Discussion
The value of \(\lim_{x \to 0} \frac{x\cos x - \sin x}{x^{2}\sin x}\) is:
Answer: A
The limit is \(-\frac{1}{3}\).
Expand numerator and denominator in powers of \(x\) near 0, using \(\cos x = 1 - x^{2}/2 + \cdots\) and \(\sin x = x - x^{3}/6 + \cdots\):
\[ \begin{aligned} x\cos x - \sin x &= x - \tfrac{x^{3}}{2} - x + \tfrac{x^{3}}{6} + \cdots \\ &= -\tfrac{x^{3}}{3} + \cdots \\ x^{2}\sin x &= x^{3} + \cdots \end{aligned} \]
The ratio therefore tends to \(-\frac{1}{3}\). The same result follows from L'Hopital's rule applied repeatedly to the 0/0 form.
\(\frac{1}{3}\) drops the sign: the numerator is negative for small positive \(x\), because \(x\cos x \lt \sin x\) there. 3 and \(-3\) invert the ratio.
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