Limits, Continuity & Differentiability
Limits involving functional equations
Grade 12
Question:
<p>Let <i>f</i> be a real valued derivable function such that <i>f</i>(<i>x</i>)<i>f</i>(<i>y</i>) = <i>f</i>(<i>x</i>)<i>y</i> + <i>x</i><i>f</i>(<i>y</i>), ∀<i>x</i>, <i>y</i> ∈ ℝ. If <i>f</i>′(0) = 2, then find \(\lim_{x \to 0} \left[\dfrac{f(x)}{\sin x}\right]\). [Note: [ ] represents greatest integer function.]</p>
Step-by-Step Solution
Key Concept: Find the function f(x) using the functional equation by substituting specific values, then use f'(0) = 2 to determine constants. Finally, evaluate the limit of f(x)/sin x as x→0.
<p><strong>Step 1: Find f(0) using the functional equation.</strong></p><p>Given: f(x)f(y) = f(x)y + xf(y) for all x, y ∈ ℝ</p><p>Substitute x = y = 0: f(0)f(0) = f(0)·0 + 0·f(0) = 0</p><p>Therefore: f(0)² = 0, so f(0) = 0</p><p><strong>Step 2: Differentiate the functional equation with respect to y.</strong></p><p>f(x)f(y) = f(x)y + xf(y)</p><p>Differentiating both sides with respect to y:</p><p>f(x)f'(y) = f(x) + xf'(y)</p><p><strong>Step 3: Substitute y = 0.</strong></p><p>f(x)f'(0) = f(x) + xf'(0)</p><p>Since f'(0) = 2:</p><p>f(x)·2 = f(x) + x·2</p><p>2f(x) = f(x) + 2x</p><p>f(x) = 2x</p><p><strong>Step 4: Verify the solution.</strong></p><p>If f(x) = 2x, then f'(x) = 2, so f'(0) = 2 ✓</p><p>Check functional equation: f(x)f(y) = 2x·2y = 4xy</p><p>f(x)y + xf(y) = 2x·y + x·2y = 2xy + 2xy = 4xy ✓</p><p><strong>Step 5: Evaluate the limit.</strong></p><p>$$\lim_{x \to 0} \frac{f(x)}{\sin x} = \lim_{x \to 0} \frac{2x}{\sin x}$$</p><p>Using the standard limit $\lim_{x \to 0} \frac{\sin x}{x} = 1$:</p><p>$$\lim_{x \to 0} \frac{2x}{\sin x} = 2 \cdot \lim_{x \to 0} \frac{x}{\sin x} = 2 \cdot 1 = 2$$</p><p><strong>Step 6: Apply the greatest integer function.</strong></p><p>$$\left[\lim_{x \to 0} \frac{f(x)}{\sin x}\right] = [2] = 2$$</p><p><strong>∴ Answer: 2</strong></p>
Correct Answer: 2