Limits, Continuity & Differentiability
Differentiability and Derivatives
Grade 12
Question:
<p>Let \(f\) be a differentiable function such that \(\displaystyle\lim_{x \to 1} \frac{f(1+x^3-x)-f(x)}{\sin(x-1)} = \lim_{x \to 0} \frac{f(1-x)-f(1)}{x} + 10\), then \(f'(1)\) is equal to:</p>
<p>(a) 5</p>
<p>(b) 4</p>
<p>(c) 2</p>
<p>(d) 1</p>
Step-by-Step Solution
Key Concept: Recognize that both limits on the right side represent derivative expressions. The left side limit must equal the sum of these derivatives, which involves evaluating f'(1) using substitution and L'Hôpital's rule.
<p><strong>Step 1: Evaluate the right side limits.</strong></p><p>The second limit on the right: $\lim_{x \to 0} \frac{f(1-x)-f(1)}{x}$</p><p>Let $u = -x$, so as $x \to 0$, $u \to 0$:</p><p>$$\lim_{u \to 0} \frac{f(1+u)-f(1)}{-u} = -\lim_{u \to 0} \frac{f(1+u)-f(1)}{u} = -f'(1)$$</p><p><strong>Step 2: Set up the equation.</strong></p><p>The given condition is:</p><p>$$\lim_{x \to 1} \frac{f(1+x^3-x)-f(x)}{\sin(x-1)} = -f'(1) + 10$$</p><p><strong>Step 3: Evaluate the left limit using L'Hôpital's rule.</strong></p><p>As $x \to 1$: numerator $\to f(1+1-1) - f(1) = f(1) - f(1) = 0$ and denominator $\to \sin(0) = 0$.</p><p>This is a $\frac{0}{0}$ form, so apply L'Hôpital's rule:</p><p>$$\lim_{x \to 1} \frac{f(1+x^3-x)-f(x)}{\sin(x-1)} = \lim_{x \to 1} \frac{f'(1+x^3-x) \cdot (3x^2-1) - f'(x)}{\cos(x-1)}$$</p><p><strong>Step 4: Substitute $x = 1$.</strong></p><p>At $x = 1$:</p><p>$$= \frac{f'(1+1-1) \cdot (3-1) - f'(1)}{\cos(0)} = \frac{f'(1) \cdot 2 - f'(1)}{1} = f'(1)$$</p><p><strong>Step 5: Solve for $f'(1)$.</strong></p><p>$$f'(1) = -f'(1) + 10$$</p><p>$$2f'(1) = 10$$</p><p>$$f'(1) = 5$$</p><p><strong>∴ Answer: A</strong></p>
Correct Answer: A