Find the limit of (e^(3x) − 1 − 3x)/x² as x → 0. Give the answer as a fraction in lowest terms.

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An exponential minus its tangent

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Find the limit of $\frac{e ^{3 x} - 1 - 3 x}{x ^{2}}$ as $x \to 0$ . Give the answer as a fraction in lowest terms.

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Hint

Expand $e^{3 x}$ to second order, or apply l'Hôpital's rule more than once.

Solution

The Taylor series gives $e^{3 x} = 1 + 3 x + \frac{9 x ^{2}}{2} + \dots$ .
So the numerator is $\frac{9 x ^{2}}{2}$ plus terms of order $x^{3}$ and higher.
Dividing by $x^{2}$ and letting $x \to 0$ leaves $\frac{9}{2}$ .

Answer: $\frac{9}{2}$