The value of `lim_(x->oo)(pi/2-tan^(- 1)x)^(1/x^2),` is

Options

A. 0
B. 1
C. -1
D. e

Answer

To solve the limit \( \lim_{x \to \infty} \left( \frac{\pi}{2} - \tan^{-1} x \right)^{\frac{1}{x^2}} \), we can follow these steps:

### Step 1: Identify the form of the limit
As \( x \to \infty \), \( \tan^{-1} x \) approaches \( \frac{\pi}{2} \). Therefore, \( \frac{\pi}{2} - \tan^{-1} x \) approaches \( 0 \). Thus, we have the limit in the form \( 0^{0} \).

**Hint:** Recognizing the form of the limit helps determine the appropriate method to solve it.

### Step 2: Take the natural logarithm
Let \( L = \lim_{x \to \infty} \left( \frac{\pi}{2} - \tan^{-1} x \right)^{\frac{1}{x^2}} \). Taking the natural logarithm, we have:
\[ \log L = \lim_{x \to \infty} \frac{1}{x^2} \log \left( \frac{\pi}{2} - \tan^{-1} x \right) \]

**Hint:** Taking the logarithm simplifies the exponentiation and allows us to work with the limit more easily.

### Step 3: Analyze the limit
We know that as \( x \to \infty \), \( \tan^{-1} x \to \frac{\pi}{2} \), thus \( \frac{\pi}{2} - \tan^{-1} x \to 0 \). Therefore, we need to analyze the behavior of \( \log \left( \frac{\pi}{2} - \tan^{-1} x \right) \).

Using the property of limits, we can express \( \frac{\pi}{2} - \tan^{-1} x \) as:
\[ \frac{\pi}{2} - \tan^{-1} x = \frac{1}{x} + o\left(\frac{1}{x}\right) \text{ as } x \to \infty \]
Thus,
\[ \log \left( \frac{\pi}{2} - \tan^{-1} x \right) \sim \log \left( \frac{1}{x} \right) = -\log x \]

**Hint:** Understanding the asymptotic behavior of functions as they approach limits can help simplify the limit.

### Step 4: Substitute back into the limit
Now substituting back, we have:
\[ \log L = \lim_{x \to \infty} \frac{-\log x}{x^2} \]
This limit is of the form \( \frac{-\infty}{\infty} \), so we can apply L'Hôpital's Rule.

**Hint:** L'Hôpital's Rule is useful for resolving indeterminate forms.

### Step 5: Apply L'Hôpital's Rule
Differentiating the numerator and denominator:
\[ \frac{d}{dx}(-\log x) = -\frac{1}{x}, \quad \frac{d}{dx}(x^2) = 2x \]
Thus, applying L'Hôpital's Rule gives:
\[ \log L = \lim_{x \to \infty} \frac{-\frac{1}{x}}{2x} = \lim_{x \to \infty} \frac{-1}{2x^2} = 0 \]

**Hint:** Differentiating can often simplify complex limits.

### Step 6: Conclude the limit
Since \( \log L = 0 \), we have:
\[ L = e^0 = 1 \]

Thus, the value of the limit is:
\[ \lim_{x \to \infty} \left( \frac{\pi}{2} - \tan^{-1} x \right)^{\frac{1}{x^2}} = 1 \]

**Final Answer:** \( \boxed{1} \)