finding a limit, calculus question

[junglewarfare]

I need to find the limit of

sin^2 3t
over
t^2

as t approaches zero. I know I'm supposed to multiply my some version of one so I can simplify it but I am stuck.

[Daisydog]

Top and bottom both approach 0. Therefore, use L'Hopital's rule (i.e., differentiate top and bottom, then take limit). After doing so, top and bottom still both approach 0. Therefore, use L'Hopital's rule again (i.e., differentiate top and bottom again). Answer should be 9.

[GMontag]

Alternatively, if you haven't gotten to L'Hopital's Rule and aren't supposed to use it (assuming this is a homework question), you can use the trigonometric expansion of sin 3t = 3cos^2tsint - sin^3t, and the fact that lim t->0 sint/t = 1.

lim t->0 (sin^2 3t)/t^2 = (lim t->0 (sin 3t)/t)^2
=(lim t->0 (3cos^2tsint - sin^3t)/t)^2
=((lim t->0 (3cos^2tsint)/t) - (lim t->0 (sin^3t)/t))^2
=(3(lim t->0 sint/t) - 0(lim t->0 sint/t))^2
=3^2
=9

[Uethym]

Simpler still...

The Taylor expansion of sin x = x - (x^3)/3! + (x^5)/5! - .... Thus as x approaches zero, all of the terms approach zero much more quickly than the first term, so sin x ~ x for small x.

Thus as t -> 0, (sin 3t)^2 ~ (3t)^2 = 9t^2, so (sin 3t)^2 / t^2 ~ 9t^2 / t^2 = 9.