Limits, Continuity & Differentiability
Limits involving periodic functions
Grade 12
Question:
<p>Let \(f(x)\) be a continuous, periodic and bounded function with period 3 such that \(\int_0^3 f(t)\,dt = 6\). Also \(g'(x) = f(x)\), such that \(g(0) = 0\). Find the value of \(\lim_{x \to 0} xg\!\left(\dfrac{1}{x}\right)\).</p>
Step-by-Step Solution
Key Concept: Since f is periodic with period 3 and ∫₀³ f(t)dt = 6, we have ∫₀ⁿ f(t)dt = 2n for any integer n. Use this with g'(x) = f(x) to find g(1/x), then apply L'Hôpital's rule to the limit xg(1/x) as x → 0.
<p><strong>Step 1:</strong> Use periodicity of f. Since f has period 3 and ∫₀³ f(t)dt = 6, for any positive integer n:</p><p>∫₀³ⁿ f(t)dt = n∫₀³ f(t)dt = 6n</p><p><strong>Step 2:</strong> Express g(1/x). Since g'(x) = f(x) and g(0) = 0:</p><p>g(1/x) = ∫₀^(1/x) f(t)dt</p><p>For x → 0⁺, let 1/x = 3m + r where m is an integer and 0 ≤ r < 3. Then:</p><p>g(1/x) = ∫₀^(3m+r) f(t)dt = 6m + ∫₀ʳ f(t)dt</p><p><strong>Step 3:</strong> Analyze the limit. As x → 0⁺, we have 1/x → ∞, so m → ∞ and r ∈ [0,3).</p><p>xg(1/x) = x[6m + ∫₀ʳ f(t)dt]</p><p>Since 1/x = 3m + r, we have x = 1/(3m + r), so:</p><p>xg(1/x) = [6m + ∫₀ʳ f(t)dt]/(3m + r)</p><p><strong>Step 4:</strong> Take the limit as m → ∞:</p><p>lim(m→∞) [6m + ∫₀ʳ f(t)dt]/(3m + r) = lim(m→∞) 6m/(3m) = 6/3 = 2</p><p>∴ Answer: <strong>2</strong></p>
Correct Answer: 2