Discussion
The value of \( \lim_{x \to 0} \dfrac{1 - \cos x}{x^{2}} \) is
Answer: D
Use the half-angle identity \( 1 - \cos x = 2\sin^{2}(x/2) \):
\[ \frac{1-\cos x}{x^{2}} \;=\; \frac{2\sin^{2}(x/2)}{x^{2}} \;=\; \frac{1}{2}\left(\frac{\sin (x/2)}{x/2}\right)^{2}. \]
The bracket tends to one, so the limit is \( 1/2 \). Two applications of L'Hopital's rule give the same value, the numerator differentiating to \( \sin x \) and then to \( \cos x \)
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