Posted by **Jackie** on Saturday, September 10, 2011 at 9:26pm.

how do you solve limit approaching 0 of (sin^2x) / (x)

- AP Calculus AB -
**bobpursley**, Saturday, September 10, 2011 at 9:30pm
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.

- AP Calculus AB -
**johnathon**, Sunday, September 11, 2011 at 12:42pm
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

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