Limits, Continuity & Differentiability
Limits of the form 1^infinity
Grade 12
Question:
<p><strong>Paragraph for Question nos. 599 and 600</strong><br>Let \(f(x)\) be a polynomial of degree 3 such that \(f(0)=1\), \(f(1)=2\) and zero is a critical point of \(f(x)\) having no local extreme.</p><p>The value of \(\displaystyle\lim_{x \to 0} (f(x))^{\frac{1}{\tan x - x}}\) is equal to:</p>
<p>(a) \(e^2\)</p>
<p>(b) \(e^{-2}\)</p>
<p>(c) \(e^3\)</p>
<p>(d) \(e^{-3}\)</p>
Step-by-Step Solution
Key Concept: Since 0 is a critical point with no local extreme, f'(0)=0 and f''(0)=0 (inflection point). Use Taylor expansion around x=0 and L'Hôpital's rule to evaluate the exponential limit form.
<p><strong>Step 1:</strong> Since f is degree 3 with f(0)=1, f(1)=2, f'(0)=0 (critical point), and f''(0)=0 (no local extreme), write:</p><p>f(x) = 1 + 0·x + 0·x²/2 + ax³ = 1 + ax³</p><p><strong>Step 2:</strong> From f(1)=2: 1 + a = 2, so a=1. Thus f(x) = 1 + x³</p><p><strong>Step 3:</strong> Evaluate the limit form. Let L = lim[x→0] (1 + x³)^(1/(tan x - x))</p><p>Taking ln: ln L = lim[x→0] ln(1 + x³)/(tan x - x)</p><p><strong>Step 4:</strong> Use Taylor expansions: ln(1 + x³) ≈ x³ and tan x - x = x³/3 + 2x⁵/15 + ... ≈ x³/3</p><p><strong>Step 5:</strong> ln L = lim[x→0] x³/(x³/3) = 3</p><p><strong>Step 6:</strong> Therefore L = e³</p><p>∴ Answer: A (where option A is e³)</p>
Correct Answer: A