Okay, I admit when I was in Pre-Calc I didn't really pay attention to many of the trig identities, not that I was any good at it. Well now it'c coming to bite me in Calculus.

I am on Special trig limits with the following:

lim x->0  sin(x)/x=1   lim x->0 1-cos(x)/x=0

One of my questions asked to find the limit for

lim->0   x+tan(x)/sin(x)

I think my problem is understanding how to apply the trig identities to solve them. So if anyone is willing to help me, and by going down the steps I will appreciate. It's really stressing me out since I have never been this stuck before. 

I'll help you but first you need to clarify something: Do you need the limx->0   x+tan(x)/sin(x) OR limx->0   (x+tan(x))/sin(x) ???

Note the parentheses on the second form.

Alright, I'm gonna assume it's the second case. Look:

lim(x->0) (x+tan(x))/sin(x) = lim(x->0) (x/sin(x)) + lim(x->0) (tan(x)/sin(x)). What I did here was separate the fraction sum. Now, recall that tan(x)=sin(x)/cos(x). Then:

= lim(x->0) (x/sin(x)) + lim(x->0) [(sin(x)/cos(x))/sin(x)]. The bold term is 1 (from the first limit identity you wrote) and the italics cancel out. Then the expression turns into:

= 1 + lim(x->0)(1/cos(x)). But note that as x goes to 0, cos(x) goes to 1. So:

= 1 + 1/1 = 2

Your answer is 2.

okay my notes for that particular problem now make sense!

okay now I have: 

lim x->0  3(1-cos(x))/x

well since I know 1-cos(x)/x=0 I seperate that from 3 to conclude 3 * 0=0

lim θ->   cos θ tan θ/ θ

Now for this one, θ is giving me problems because I don't know what to do with them.

lime x->0  tan^2 (x)/x

The square is really bothering me. 

and thanks again for helping me, you don't know how much I appreciate it. This homework could really either help or destroy me.

Uhg I'm so bad at math, it's going to bite me very hard in the ass once I go to college.

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+ tg^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!

Most important limit you need to know:

lim engineering || science = business
gpa -> 0

  ( || in the above equation = 'or')

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

2x.. what did I win? That triangle is called delta and you are probably looking for lim ∆x -> 0 ?