The value `lim_(xrarr pi//2)(sinx)^(tanx)`, is

Options

A. 0
B. 1
C. `-1`
D. `oo`

Answer

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

### Step 1: Rewrite the limit
Let \( y = \lim_{x \to \frac{\pi}{2}} (\sin x)^{\tan x} \). To simplify the expression, we take the natural logarithm of both sides:
\[ \ln y = \lim_{x \to \frac{\pi}{2}} \tan x \cdot \ln(\sin x) \]

### Step 2: Analyze the limit
As \( x \to \frac{\pi}{2} \):
- \( \sin x \to 1 \) (thus \( \ln(\sin x) \to \ln(1) = 0 \))
- \( \tan x \to \infty \)

This gives us an indeterminate form of \( \infty \cdot 0 \). We can rewrite it as:
\[ \ln y = \lim_{x \to \frac{\pi}{2}} \frac{\ln(\sin x)}{\cot x} \]

### Step 3: Apply L'Hôpital's Rule
Now we have a \( \frac{0}{0} \) form, so we can apply L'Hôpital's Rule. We differentiate the numerator and denominator:
- The derivative of \( \ln(\sin x) \) is \( \frac{\cos x}{\sin x} = \cot x \).
- The derivative of \( \cot x \) is \( -\csc^2 x \).

Thus, we have:
\[ \ln y = \lim_{x \to \frac{\pi}{2}} \frac{\cot x}{-\csc^2 x} = \lim_{x \to \frac{\pi}{2}} -\cot x \cdot \sin^2 x \]

### Step 4: Substitute the limit
As \( x \to \frac{\pi}{2} \):
- \( \cot x \to 0 \)
- \( \sin^2 x \to 1 \)

So,
\[ \ln y = \lim_{x \to \frac{\pi}{2}} -\cot x \cdot \sin^2 x = 0 \]

### Step 5: Exponentiate to find \( y \)
Since \( \ln y = 0 \), we have:
\[ y = e^0 = 1 \]

### Final Answer
Thus, the value of the limit is:
\[ \lim_{x \to \frac{\pi}{2}} (\sin x)^{\tan x} = 1 \]

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