Find the Limit?

Question

lim tan6x/sin2t
x-> 0

i know the answers 3 by using the table method but i need to know how to work it out using trig functions? Any help would be sweet.
cheers

lim tan6x/sin2x
x-> 0

Answer 1

when u typed sin 2t did you mean sin 2x? i'll assume so...

tan6x = sin6x / cos6x

weird identity:
sin 3t = 3 sin t – 4 sin^3( t)
and cos 3t = 4 cos^3( t) –3 cos t

^^ the 3t's can be 6t's if the t's are substituted as 2t's
so the limit is
(3sin2x-4sin^3(2x))/((sin2x)*4cos^3(2x)-3cos(2x))
factor out sin2x on top and that'll cancel with the sin2x on bottom leaving:
(3-4sin^3(0))/(4*1-3*1)=3/1=3 !!

Answer 2

Use L'Hospital's Rule. The limit is the same as
lim(x->0)[6sec^2(6x)]/[2cos(2x)] = 6/2 = 3.