The function `f(x)=(sinx)^(tan^(2)x)` is not defined at `x=(pi)/(2)`. The value of `f((pi)/(2))` such that f is continuous at `x=(pi)/(2)` is

Options

A. `sqrte`
B. `(1)/(sqrte)`
C. 2
D. None of these

Answer

To find the value of \( f\left(\frac{\pi}{2}\right) \) such that the function \( f(x) = (\sin x)^{\tan^2 x} \) is continuous at \( x = \frac{\pi}{2} \), we will analyze the limit of \( f(x) \) as \( x \) approaches \( \frac{\pi}{2} \).

### Step 1: Identify the form of \( f\left(\frac{\pi}{2}\right) \)
We start by evaluating \( f\left(\frac{\pi}{2}\right) \):
\[ f\left(\frac{\pi}{2}\right) = (\sin\left(\frac{\pi}{2}\right))^{\tan^2\left(\frac{\pi}{2}\right)} = 1^{\infty} \]
This is an indeterminate form.

**Hint:** Recognize that \( 1^{\infty} \) is an indeterminate form that requires further analysis using limits.

### Step 2: Use the limit definition for continuity
Since \( f \) is continuous at \( x = \frac{\pi}{2} \), we need to find:
\[ \lim_{x \to \frac{\pi}{2}} f(x) \]
This can be rewritten as:
\[ \lim_{x \to \frac{\pi}{2}} (\sin x)^{\tan^2 x} \]

**Hint:** Use the property of limits to express the function in a more manageable form.

### Step 3: Rewrite the function using exponentials
We can express the limit in terms of the exponential function:
\[ (\sin x)^{\tan^2 x} = e^{\tan^2 x \ln(\sin x)} \]
Thus, we need to evaluate:
\[ \lim_{x \to \frac{\pi}{2}} \tan^2 x \ln(\sin x) \]

**Hint:** Focus on simplifying \( \tan^2 x \ln(\sin x) \) as \( x \) approaches \( \frac{\pi}{2} \).

### Step 4: Analyze the limit
As \( x \to \frac{\pi}{2} \):
- \( \sin x \to 1 \) which means \( \ln(\sin x) \to \ln(1) = 0 \).
- \( \tan^2 x \to \infty \).

This results in the form \( \infty \cdot 0 \), which is also indeterminate. We can rewrite it as:
\[ \lim_{x \to \frac{\pi}{2}} \frac{\ln(\sin x)}{\cot^2 x} \]

**Hint:** Rewrite the expression to apply L'Hôpital's Rule.

### Step 5: Apply L'Hôpital's Rule
Since both the numerator and denominator approach 0 as \( x \to \frac{\pi}{2} \), we can apply L'Hôpital's Rule:
\[ \lim_{x \to \frac{\pi}{2}} \frac{\ln(\sin x)}{\cot^2 x} = \lim_{x \to \frac{\pi}{2}} \frac{\frac{d}{dx}[\ln(\sin x)]}{\frac{d}{dx}[\cot^2 x]} \]
Calculating the derivatives:
- The derivative of \( \ln(\sin x) \) is \( \frac{\cos x}{\sin x} = \cot x \).
- The derivative of \( \cot^2 x \) is \( -2 \cot x \csc^2 x \).

Thus, we have:
\[ \lim_{x \to \frac{\pi}{2}} \frac{\cot x}{-2 \cot x \csc^2 x} = \lim_{x \to \frac{\pi}{2}} \frac{1}{-2 \csc^2 x} \]

**Hint:** Evaluate the limit as \( x \) approaches \( \frac{\pi}{2} \).

### Step 6: Evaluate the limit
As \( x \to \frac{\pi}{2} \), \( \csc^2 x \to 1 \):
\[ \lim_{x \to \frac{\pi}{2}} \frac{1}{-2 \csc^2 x} = \frac{1}{-2} = -\frac{1}{2} \]

### Step 7: Final result
Substituting back into our exponential form:
\[ \lim_{x \to \frac{\pi}{2}} f(x) = e^{-\frac{1}{2}} = \frac{1}{\sqrt{e}} \]

Thus, the value of \( f\left(\frac{\pi}{2}\right) \) such that \( f \) is continuous at \( x = \frac{\pi}{2} \) is:
\[ \boxed{\frac{1}{\sqrt{e}}} \]