Exercise · ZW-SPECIAL-TRIG-LIMITS-011 · ★★★ Advanced

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

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  • C. \(2\)

Explanation

Write \(\dfrac{\sin x^{2}}{1-\cos x}=\dfrac{\sin x^{2}}{x^{2}}\cdot\dfrac{x^{2}}{1-\cos x}\). The first factor tends to 1 and, since \(\dfrac{1-\cos x}{x^{2}}\to\dfrac{1}{2}\), the second tends to 2. The limit is 2.

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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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