Question:

<p><strong>Step 1: Analyze Option (a):</strong> $\lim_{x \to \infty} \csc^{-1}\left(\frac{x}{x+7}\right)$</p><p>As $x \to \infty$: $\frac{x}{x+7} = \frac{1}{1+\frac{7}{x}} \to 1$</p><p>Since $\csc^{-1}(1) = \frac{\pi}{2}$, this limit exists and equals $\frac{\pi}{2}$. ✓</p><p><strong>Step 2: Analyze Option (b):</strong> $\lim_{x \to 1} \sec^{-1}(\sin^{-1} x)$</p><p>As $x \to 1$: $\sin^{-1}(1) = \frac{\pi}{2}$</p><p>Then $\sec^{-1}\left(\frac{\pi}{2}\right)$ is defined since $\frac{\pi}{2} \approx 1.57 > 1$ (within domain of $\sec^{-1}$)</p><p>This limit exists. ✓</p><p><strong>Step 3: Analyze Option (c):</strong> $\lim_{x \to 0^+} x^{\frac{1}{x}}$</p><p>Rewrite: $x^{\frac{1}{x}} = e^{\frac{\ln x}{x}}$</p><p>As $x \to 0^+$: $\ln x \to -\infty$ and $x \to 0^+$, so $\frac{\ln x}{x} \to -\infty$</p><p>Therefore $e^{-\infty} = 0$. The limit exists and equals $0$. ✓</p><p><strong>Step 4: Analyze Option (d):</strong> $\lim_{x \to 0} \tan\left(\frac{\pi}{8} + x\right)^{\cot x}$</p><p>Base: As $x \to 0$, $\tan\left(\frac{\pi}{8} + x\right) \to \tan\left(\frac{\pi}{8}\right) = \sqrt{2} - 1 \approx 0.414 < 1$</p><p>Exponent: As $x \to 0^+$, $\cot x \to +\infty$; as $x \to 0^-$, $\cot x \to -\infty$</p><p>From right: $(0.414)^{+\infty} \to 0$</p><p>From left: $(0.414)^{-\infty} \to +\infty$</p><p>The left and right limits differ, so the limit does not exist. ✗</p><p><strong>∴ Answer:</strong> D</p>