apply 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:
\[ \frac{d}{dx}(\log_e \sin x) = \frac{1}{\sin x} \cdot \cos x = \frac{\cos x}{\sin x} = \cot x \]
- The derivative of the denominator \( \cot x \) is:
\[ \frac{d}{dx}(\cot x) = -\csc^2 x \]

Now we can rewrite the limit:
\[ \lim_{x \to \frac{\pi}{2}} \frac{\cot x}{-\csc^2 x} \]

### Step 4: Simplify the limit
Substituting the derivatives back into the limit gives:
\[ \lim_{x \to \frac{\pi}{2}} \frac{\cot x}{-\csc^2 x} = \lim_{x \to \frac{\pi}{2}} \frac{\cos x / \sin x}{-\frac{1}{\sin^2 x}} = \lim_{x \to \frac{\pi}{2}} -\cos x \sin x \]

### Step 5: Evaluate the limit
As \( x \to \frac{\pi}{2} \):
- \( \cos x \to 0 \)
- \( \sin x \to 1 \)

Thus:
\[ \lim_{x \to \frac{\pi}{2}} -\cos x \sin x = -0 \cdot 1 = 0 \]

### Final Result
The final result is:
\[ \lim_{x \to \frac{\pi}{2}} \tan x \log_e \sin x = 0 \]