Lim x (tends to) -1      (cos2 - cos2x ) / x² - mod x

as x tends to -1 the denomiator will become x²+x.

Now you can either use L'hospital or cos A+cos B for the numerator  to evaluate the limit.

is it 2 cos(2)

the answer is 2 sin 2

limit xtendin to -1 (cos2-cos2x )/x2-mod x

so

limit h tendin to 0+ (cos2-cos2(1-h)/(1-h)^2-mod of(1-h)

so u get

limit x tendin to 0+ {cos2/h^2-h -cos2(1-h)/h^2-h}

then apply l'pithal rule so u get  -(-2)*sin2(1-h)/2h-1

nw substitue limit

u get

2sin2

corrrect em if i m wrng !!

YES THE ANS IS 2SIN2

apply cos a+cos b formula and the apply L HOSPITALS RULE as it is 0/0 form to get the ans

hope u got it.