Exercise · ZW-LHOPITALS-RULE-018 · ★★★ Advanced

Compute \(\displaystyle\lim_{x\to0}\frac{\tan x-x}{x^3}\).

Choose one option.

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  • D. \(\tfrac13\)

Explanation

It is \(0/0\). Differentiating: \(\frac{\sec^2x-1}{3x^2}=\frac{\tan^2x}{3x^2}\to\frac13\), since \(\frac{\tan x}{x}\to1\).

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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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