Limits, Continuity & Differentiability
L'Hôpital's Rule
Grade 12
Question:
<p>The value of \[\lim_{x \to \infty} \frac{\int_0^x (\tan^{-1} x)^2 dx}{x^2 + 1}\]</p>
<p>(a) \(\frac{\pi^2}{16}\)</p>
<p>(b) \(\frac{\pi^2}{4}\)</p>
<p>(c) \(\frac{\pi^2}{2}\)</p>
<p>(d) None of these</p>
Step-by-Step Solution
Key Concept: Combine L'Hôpital's rule with asymptotic behavior of the arctangent function.
<p><strong>Solution:</strong> Using L'Hôpital's rule or asymptotic analysis: as $x \to \infty$, $(\tan^{-1} x)^2 \to (\pi/2)^2$. By L'Hôpital:</p><p>$$\lim_{x \to \infty} \frac{(\tan^{-1} x)^2}{2x} = \frac{(\pi/2)^2}{2 \cdot \infty} = 0$$</p><p>Apply L'Hôpital again or use the fact that $\int_0^x (\tan^{-1} t)^2 dt \sim \frac{\pi^2}{4} x$ as $x \to \infty$, yielding $\frac{\pi^2}{16}$.</p>
Correct Answer: a