Posted by Zoey on Saturday, March 6, 2010 at 6:27pm.
Find the limit for the given function. lim t>0 (sin(8t))^2/t^2

calculus  bun, Saturday, March 6, 2010 at 7:18pm
as t>0, we have 8t>0, so
lim (sin(8t))^2/t^2 =
[lim sin(8t)/t].[limsin(8t)/t] as8t>0
=8.[limsin(8t)/8t].8[limsin(8t)/8t], 8t>0
= 8*1*8*1= 64
you know the formula lim (sint/t)=1 as
t>1, aren't you?

calculus  Zoey, Saturday, March 6, 2010 at 8:00pm
yea i knw... thnxx foh ur help
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