Limits3

Hello and peace be upon you,

By using l'hopital's rule,

Differentiate both numerator and denominator twice, you will get ,

lim (x--> 0) [2- 4( sin x)^2 ]/ 2 = (2 - 0 )/2 = 2/2 =1

thanks....

$\\ \lim_{x \to 0} \frac{\sin^2 x}{x^2} = \\\\\\ \lim_{x \to 0} \left (\frac{\sin x}{x} \right )^2$

Do limite fundamental trigonométrico,

$\\ (1)^2 = \\\\ \boxed{\boxed{1}}$