(2·COS(x)^2 + COS(x) - 3)/SIN(x)^2
COS(x) = Χ
SIN(x) = Υ
Χ^2 + Υ^2 = 1
(2·Χ^2 + Χ - 3)/(1 - Χ^2)= (Χ - 1)·(2·Χ + 3)/((Χ + 1)·(1 - Χ))=
=- (2·Χ + 3)/(Χ + 1)
per x--->0 : X--->1
LIM(- (2·Χ + 3)/(Χ + 1)) = -5/2
Χ--> 1
$ \displaystyle\lim_{x \to 0} \frac{2cos^2 x+cosx-3}{sin^2x} = $
$ = \displaystyle\lim_{x \to 0} \frac{2cos^2 x+cosx-3 cos^2 x -3 sin^2 x}{sin^2x} = $
$ = \displaystyle\lim_{x \to 0} \frac{cosx - cos^2 x -3 sin^2 x}{sin^2x} = $
$ = \displaystyle\lim_{x \to 0} \frac{cosx(1 - cos x)}{sin^2x} -3 = $
$ = \displaystyle\lim_{x \to 0} \frac{cosx(1 - cos x)}{x^2}\frac{x^2}{sin^2 x}-3 = 1 \cdot \frac{1}{2} \cdot 1 - 3 = -\frac{5}{2} $