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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