Trig Limits: find lim (x -> 0) (sin^2 (3x)) / (5x^2)

[xcal]

Hi, I trying to self study some calculus but I don't remember doing limits with a lot of trig in them.
How can I find

lim (sin^2 3x)/(5x^2)
x->0

I don't know what to do with the ^2. I know that lim (sinx)/x = 1 as x->0. Substitution with u doesn't seem to be working either.

 

[skeeter]

Re: Trig Limits

how about a bit of algebraic "manipulation" ? ...

\(\displaystyle \frac{\sin^2(3x)}{5x^2} = \frac{9}{5} \cdot \frac{\sin^2(3x)}{9x^2} = \frac{9}{5} \cdot \frac{\sin(3x)}{3x} \cdot \frac{\sin(3x)}{3x}\)

 

[xcal]

Re: Trig Limits

thanks