Find the limit for the given function. lim t-->0 (sin(8t))^2/t^2

Answer 1

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?

Answer 2

yea i knw... thnxx foh ur help