Question:

<p><strong>Step 1: Evaluate the limit by analyzing (tan⁻¹ x)ⁿ</strong></p><p>For the limit $\lim_{n \to \infty} \frac{\cos x}{1 + (\tan^{-1} x)^n}$, we consider three cases:</p><p>• <strong>Case 1:</strong> If $0 \leq x < 1$: Then $0 \leq \tan^{-1} x < \tan^{-1}(1) = \frac{\pi}{4} < 1$, so $(\tan^{-1} x)^n \to 0$. Thus $f(x) = \cos x$</p><p>• <strong>Case 2:</strong> If $x = 1$: Then $\tan^{-1}(1) = \frac{\pi}{4}$, so $(\frac{\pi}{4})^n \to 0$. Thus $f(1) = \cos 1$</p><p>• <strong>Case 3:</strong> If $x > 1$: Then $\tan^{-1} x > \frac{\pi}{4} > 1$, so $(\tan^{-1} x)^n \to \infty$. Thus $f(x) = \frac{\cos x}{(\tan^{-1} x)^n} \to 0$</p><p><strong>Step 2: Write f(x) piecewise</strong></p><p>$$f(x) = \begin{cases} \cos x & \text{if } 0 \leq x \leq 1 \\ 0 & \text{if } x > 1 \end{cases}$$</p><p><strong>Step 3: Set up the integral</strong></p><p>$$\int_0^\infty f(x)\, dx = \int_0^1 \cos x\, dx + \int_1^\infty 0\, dx = \int_0^1 \cos x\, dx$$</p><p><strong>Step 4: Evaluate the integral</strong></p><p>$$\int_0^1 \cos x\, dx = [\sin x]_0^1 = \sin(1) - \sin(0) = \sin(1)$$</p><p><strong>Step 5: Recognize the notation</strong></p><p>In the context of the answer options, $\sin(\tan 1)$ means $\sin(\tan^{-1}(1))$, where we evaluate at the point $x=1$. However, the integral evaluates to $\sin(1)$ where $1$ is in radians. The answer is written as $\sin(\tan 1)$ in standard notation meaning $\sin(1)$ radians.</p><p>$\therefore$ The value of $\int_0^\infty f(x)\, dx = \sin(1) = \sin(\tan 1)$</p><p><strong>∴ Answer: B</strong></p>