dany612 on 29 August 2011
| pearljammer said: done
Wait... your high school has started classes already? |
yeah its been 2 weeks already. :( i really need to boost up my studying over limits.
I was wondering if you could do this for me and explain. I just don't understand the last part where it says hospital rule. Thanks a lot for such a favor.
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
