Find the limit for the given function? #lim_(t->0) text( ) ((text(sin)(8 t))^2)/t^2#

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Answer 1 · 2017-02-27T19:09:10

#64#

#sin(8t)/t = 8 sin(8t)/(8 t)# and

#(sin(8t)/t)^2=64(sin(8t)/(8 t))^2#

but

#lim_(y->0)sin y/y = 1# so

#lim_(t->0) (sin(8t)/t)^2 =64#