AP Calculus AB
posted by Jackie .
how do you solve limit approaching 0 of (sin^2x) / (x)

wouldn't it be the same as Lim sinx * lim sinx/x=0*1=0?
lim ab= lim a * lim b in my part of the woods. 
the limit of ab does not always = the limit of a * limit b
use lpatls rule i am probboly spelling that wrong but it says that the limit at x approches a of f/g = the limit as x approches a of f'/g' this is only true if the function is of the indeterminate for 0/0 or infinity/infinity
you have 0/0 that is lim of sin(2x) as x approches 0 is 0 and the lim of x as x approches 0 is 0
f=sin(2x) g=x f'=cos(2x)*2 g'=1
the lim as x approches 0 of sin(2x)/(x)=
the lim as x approches 0 of cos(2x)
cos(0)=1