Limits, Continuity & Differentiability
Limits
Grade 12
Question:
<p>\(\lim_{x \to 0} \frac{2x^2 - \log(1+x)}{x^2}\) is equal to</p>
<p>(a) \(\frac{3}{2}\)</p>
<p>(b) \(\frac{2}{3}\)</p>
<p>(c) 0</p>
<p>(d) 1</p>
<p>(e) none of these</p>
Step-by-Step Solution
Key Concept: Use Taylor series expansion of log(1+x) = x - x²/2 + x³/3 - ... to convert the logarithmic expression into polynomial form, allowing direct limit evaluation.
<p><strong>Step 1:</strong> Recognize the indeterminate form 0/0 as x → 0.</p><p><strong>Step 2:</strong> Apply Taylor series expansion: log(1+x) = x - x²/2 + x³/3 - x⁴/4 + ...</p><p><strong>Step 3:</strong> Substitute into the numerator:</p><p>2x² - log(1+x) = 2x² - (x - x²/2 + x³/3 - ...) = 2x² - x + x²/2 - x³/3 + ...</p><p><strong>Step 4:</strong> Rearrange: 2x² - log(1+x) = -x + (2 + 1/2)x² - x³/3 + ... = -x + (5/2)x² - x³/3 + ...</p><p><strong>Step 5:</strong> Divide by x²:</p><p>$$\frac{2x^2 - \log(1+x)}{x^2} = \frac{-x + (5/2)x^2 - x^3/3 + ...}{x^2} = \frac{-x}{x^2} + \frac{5}{2} - \frac{x}{3} + ...$$</p><p><strong>Step 6:</strong> As x → 0, the terms with x in the denominator vanish:</p><p>∴ Answer: <strong>5/2</strong></p>
Correct Answer: A