Calculus HELP? Please

[Vertigo-X]

chapset said:

lim θ->0   (cos θ tan θ)/ θ = lim θ->0 [cos θ(sonθ/cosθ)]/θ = lim  θ->0 (sin θ)/ θ = using the hospital rule ( which mean you use a derivative on top and in the bottom) lim  θ->0 (cos θ)/1= 1/1 = 1

θ is just an other variable here

for the second one I am not 100% sure for this one

I used those two trigo identity: 1+ tan^2(x)= sec^2(x)

                                                        sec(x)= 1/cos(x) so sec^2(x)= 1^2/cos^2(x)= 1/cos^2(x)

lim x->0 tan^2(x)/x = lim x->0 (sec^2(x)-1)/x

lim x->0 [sec^2(x)]/x  - 1/x = lim x->0 [1/cos^2(x)]/x - 1/x = lim x->0 1/xcos^2(x) - 1/x  using the hospital rule you get lim x->0 0/something - 0/1 = 0

Not to be a nag, but it's L'Hopital's rule. I also fixed your trigonometric identity (it was tg^2(x) before).

 

Good ol' fashioned calculus!