# Math Help - How do I evaluate this limit?

1. ## How do I evaluate this limit?

lim $\frac{sin^2(4x)}{x^2}$
x->0

I know by finding derivative of the function I get:

$\frac{4x^2cos^2(4x) - 2xsin^2(4x)}{x^4}$
(hopefully that's right so far)

what do I do next? is there a trig I identity that I can use to simplify or something?

thanks!

2. Split it up and use the famous limit $\lim_{x\to 0}\frac{sin(x)}{x}=1$

$\lim_{x\to 0}\frac{sin(4x)}{x}\cdot\lim_{x\to 0}\frac{sin(4x)}{x}$

Now, using one of them multiply top and bottom by 4:

$\lim_{x\to 0}\frac{4sin(4x)}{4x}$

Let t=4x:

$4\lim_{t\to 0}\frac{sin(t)}{t}=4$