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 \(\lim_{x \to 0} x g\!\left(\dfrac{1}{x}\right)\).</p>
Step-by-Step Solution
Key Concept: Since f is periodic with period 3 and g'(x) = f(x), we can express g(x) as an integral of the periodic function. The key is to use the periodicity of f and the definition of the limit to relate g(1/x) to integrals over complete periods.
<p><strong>Step 1:</strong> Since g'(x) = f(x) and g(0) = 0, we have:</p><p>g(x) = ∫₀ˣ f(t)dt</p><p><strong>Step 2:</strong> We need to find lim(x→0) x·g(1/x). As x → 0, we have 1/x → ∞, so:</p><p>lim(x→0) x·g(1/x) = lim(x→0) x·∫₀^(1/x) f(t)dt</p><p><strong>Step 3:</strong> Let u = 1/x, so x = 1/u and as x → 0⁺, u → ∞. The limit becomes:</p><p>lim(u→∞) (1/u)·∫₀ᵘ f(t)dt = lim(u→∞) (1/u)·∫₀ᵘ f(t)dt</p><p><strong>Step 4:</strong> This is the Cesàro mean of the integral. For a periodic function with period 3:</p><p>Since f is periodic with period 3, we can write: ∫₀ᵘ f(t)dt = ⌊u/3⌋·∫₀³ f(t)dt + ∫₀^(u mod 3) f(t)dt</p><p><strong>Step 5:</strong> As u → ∞, let u = 3n + r where n → ∞ and 0 ≤ r < 3:</p><p>∫₀ᵘ f(t)dt = n·∫₀³ f(t)dt + ∫₀ʳ f(t)dt = n·6 + ∫₀ʳ f(t)dt</p><p><strong>Step 6:</strong> Therefore:</p><p>lim(u→∞) (1/u)·∫₀ᵘ f(t)dt = lim(u→∞) (1/u)[n·6 + ∫₀ʳ f(t)dt]</p><p>Since u = 3n + r, we have n = (u-r)/3, and:</p><p>= lim(u→∞) (1/u)[(u-r)·6/3 + ∫₀ʳ f(t)dt] = lim(u→∞) (1/u)[2(u-r) + bounded term]</p><p>= lim(u→∞) [2 - 2r/u + O(1/u)] = 2</p><p><strong>∴ Answer:</strong> 2</p>
Correct Answer: 2