Using L'Hopital's rule, we know that #lim_(x->a)(f(x))/(g(x))=>(f'(a))/(g'(a))#
#lim_(x->-1)(cos(2)-cos(2x))/(x^2-|x|)=>(d/dx[cos(2)-cos(2x)])/(d/dx[x^2-abs(x)])#
#(d/dx[cos(2)-cos(2x)])/(d/dx[x^2-abs(x)])=(d/dx[cos(2)]-d/dx[cos(2x)])/(d/dx[x^2]-d/dx[abs(x)])#
#color(white)((d/dx[cos(2)-cos(2x)])/(d/dx[x^2-abs(x)]))=(0-2(-sin(2x)))/(2x-x/abs(x))#
#color(white)((d/dx[cos(2)-cos(2x)])/(d/dx[x^2-abs(x)]))=(2sin(2x))/(2x-x1/abs(x))#
#lim_(x->-1)(cos(2)-cos(2x))/(x^2-|x|)=>(f'(x))/(g'(x))=(2sin(2x))/(2x-x/abs(x))#
#color(white)(lim_(x->-1)(cos(2)-cos(2x))/(x^2-|x|))=(f'(-1))/(g'(-1))=(2sin(2(-1)))/(2(-1)-(-1)/abs(-1))#
#color(white)(lim_(x->-1)(cos(2)-cos(2x))/(x^2-|x|)=(f'(-1))/(g'(-1)))=(2sin(-2))/(2(-1)-(-1)/abs(-1))#
#sin(-a)=-sin(a)#
#color(white)(lim_(x->-1)(cos(2)-cos(2x))/(x^2-|x|)=(f'(-1))/(g'(-1)))=(-2sin(2))/(-2-(-1)/1)#
#color(white)(lim_(x->-1)(cos(2)-cos(2x))/(x^2-|x|)=(f'(-1))/(g'(-1)))=(-2sin(2))/(-2+1)#
#color(white)(lim_(x->-1)(cos(2)-cos(2x))/(x^2-|x|)=(f'(-1))/(g'(-1)))=(-2sin(2))/(-1)#
#color(white)(lim_(x->-1)(cos(2)-cos(2x))/(x^2-|x|)=(f'(-1))/(g'(-1)))=2sin(2)#
A graph of #color(red)(y=(cos(2)-cos(2x))/(x^2-abs(x)))#, #color(blue)(x=1)# and #color(green)(y=2sin(2))#:
