Question:

Step 1: Analyze the limit behavior of $f(x) = \lim_{n \to \infty} \dfrac{\cos x}{1 + (\tan^{-1} x)^n}$. We need to understand how $(\tan^{-1} x)^n$ behaves as $n \to \infty$. Recall that $\tan^{-1} x \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ for all real $x$. Step 2: Determine $f(x)$ when $|\tan^{-1} x| < 1$. When $|\tan^{-1} x| < 1$, we have $x \in (-\tan 1, \tan 1)$. In this case, as $n \to \infty$: $$(\tan^{-1} x)^n \to 0$$ Therefore: $$f(x) = \lim_{n \to \infty} \dfrac{\cos x}{1 + 0} = \cos x$$ Step 3: Determine $f(x)$ when $|\tan^{-1} x| > 1$. When $|\tan^{-1} x| > 1$, we have $x > \tan 1$ or $x < -\tan 1$. For $x > \tan 1$, we have $\tan^{-1} x > 1 > 0$, so: $$(\tan^{-1} x)^n \to \infty \text{ as } n \to \infty$$ Therefore: $$f(x) = \lim_{n \to \infty} \dfrac{\cos x}{1 + (\tan^{-1} x)^n} = \lim_{n \to \infty} \dfrac{\cos x}{(\tan^{-1} x)^n} = 0$$ Step 4: Determine $f(x)$ when $\tan^{-1} x = 1$. When $\tan^{-1} x = 1$, we have $x = \tan 1$. At this point: $$f(\tan 1) = \dfrac{\cos(\tan 1)}{1 + 1^n} = \dfrac{\cos(\tan 1)}{2}$$ This is a single point and contributes zero to the integral. Step 5: Summarize the piecewise form of $f(x)$. For the integral over $[0, \infty)$, we have: $$f(x) = \begin{cases} \cos x & \text{if } 0 \le x < \tan 1 \\ 0 & \text{if } x > \tan 1 \end{cases}$$ Step 6: Evaluate the integral. $$\int_0^{\infty} f(x)\, dx = \int_0^{\tan 1} \cos x\, dx + \int_{\tan 1}^{\infty} 0\, dx$$ $$= \int_0^{\tan 1} \cos x\, dx$$ $$= [\sin x]_0^{\tan 1}$$ $$= \sin(\tan 1) - \sin(0)$$ $$= \sin(\tan 1)$$ **Final Answer:** The value of $\int_0^{\infty} f(x)\, dx = \sin(\tan 1)$, which corresponds to **Option 2: (b) sin(tan 1)**.