Question

Can I help you?

Answer 1

uh, why do you make a new question everytime ?
there are many open question, Please go to them and help wherever you can.
do not expect people to come to your post and ask their questions.
and please do not make such questions anymore.

Answer 2

b/c they are 99.99% answered already

Answer 3

does this count as advertising...?

Answer 4

@Jack1, what?

Answer 5

The one below you isn't answered

Answer 6

so you are not creative? make one up! do it now!

Answer 7

yes you can absolutely not

Answer 8

I am trying to take the lim of \[\lim_{n \rightarrow \infty } \frac{ n^{12} }{ n^{13}-(n-3)^{13} }\] and I cannot use derivatives, I am trying to use the squeeze theorem

Answer 9

@soccerman1234 ha! love the +ve negative ;D

Answer 10

wow everyones here lmao XD

Answer 11

What do you post here, what exactly are you trying to achieve?

Answer 12

arykk, do u have that in your open question ? i think i can help ya

Answer 13

OK you just span my thread and totally wasted it...!

Answer 14

just a thought... we could turn this into our chat for a while... ;D

Answer 15

Solomon, there are tons of questions that need to be answered. It's not really fair to try to get people to come to you instead of going out to get people, I understand you're trying to help, but there's a system and it does work. c:

Answer 16

naah, this wastes time of helpers.
i closed it and solomon, you are not gonna make a new question on this again, ok ?

Answer 17

Should I really?

Answer 18

OK!

Answer 19

Later, not now!

Answer 20

Bye! I am out of this thread!

Answer 21

i'll miss u solomon... come back?

Answer 22

xD

Answer 23

soccerman1234 , and Jack1, smart medal exchange!

Answer 24

@Jack1, sarcasm....
I don't think that "Bye! I am out of this thread!" deserves that comment!
whatever....

Answer 25

@SolomonZelman
cosx+sec^2x-1

Answer 26

sec^2x-1= tan^2 x

Answer 27

Thank You for that tip!

Answer 28

Why did you close all your questions?

Answer 29

The mod did, not me.