Limits, Continuity & Differentiability
Limit as Derivative
Grade 12
Question:
<p>If $f(x) = \cot^{-1}\left(\frac{3x - x^3}{1 - 3x^2}\right)$ and $g(x) = \cos^{-1}\left(\frac{1-x^2}{1+x^2}\right)$, then $$\lim_{x \to a} \frac{f(x) - f(a)}{g(x) - g(a)}, \quad 0 < a < \frac{1}{2}$$ is:</p>
<p>(a) $\frac{3}{2(1+a^2)}$</p>
<p>(b) $\frac{3}{2}$</p>
<p>(c) $-\frac{3}{2(1+a^2)}$</p>
<p>(d) $-\frac{3}{2}$</p>
Step-by-Step Solution
Key Concept: Recognize that f(x) and g(x) can be simplified using inverse trigonometric identities, then apply L'Hôpital's rule or compute derivatives directly to evaluate the limit of the form f(x)-f(a)/g(x)-g(a).
<p><strong>Step 1: Simplify f(x) using the triple angle formula for cotangent.</strong></p><p>Note that cot(3θ) = (3cot θ - cot³θ)/(1 - 3cot²θ). If we set cot θ = x, then:</p><p>cot(3θ) = (3x - x³)/(1 - 3x²)</p><p>Therefore, cot⁻¹((3x - x³)/(1 - 3x²)) = 3cot⁻¹(x) (for appropriate domain)</p><p>So <strong>f(x) = 3cot⁻¹(x)</strong></p><p><strong>Step 2: Simplify g(x) using the double angle formula for cosine.</strong></p><p>Note that cos(2φ) = (1 - tan²φ)/(1 + tan²φ). If we set tan φ = x, then:</p><p>cos(2φ) = (1 - x²)/(1 + x²)</p><p>Therefore, cos⁻¹((1 - x²)/(1 + x²)) = 2tan⁻¹(x) (for appropriate domain)</p><p>So <strong>g(x) = 2tan⁻¹(x)</strong></p><p><strong>Step 3: Compute the derivatives.</strong></p><p>f'(x) = 3 · d/dx[cot⁻¹(x)] = 3 · (-1/(1 + x²)) = -3/(1 + x²)</p><p>g'(x) = 2 · d/dx[tan⁻¹(x)] = 2 · 1/(1 + x²) = 2/(1 + x²)</p><p><strong>Step 4: Apply L'Hôpital's rule.</strong></p><p>Since both f(x) - f(a) → 0 and g(x) - g(a) → 0 as x → a:</p><p>$$\lim_{x \to a} \frac{f(x) - f(a)}{g(x) - g(a)} = \frac{f'(a)}{g'(a)} = \frac{-3/(1 + a²)}{2/(1 + a²)} = \frac{-3}{2}$$</p><p><strong>∴ Answer: B</strong></p>
Correct Answer: B