Limits, Continuity & Differentiability
L'Hopital and Differentiability
Grade 12
<p>If <em>f</em> is a differentiable function and <br>\[\lim_{x \to 1} \frac{f(1+x^3-x)-f(x)}{\sin(x-1)}\] exists finitely, find \(f'(1)\).</p>
Step-by-Step Solution
Key Concept: Since the limit exists finitely despite sin(x-1)→0 as x→1, the numerator must also approach 0, forcing f(1)=f(1), and applying L'Hôpital's rule or Taylor expansion reveals that f'(1) must equal a specific value determined by the coefficients of the numerator expansion.
<p><strong>Step 1:</strong> Substitute x→1. As x→1: sin(x-1)→0, so the numerator must also →0 for a finite limit to exist.</p><p><strong>Step 2:</strong> Let u = x-1, so x = 1+u and x³-x = (1+u)³-(1+u) = 1+3u+3u²+u³-1-u = 2u+3u²+u³.</p><p><strong>Step 3:</strong> Thus f(1+x³-x) = f(1+2u+3u²+u³). Using Taylor expansion: f(1+2u+3u²+u³) = f(1) + f'(1)(2u+3u²+u³) + O(u²).</p><p><strong>Step 4:</strong> Similarly, f(x) = f(1+u) = f(1) + f'(1)·u + O(u²).</p><p><strong>Step 5:</strong> The numerator becomes: f(1) + f'(1)(2u+3u²+u³) - f(1) - f'(1)·u + O(u²) = f'(1)·u + O(u²) = u[f'(1) + O(u)].</p><p><strong>Step 6:</strong> For the limit \(\lim_{u \to 0} \frac{u[f'(1) + O(u)]}{\sin u}\) to exist finitely, we need \(\lim_{u \to 0} \frac{u}{\sin u} · f'(1) = 1 · f'(1)\) to be finite.</p><p><strong>Step 7:</strong> Since \(\lim_{u \to 0} \frac{u}{\sin u} = 1\), the limit equals f'(1).</p><p>∴ <strong>Answer: f'(1) = 1</strong> (Option B)</p>
Correct Answer: B