Limits, Continuity & Differentiability
Continuity
Grade 12
Question:
<p>The function <span>\( f : R \sim \{0\} \to R \)</span> given by <span>\[ f(x) = \frac{1}{x} - \frac{2}{e^{2x} - 1} \]</span> can be made continuous at <span>\( x = 0 \)</span> by defining <span>\( f(0) \)</span> as</p>
<p>2</p>
<p>\(-1\)</p>
<p>0</p>
<p>1</p>
Step-by-Step Solution
Key Concept: Use Taylor expansion of exponential function around x=0 to find the limit of f(x) as x→0, then assign f(0) equal to this limit value to make the function continuous.
<p><strong>Step 1:</strong> To make f continuous at x=0, we need f(0) = lim(x→0) f(x)</p><p><strong>Step 2:</strong> Expand e^(2x) using Taylor series: e^(2x) = 1 + 2x + (2x)²/2! + (2x)³/3! + ... = 1 + 2x + 2x² + (4x³/3) + ...</p><p><strong>Step 3:</strong> Therefore e^(2x) - 1 = 2x + 2x² + (4x³/3) + ... = 2x(1 + x + (2x²/3) + ...)</p><p><strong>Step 4:</strong> So 2/(e^(2x) - 1) = 2/[2x(1 + x + (2x²/3) + ...)] = 1/[x(1 + x + (2x²/3) + ...)]</p><p><strong>Step 5:</strong> Expanding 1/(1 + x + (2x²/3) + ...) ≈ 1 - x - (2x²/3) + ... using geometric series</p><p><strong>Step 6:</strong> Thus 2/(e^(2x) - 1) = (1/x)[1 - x - (2x²/3) + ...] = 1/x - 1 - (2x/3) + ...</p><p><strong>Step 7:</strong> Therefore f(x) = 1/x - [1/x - 1 - (2x/3) + ...] = 1 + (2x/3) + O(x²)</p><p><strong>Step 8:</strong> Taking limit: lim(x→0) f(x) = 1</p><p>∴ Answer: D (f(0) = 1)</p>
Correct Answer: D