The limit!?

Question

Evaluate each limit, if it exists:

lim [sin^2 (3x)] / [4x^2] = ?
x->0

Can anyone help me, thank a lot!

Answer 1

Rewrite as sin^2(3x) / (3x)^2 * (3x)^2 / 4x^2.
Using lim sin(t)/t = 1 as t goes to 0 and making the change of variable t=3x, you have lim sin^(3x) / (3x)^2 = 1.
And you're left with (3x)^2 / 4x^2 = 9/4

Answer 2

As x --> 0, lim sin(x)/x = 1 (this is the lim that serves to prove that sin' = cos), so

lim [sin^2(3x)] / [4x^2] = (9/4) lim [sin(3x)/3x]^2 = 9/4.

(Alternatively, note that at x = 0 this is a 0/0 indeterminacy, and apply de l'Hopital's Rule.)

Answer 3

To find this limit we can use l'Hôpital's rule

lim (sin^2 (3x))/(4x^2) (take derivative for this expression)
x->0

= lim ((6cos(3x) sin(3x))/(8x ) (derivative)
x->0

= lim ( - 6sin^2 (3x) + 6cos^2(3x))/8
x->0

= -6/8 ( lim sin^2(3x) -cos^2(3x))
x->0

= -6/8 ( lim sin^2(3x) - lim cos^2(3x))
x->0 x->0

= -6/8 ( 0 -1)

= +6/8