Limits, Continuity & Differentiability
L'Hospital's Rule with integrals
Grade 12
Question:
<p>Find <span>\(L = \lim_{x \to \infty} \dfrac{\left(\int_0^x e^{x^2}\, dx\right)^2}{\int_0^x e^{2x^2}\, dx}\)</span>.</p>
<p>A) 1</p>
<p>B) 2</p>
<p>C) \(\infty\)</p>
<p>D) 0</p>
Step-by-Step Solution
Key Concept: Use L'Hôpital's rule on the ∞/∞ form by differentiating numerator and denominator with respect to x, recognizing that d/dx[∫₀ˣ f(t)dt] = f(x) by the Fundamental Theorem of Calculus.
<p><strong>Step 1:</strong> Verify the form is ∞/∞. As x → ∞, both ∫₀ˣ e^(t²)dt and ∫₀ˣ e^(2t²)dt grow without bound, so L'Hôpital's rule applies.</p><p><strong>Step 2:</strong> Apply L'Hôpital's rule by differentiating numerator and denominator with respect to x:<br/>Numerator: d/dx[(∫₀ˣ e^(t²)dt)²] = 2(∫₀ˣ e^(t²)dt) · e^(x²)<br/>Denominator: d/dx[∫₀ˣ e^(2t²)dt] = e^(2x²)</p><p><strong>Step 3:</strong> The limit becomes:<br/>L = lim_{x→∞} [2(∫₀ˣ e^(t²)dt) · e^(x²)]/e^(2x²) = lim_{x→∞} 2(∫₀ˣ e^(t²)dt)/e^(x²)</p><p><strong>Step 4:</strong> Apply L'Hôpital's rule again (still ∞/∞):<br/>L = lim_{x→∞} 2e^(x²)/(2xe^(x²)) = lim_{x→∞} 1/x = 0</p><p><strong>Step 5:</strong> Alternatively, use the asymptotic behavior: ∫₀ˣ e^(t²)dt ~ e^(x²)/(2x) as x → ∞, so the numerator behaves like e^(2x²)/(4x²) while denominator is e^(2x²), giving L = lim 1/(4x²) = 0.</p><p>∴ Answer: <strong>D (L = 0)</strong></p>
Correct Answer: D