The value of limx→02(1-cosxcos2xcos3x3⋯cos10x10x2) is _______ . [2024]
(55)
limx→02(1-cosxcos2xcos3x3…cos10x10x2)
Let f=cosxcos2xcos3x3…cos10x10
f=cosx(cos2x)1/2(cos3x)1/3…(cos10x)1/10
Taking log on both sides, we get
logf=log cosx+12cos2x+13cos3x+…+110cos10x
Differentiating w.r.t. x
1fdfdx=-tanx-tan2x…-tan10x
dfdx=-f(tanx+tan2x…+tan10x)
Using L'Hospital's Rule
limx→02(f(tanx+tan2x…+tan10x))2x (∵ f=1)
=1+2+…+10=55