Calculus HELP? Please

[dany612]

Vertigo-X said:

dany612 said:

Vertigo-X said:

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!

thanks vertigo and chapset. Okay I have one question this time it involves rationalization, something which I understand more fully. Anyhow this question has this triangle shape as a coefficent which I assume is Delta but I don't recall it's revalance to this:

find limit:

((x+∆x)^2-x^2)/∆x

delta x is simply the small difference between one point along a curve and the next. As in ∆x = x2-x1

 

I suppose I could take a stab at it.

 

lim ∆x->0 [(x+∆x)^2-x^2]/∆x

lim ∆x->0 [(x^2+2x∆x-∆x^2)-x^2]/∆x

lim ∆x->0 [(x^2-x^2)+2x∆x-∆x^2]/∆x

lim ∆x->0 [2x-∆x]∆x/∆x

lim ∆x->0 2x-∆x = 2x

 

f(x) = x^2

lim ∆x->0 [f(x + ∆x) - f(x)]/∆x

AKA the derivative!

y = nx^m

dy/dx = (n*m)x^(m-1)

 

The derivative is the easy way to do it. :)

Okay you lost me in the bolded. So what would be the limit?