Question

lim=0 (1-cosx√cos2x/x²) without l hospital rule​

Answer 1

Answer:

Let us denote limx->0 as just x->0 for easy typing.

z=x->0[(1-cosx√cos2x)/x²]

When we put zero it is of (0/0) form so apply

L-hospital rule.

z=x->0[D(1-cosx√cos2x)/D(x²)]

z=x->0[0-D(cosx√(cos2x))/(2x)]

z=x->0[cosxD(√(cos2x))+√(cos2x)D(cosx)]/(2x)

D(√cos2x)=-sin2x/√cos2x

D(√cos2x)=-sin2x.√sec2x

z=x->0[cosx[-sin2x√sec2x]-(sinx/(2x))√cos2x]

z=x->0[-sin2xcosx√sec2x]-x->0[(sinx/(2x))√cos2x]

As x->0[sinax/x]=a

x->0[cosax]=1

x->0[secax]=1

x->0[sinax]=0

z=((-0)(1)√1)-(1/2)√1

z=0–1/2

z=-1/2