Limits, Continuity & Differentiability
L'Hôpital's Rule
Grade 12
Question:
<p>The graph of function \(y = f(x)\) has a unique tangent at \((e^a, 0)\) through which the graph passes. Then \(\lim_{x \to e^a} \frac{\log(1 + 7f(x)) - \sin(f(x))}{3f(x)}\) equals</p>
<p>(a) \(\frac{4}{3}\)</p>
<p>(b) 2</p>
<p>(c) \(\frac{7}{3}\)</p>
<p>(d) None of these</p>
Step-by-Step Solution
Key Concept: Use L'Hôpital's rule for 0/0 form and simplify using properties of logarithm and trigonometric functions.
<p><strong>Solution:</strong> Using L'Hôpital's rule:</p><p>$\lim_{x \to e^a} \frac{\log(1 + 7f(x)) - \sin(f(x))}{3f(x)}$</p><p>$= \lim_{x \to e^a} \frac{7f'(x)\{1 + 7f(x)\}^{-1} - \cos(f(x)) \cdot f'(x)}{3f'(x)}$</p><p>$= \lim_{x \to e^a} \frac{7 - \cos(f(x))\{1 + 7f(x)\}}{3\{1 + 7f(x)\}}$</p><p>$= \frac{7 - 1}{3} = 2$</p><p>∴ Answer is (b) 2</p>
Correct Answer: B