Limits, Continuity & Differentiability
Limits of the form 1^∞ using L'Hospital's Rule
Grade 12
Question:
<p>Given \(f'(3) + f'(2) = 0\). Let
\[y = \lim_{x \to 0} \left[\frac{1 + f(3+x) - f(3)}{1 + f(2-x) - f(2)}\right]^{1/x}\]
Find the value of \(y\).</p>
<p>\(1\)</p>
<p>\(e\)</p>
<p>\(e^2\)</p>
<p>\(e^{-1}\)</p>
Step-by-Step Solution
Key Concept: Recognize the limit form involves derivatives in the exponent: numerator→f'(3) and denominator→-f'(2) as x→0. Use the standard limit form [1+u]^(1/x) = e^(u/x) when u→0, converting the indeterminate form to e^(f'(3)+f'(2)).
<p><strong>Step 1: Identify the limit form</strong></p><p>As x→0:</p><p>• Numerator: 1 + f(3+x) - f(3) → 1 + f'(3)·x + o(x) ≈ 1 + f'(3)x</p><p>• Denominator: 1 + f(2-x) - f(2) → 1 + f'(2)·(-x) + o(x) ≈ 1 - f'(2)x</p><p>This gives form [1]^∞, requiring logarithmic analysis.</p><p><strong>Step 2: Apply logarithm</strong></p><p>Let L = lim(x→0) [1 + f'(3)x]/[1 - f'(2)x])^(1/x)</p><p>ln L = lim(x→0) (1/x)·ln[(1 + f'(3)x)/(1 - f'(2)x)]</p><p><strong>Step 3: Expand using Taylor series</strong></p><p>ln[(1 + f'(3)x)/(1 - f'(2)x)] = ln(1 + f'(3)x) - ln(1 - f'(2)x)</p><p>= [f'(3)x - (f'(3)x)²/2 + ...] - [-f'(2)x - (f'(2)x)²/2 - ...]</p><p>= [f'(3) + f'(2)]x + O(x²)</p><p><strong>Step 4: Evaluate the limit</strong></p><p>ln L = lim(x→0) (1/x)·[f'(3) + f'(2)]x = f'(3) + f'(2) = 0</p><p><strong>Step 5: Solve for y</strong></p><p>L = e⁰ = 1</p><p>∴ y = <strong>1</strong> (Answer: A)</p>
Correct Answer: A