Discussion
\(\displaystyle\lim_{x \to 0}\left(\frac{1-\cos x}{x^{2}}\right)\) is
Answer: B
The limit is 1/2.
Both numerator and denominator tend to 0, so apply L'Hospital's rule twice:
\[ \begin{aligned} \lim_{x \to 0}\frac{1-\cos x}{x^{2}} &= \lim_{x \to 0}\frac{\sin x}{2x} \\ &= \lim_{x \to 0}\frac{\cos x}{2} = \frac{1}{2} \end{aligned} \]
The same result follows from the series \(\cos x = 1 - x^{2}/2 + \dots\), which makes the numerator \(x^{2}/2\) for small x.
1 is the value of \(\lim \sin x / x\), a different standard limit; 1/4 and 2 come from mishandling the factor of 2.
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