Limits, Continuity & Differentiability
L'Hôpital's rule with integral functions
Grade None
Question:
<p>\(\lim_{x\to\infty} \dfrac{\left(\displaystyle\int_0^x e^{x^2}\,dx\right)^2}{\displaystyle\int_0^x e^{2x^2}\,dx} = \)</p>
Step-by-Step Solution
Key Concept: Use L'Hôpital's rule on the indeterminate form ∞/∞ by differentiating numerator and denominator with respect to x. The numerator derivative involves the chain rule on a squared integral, giving 2·(∫₀ˣ e^(x²)dx)·e^(x²).
<p><strong>Step 1:</strong> Recognize the form as ∞/∞ as x→∞. Apply L'Hôpital's rule.</p><p><strong>Step 2:</strong> Differentiate numerator: d/dx[(∫₀ˣ e^(x²)dx)²] = 2·(∫₀ˣ e^(x²)dx)·e^(x²) (by chain rule and FTC)</p><p><strong>Step 3:</strong> Differentiate denominator: d/dx[∫₀ˣ e^(2x²)dx] = e^(2x²) (by FTC)</p><p><strong>Step 4:</strong> The limit becomes: lim(x→∞) [2·(∫₀ˣ e^(x²)dx)·e^(x²)]/e^(2x²)</p><p><strong>Step 5:</strong> Simplify: lim(x→∞) 2·(∫₀ˣ e^(x²)dx)·e^(x²)/e^(2x²) = lim(x→∞) 2·(∫₀ˣ e^(x²)dx)/e^(x²)</p><p><strong>Step 6:</strong> This is again ∞/∞. Apply L'Hôpital's rule again: lim(x→∞) 2·e^(x²)/2x·e^(x²) = lim(x→∞) 1/x = 0</p><p>Alternatively, using the asymptotic analysis: ∫₀ˣ e^(x²)dx ~ e^(x²)/(2x) for large x, so the original limit = lim(x→∞) [e^(x²)/(2x)]²/[e^(2x²)/(2x)] = lim(x→∞) e^(2x²)/(4x²)·(2x)/e^(2x²) = lim(x→∞) 1/(2x) = 0</p><p>∴ Answer: <strong>0</strong></p>
Correct Answer: 0