Question

lim x tends to 0... 1-cos2x+tan^2x/xsinx

Answer 1

Take x common and 1-cos2x=2(sin^2 x). hence x [tan2x - 2tanx]/{2(sin^2 x)}^2 is the equation)
tan2x = 2tanx/(1 - tan^2 x). On substituting and simplifying we get x[2tanx . tan^2 x] / (1 - tan^2 x)(4 sin^4 x)
Write tanx = sinx/cosx. On substituting and simplifying we get 2x / 4sinxcosx (cos^2 x - sin^2 x)
Write cos^2 x = 1 - sin^2 x and take numerator x to denominator, we get 2 / [4 (sinx/x)cosx - 8 (sinx/x)cosx sin^2 x]
Lim x -> 0 sinx/x = 1, therefore 2/(4-0) = 2/4 = 1/2