trig limit thing? (1 Viewer)

[akuchan]

Find:

lim 1 - cos2x
x->0 x^2

answer is 2... i hate these questions

 

[Mattamz]

lim x->0 (1-cos2x)/(x^2)
= lim x->0 (1-[(cosx)^2-(sinx)^2])/(x^2) , since cos2x = (cosx)^2-(sinx)^2
=lim x->0 [2(sinx)^2]/(x^2)
=lim x->0 2*(sinx/x)(sinx/x)
= 2*1*1 , since lim x->0 sinx/x = 1
=2

 

[SoulSearcher]

Could also use the identity that cos2x = 1 - 2sin²x, works out the same though.

 

[Mattamz]

Could also use the identity that cos2x = 1 - 2sin²x, works out the same though.

it works the same because it is the same identity...
ie
cos2x
= (cosx)^2 - (sinx)^2
= 1 - 2(sinx)^2 , since (sinx)^2 + (cosx)^2 = 1