Determine the value of the limit as x approaches 0 of (e^x - 1 - x) / x^2.

A. 1/2 — Using the Taylor series expansion for e^x = 1 + x + x^2/2! + x^3/3! + ..., the numerator becomes x^2/2 + x^3/6 + ... Dividing by x^2 and taking the limit as x approaches 0 gives 1/2.

JEE Main · MathematicsMedium

$lim_{x \to 0} \frac{e ^{x} - 1 - x}{x ^{2}}$

A. $1/2$
B. $1$
C. $0$
D. $2$

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Correct answer: A — $1/2$

Solution

\frac{( 1 + x + \frac{x ^{2}}{2} + \frac{x ^{3}}{6} + \dots ) - 1 - x}{x ^{2}} = \frac{\frac{x ^{2}}{2} + \frac{x ^{3}}{6} + \dots}{x ^{2}} = \frac{1}{2} + \frac{x}{6} + \dots \to \frac{1}{2} .

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$lim_{x \to 0} \frac{e ^{x} - 1 - x}{x ^{2}}$