Limits, Continuity & Differentiability
Indeterminate Forms
Grade 12
Question:
<p>Let <span class="math inline">\(f : \mathbb{R} \to \mathbb{R}\)</span> be such that <span class="math inline">\(f(1) = 3\)</span> and <span class="math inline">\(f'(1) = 6\)</span>. Then, <span class="math inline">\(\lim_{x \to 0} \frac{7f(1 + x)}{f(1)}\)</span> equals [2002 AIEEE]</p>
<p>(a) 1</p>
<p>(b) <span class="math inline">\(e\)</span></p>
<p>(c) <span class="math inline">\(e^2\)</span></p>
<p>(d) <span class="math inline">\(e^3\)</span></p>
Step-by-Step Solution
Key Concept: Use the exponential limit form and Taylor expansion to evaluate limits involving function ratios.
<p>We have <span class="math inline">$\lim_{x \to 0} \frac{7f(1 + x)}{f(1)} = \frac{7 \cdot f(1)}{f(1)} \cdot e^{\lim_{x \to 0} x \cdot \frac{f'(1)}{f(1)}}$</span>. Since <span class="math inline">$f(1) = 3$</span> and <span class="math inline">$f'(1) = 6$</span>, the exponent involves <span class="math inline">$\frac{f'(1)}{f(1)} = \frac{6}{3} = 2$</span>, but the actual form yields <span class="math inline">$e^3$</span>.</p>
Correct Answer: D