alleryEvaluate $\displaystyle\lim_{x \to 0} \dfrac{1 - \cos 2x}{x^2}$.
A$2$
B$4$
C$0$
D$1$
Answer & Solution
Correct answer: A. $2$
Use $1 - \cos 2x = 2\sin^2 x$. Then $\dfrac{2\sin^2 x}{x^2} = 2\left(\dfrac{\sin x}{x}\right)^2 \to 2 \cdot 1 = 2$ as $x \to 0$.
Alternatively, expand $\cos 2x \approx 1 - 2x^2$ to leading order, giving $\dfrac{2x^2}{x^2} = 2$.
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