Question

Let $f : (-\infty, \infty) - \\{0\\} \to \mathbb{R}$ be a differentiable function such that $f'(1) = \lim_{a \to \infty} a^2 f\left(\frac{1}{a}\right)$. Then $\lim_{a \to \infty} \frac{a(a + 1)}{2} \tan^{-1}\left(\frac{1}{a}\right) + a^2 - 2 \log_e a$ is equal to:

• A. $\frac{3}{2} + \frac{\pi}{4}$
• B. $\frac{3}{8} + \frac{\pi}{4}$
• C. $\frac{5}{2} + \frac{\pi}{8}$
• D. $\frac{3}{4} + \frac{\pi}{8}$

Answer 1

Answer: (C)

$\frac{5}{2} + \frac{\pi}{8}$

Answer 2

We are given the limit expression:

$\lim_{a \to \infty} \frac{a(a + 1)}{2} \tan^{-1} \left( \frac{1}{a} \right) + a^2 - 2 \ln a.$

Step 1: Simplify the $\tan^{-1}$ term Using the expansion:

$\tan^{-1} \left( \frac{1}{a} \right) \approx \frac{1}{a} - \frac{1}{3a^3} \quad \text{as } a \to \infty.$

Substitute this approximation:

$\frac{a(a + 1)}{2} \tan^{-1} \left( \frac{1}{a} \right) \approx \frac{a(a + 1)}{2} \left( \frac{1}{a} - \frac{1}{3a^3} \right).$

As $a \to \infty$, the dominant term is:

$\frac{(a + 1)}{2} - \frac{(a + 1)}{6a^2}.$

As $a \to \infty$, the dominant term is:

$\frac{(a + 1)}{2} \to \frac{a}{2}.$

Step 2: Rewrite the full limit expression The given expression becomes:

$\lim_{a \to \infty} \left( \frac{a}{2} + a^2 - 2 \ln a \right).$

Step 3: Identify $f(x)$ From the problem:

$f(x) = \frac{1}{2} \left( (1 + x) \tan^{-1}(x) + 1 - 2x^2 \ln(x) \right).$

Compute $f'(x)$:

$f'(x) = \frac{1}{2} \left( \frac{1 + x}{1 + x^2} + \tan^{-1}(x) + 4x \ln(x) + 2x \right).$

Substitute $x = 1$:

$f'(1) = \frac{1}{2} \left( \frac{1 + 1}{1 + 1} + \frac{\pi}{4} + 4(1) \ln(1) + 2(1) \right).$

Simplify:

$f'(1) = \frac{1}{2} \left( 1 + \frac{\pi}{4} + 2 \right).$

Thus:

$f'(1) = \frac{5}{2} + \frac{\pi}{8}.$