$\lim_{x \to 0} \frac{1-cos2x}{x^2}$ ko dùng lhopital
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$\displaystyle\lim_{x \to 0} \dfrac{1-\cos 2x}{x^2}\\ =\displaystyle\lim_{x \to 0} \dfrac{1-(1-2\sin^2x)}{x^2}\\ =\displaystyle\lim_{x \to 0} \dfrac{2\sin^2x}{x^2}\\ =2\displaystyle\lim_{x \to 0} \left(\dfrac{\sin x}{x }\right)^2\\ =2$
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