Exercise · ZW-SPECIAL-TRIG-LIMITS-010 · ★★☆ Intermediate

Compute \(\lim_{x\to 0}\,\frac{\tan {2x}}{\sin {5x}}\).

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  • D. \(\frac{2}{5}\)

Explanation

Since \(\tan 2x=\dfrac{\sin 2x}{\cos 2x}\), the quotient is \(\dfrac{\sin 2x}{\sin 5x}\cdot\dfrac{1}{\cos 2x}\to\dfrac{2}{5}\cdot1\cdot1=\dfrac{2}{5}\).

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Review the concept: Special trigonometric limits

Two basic limits in radians: \(\frac{\sin x}{x}\) tends to \(1\) and \(\frac{1-\cos x}{x}\) tends to \(0\) as \(x\) tends to \(0\).

\[\begin{gathered}\lim_{x\to0}\frac{\sin x}{x}=1\\ \lim_{x\to0}\frac{1-\cos x}{x}=0\end{gathered}\]

More practice problems: special trigonometric limits

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