Exercise · ZW-LHOPITALS-RULE-004 · ★☆☆ Basic

Compute \(\lim_{x\to 0}\,\frac{e^{2 x} - 1}{3 x}\).

Choose one option.

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  • D. \(\frac{2}{3}\)

Explanation

Substituting gives the indeterminate form 0/0, so L'Hôpital's rule applies by differentiating numerator and denominator: \(\frac{2 e^{2 x}}{3}\). The resulting limit is \(\frac{2}{3}\).

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Review the concept: L'Hôpital's rule

Rule that computes limits of quotients in indeterminate form \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\) by differentiating numerator and denominator separately.

\[\lim_{x\to a}\frac{f(x)}{g(x)}=\lim_{x\to a}\frac{f'(x)}{g'(x)}\]

More practice problems: l'hôpital's rule

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