Evaluate $\displaystyle\lim_{x o 0} \dfrac{1 - \cos 2x}{x^2}$.

Question

Evaluate $\displaystyle\lim_{x 	o 0} \dfrac{1 - \cos 2x}{x^2}$.

Accepted Answer

A. $2$

Use $1 - \cos 2x = 2\sin^2 x$. Then $\dfrac{2\sin^2 x}{x^2} = 2\left(\dfrac{\sin x}{x}ight)^2 	o 2 \cdot 1 = 2$ as $x 	o 0$.
Alternatively, expand $\cos 2x \approx 1 - 2x^2$ to leading order, giving $\dfrac{2x^2}{x^2} = 2$.

Answer

$4$

Answer

$0$

Answer

$1$

Evaluate $\displaystyle\lim_{x \to 0} \dfrac{1 - \cos 2x}{x^2}$.