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Q. $ \displaystyle\lim_{x \to 0} \frac{1 - \cos \, 2x}{x^2} $ is

Limits and Derivatives

Solution:

$ \displaystyle\lim_{x \to 0} \frac{1 - \cos \, 2x}{x^2} = \displaystyle\lim_{x \to 0} \frac{2 \, \sin^2 \, x}{x^2} $
$ = 2 \displaystyle\lim_{x \to 0} \left(\frac{\sin \, x}{x} . \frac{\sin \, x}{x} \right)$
$ = 2 \displaystyle\lim_{x \to 0} \frac{\sin \, x}{x} . \displaystyle\lim_{x \to 0} \frac{\sin \, x}{x} = 2 (1) (1) = 2$