Allied & ICAR Exam Bank • ALLIED_2638 NAT • 2 Marks

Section 1: Engineering Mathematics

Calculus

Subtopic: Limit, continuity and differentiability

Question Statement:

Evaluating $\lim_{x \to 0} \frac{e^x - e^{-x} - 2x}{x^3}$ using L'Hôpital's rule or series expansion gives $\frac{1}{k}$. The integer $k$ is:

Answer Key:

3

Step-by-Step Solution & Explanation

Expanding: $e^x - e^{-x} = 2 [x + \frac{x^3}{6} + \dots] = 2x + \frac{x^3}{3} + \dots$ Subtracting $2x$ leaves $\frac{x^3}{3} + \dots$ Dividing by $x^3$ yields $1/3$. Thus $k = 3$.
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