Question

Evaluate the limit: \[ \lim_{n \to \infty} \left[ \frac{1}{n^2} \sec^2 \frac{1}{n^2} + \frac{2}{n^2} \sec^2 \frac{4}{n^2} + \frac{3}{n^2} \sec^2 \frac{9}{n^2} + .s + \frac{1}{n^2} \sec^2 1 \right] = \]

• A. \( \tan^{-1} 1 \)
• B. \( \frac{1}{2} \tan^{-1} 1 \)
• C. \( \frac{1}{2} \tan 1 \)
• D. \( \frac{1}{2} \sec 1 \)

Hint:

In limit problems involving sums with increasing index terms and a function applied to squares, look for Riemann sums and convert to definite integrals using suitable substitutions.

Answer 1

The Correct Option is C

We are given: \[ \lim_{n \to \infty} \sum_{k=1}^{n} \frac{k}{n^2} \sec^2 \left( \frac{k^2}{n^2} \right) \] Step 1: Convert to Riemann sum form. Let: \[ x = \frac{k}{n} \Rightarrow \frac{k^2}{n^2} = x^2,
\frac{1}{n} \text{ is the width of subinterval} \] So the expression becomes: \[ \lim_{n \to \infty} \sum_{k=1}^{n} \left( \frac{k}{n} . \frac{1}{n} \sec^2 \left( \left( \frac{k}{n} \right)^2 \right) \right) = \lim_{n \to \infty} \sum_{k=1}^{n} x \sec^2(x^2) . \frac{1}{n} \] This is a Riemann sum: \[ \int_0^1 x \sec^2(x^2) \, dx \] Step 2: Solve the definite integral. Let \( u = x^2 \Rightarrow du = 2x dx \Rightarrow x dx = \frac{1}{2} du \) Then, \[ \int_0^1 x \sec^2(x^2) \, dx = \int_0^1 \sec^2(x^2) . x \, dx = \frac{1}{2} \int_0^1 \sec^2 u \, du = \frac{1}{2} \tan u \Big|_0^1 = \frac{1}{2} \tan 1 \] \[ \boxed{ \lim_{n \to \infty} \left[ .s \right] = \frac{1}{2} \tan 1 } \]