Limits, Continuity & Differentiability
Limits Using Exponential Expansions
Grade 12
<p>$\lim_{x \to 0} \frac{2e^{\sin x} - e^{-\sin x} - 1}{x^2 + 2x}$ equals:</p>
Step-by-Step Solution
Key Concept: When direct substitution gives 0/0, use Taylor series expansions of e^(sin x) and e^(-sin x) around x=0, keeping terms up to x² since the denominator is O(x²).
<p><strong>Step 1: Check the form</strong><br>At x → 0: numerator = 2e⁰ - e⁰ - 1 = 2 - 1 - 1 = 0 and denominator = 0. This is 0/0, so we can use Taylor series.</p><p><strong>Step 2: Expand sin x</strong><br>sin x = x - x³/6 + O(x⁵)</p><p><strong>Step 3: Expand e^(sin x) using e^u = 1 + u + u²/2 + u³/6 + ...</strong><br>e^(sin x) = 1 + sin x + (sin x)²/2 + (sin x)³/6 + ...<br>= 1 + (x - x³/6) + x²/2 + x³/6 + O(x⁴)<br>= 1 + x + x²/2 + O(x³)</p><p><strong>Step 4: Expand e^(-sin x)</strong><br>e^(-sin x) = 1 - sin x + (sin x)²/2 - (sin x)³/6 + ...<br>= 1 - (x - x³/6) + x²/2 - x³/6 + O(x⁴)<br>= 1 - x + x²/2 + O(x³)</p><p><strong>Step 5: Calculate the numerator</strong><br>2e^(sin x) - e^(-sin x) - 1<br>= 2(1 + x + x²/2) - (1 - x + x²/2) - 1 + O(x³)<br>= 2 + 2x + x² - 1 + x - x²/2 - 1 + O(x³)<br>= 3x + x²/2 + O(x³)</p><p><strong>Step 6: Divide by the denominator</strong><br>lim(x→0) [3x + x²/2 + O(x³)] / [x² + 2x]<br>= lim(x→0) [3x + x²/2] / [x(x + 2)]<br>= lim(x→0) [x(3 + x/2)] / [x(x + 2)]<br>= lim(x→0) (3 + x/2) / (x + 2)</p><p><strong>Step 7: Substitute x = 0</strong><br>= 3/2</p><p><strong>∴ Answer: a</strong></p>
Correct Answer: a