Find lim _(xarrow (pi )/(2))(3cos^2(x))/(2-2sin(x)) Choose 1 answer: A ) 3 B (3)/(2) (3)/(4) The limit doesn't exist

To find the limit of the given expression as x approaches π/2, we can use the fact that the limit of a rational function is the ratio of the limits of the numerator and denominator, provided that the limit of the denominator is not zero.<br /><br />In this case, the numerator is 3cos^2(x) and the denominator is 2-2sin(x). As x approaches π/2, cos(x) approaches 0 and sin(x) approaches 1. Therefore, the numerator approaches 3(0)^2 = 0 and the denominator approaches 2-2(1) = 0.<br /><br />Since the limit of the denominator is zero, we cannot directly conclude that the limit of the entire expression is zero. Instead, we can use L'Hopital's rule, which states that if the limit of a rational function is an indeterminate form (such as 0/0 or ∞/∞), then the limit of the derivative of the numerator divided by the derivative of the denominator is the same as the original limit.<br /><br />Taking the derivatives of the numerator and denominator, we get:<br /><br />d/dx [3cos^2(x)] = 6cos(x)(-sin(x))<br />d/dx [2-2sin(x)] = -2cos(x)<br /><br />Now, we can evaluate the limit of the ratio of these derivatives as x approaches π/2:<br /><br />lim(x→π/2) [6cos(x)(-sin(x)) / -2cos(x)] = lim(x→π/2) [3sin(x)] = 3<br /><br />Therefore, the correct answer is A) 3.