Limits, Continuity & Differentiability
Limits of the form 1^infinity
Grade 12
Question:
<p>If \(\lim_{x \to 0}\left(1 + \int_0^{\sqrt{a^x-1}} (\sin(2\,\text{arc}\tan t))(1+t^2)^{\ln a}\,dt\right)^{\!\frac{1}{x}} = 5\), then find the value of \(a\), where \(a \in N\) and \(a > 1,\; x > 0\).</p>
Step-by-Step Solution
Key Concept: Recognize this as a 1^∞ indeterminate form requiring L'Hôpital's rule. The exponent 1/x → ∞ as x → 0⁺, so convert to exponential form: e^(ln(expression)/x) and apply L'Hôpital's on the exponent.
<p><strong>Step 1: Identify the form and convert</strong></p><p>As x → 0⁺: a^x → 1, so √(a^x - 1) → 0, making the integral → 0. The limit becomes 1^∞.</p><p>Write: L = lim_{x→0⁺} exp[(1/x) ln(1 + I(x))], where I(x) = ∫₀^√(a^x-1) sin(2 arctan t)(1+t²)^(ln a) dt</p><p><strong>Step 2: Simplify the integrand</strong></p><p>Using sin(2 arctan t) = 2t/(1+t²):</p><p>I(x) = ∫₀^√(a^x-1) [2t/(1+t²)](1+t²)^(ln a) dt = ∫₀^√(a^x-1) 2t(1+t²)^(ln a - 1) dt</p><p><strong>Step 3: Apply L'Hôpital's rule</strong></p><p>lim_{x→0⁺} [ln(1 + I(x))]/x = lim_{x→0⁺} [I'(x)]/1 (using L'Hôpital's)</p><p>By Leibniz rule: I'(x) = 2√(a^x-1) · (1 + (a^x-1))^(ln a - 1) · [d/dx√(a^x-1)]</p><p>Since d/dx√(a^x-1) = (a^x ln a)/(2√(a^x-1))</p><p>As x → 0⁺: √(a^x-1) → 0, (1 + (a^x-1))^(ln a - 1) → 1</p><p>I'(x) ≈ 2√(a^x-1) · (a^x ln a)/(2√(a^x-1)) = a^x ln a → ln a</p><p><strong>Step 4: Solve for a</strong></p><p>L = e^(ln a) = a = 5</p><p>∴ Answer: a = 5</p>
Correct Answer: 5