<p>Let \(I_n = \int_0^{\pi/4} \tan^n x\,dx\). Find \(\lim_{n \to \infty} n(I_n + I_{n+2})\).</p>
Step-by-Step Solution
Key Concept: Use the reduction formula $I_n + I_{n+2} = \int_0^{\pi/4} \tan^n x(1 + \tan^2 x)dx = \int_0^{\pi/4} \tan^n x \sec^2 x\, dx$, which yields a simple antiderivative $\frac{\tan^{n+1} x}{n+1}\big|_0^{\pi/4} = \frac{1}{n+1}$.
<p><strong>Step 1:</strong> Establish the reduction formula.</p><p>$$I_n + I_{n+2} = \int_0^{\pi/4} \tan^n x\, dx + \int_0^{\pi/4} \tan^{n+2} x\, dx = \int_0^{\pi/4} \tan^n x(1 + \tan^2 x)\, dx$$</p><p><strong>Step 2:</strong> Simplify using the trigonometric identity.</p><p>Since $1 + \tan^2 x = \sec^2 x$:</p><p>$$I_n + I_{n+2} = \int_0^{\pi/4} \tan^n x \sec^2 x\, dx$$</p><p><strong>Step 3:</strong> Integrate using substitution $u = \tan x$, so $du = \sec^2 x\, dx$.</p><p>$$I_n + I_{n+2} = \int_0^1 u^n\, du = \left[\frac{u^{n+1}}{n+1}\right]_0^1 = \frac{1}{n+1}$$</p><p><strong>Step 4:</strong> Evaluate the limit.</p><p>$$\lim_{n \to \infty} n(I_n + I_{n+2}) = \lim_{n \to \infty} \frac{n}{n+1} = \lim_{n \to \infty} \frac{1}{1 + \frac{1}{n}} = 1$$</p><p>∴ Answer: <strong>1</strong></p>
Correct Answer: A