Question Bank • QB_EM_ADV_026 NAT • 2 Marks

Section 1: Engineering Mathematics

Calculus

Subtopic: Limit, continuity and differentiability

Question Statement:

Evaluate the limit: $\lim_{x \to 0} \frac{e^x - 1 - x - \frac{x^2}{2}}{x^3}$ The value of the limit is ________ (round off to three decimal places).

Answer Key:

0.167

Step-by-Step Solution & Explanation

**Method 1: Maclaurin Series Expansion** Expanding $e^x$ around $x = 0$: $e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \dots$ $e^x - 1 - x - \frac{x^2}{2} = \frac{x^3}{6} + \frac{x^4}{24} + \dots$ Dividing by $x^3$: $\lim_{x \to 0} \left( \frac{1}{6} + \frac{x}{24} + \dots \right) = \frac{1}{6} \approx 0.1667$ **Method 2: L'Hôpital's Rule (3 Successive Applications)** $\lim_{x \to 0} \frac{e^x - 1 - x - x^2/2}{x^3} = \lim_{x \to 0} \frac{e^x - 1 - x}{3x^2} = \lim_{x \to 0} \frac{e^x - 1}{6x} = \lim_{x \to 0} \frac{e^x}{6} = \frac{1}{6} \approx 0.167$
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