1. ## calculus limits...trig

I can't seem to find out where to start or finish for this limit:

Lim 1-cos 3x / cos(squared) 3x - 1
x->0

2. Just factorise the bottom.

3. ## calculus limits...trig

sorry, that would make it a lot easier, but the bottom 3x was suppose to be a 5x. Thats what makes it so complicated. NEED HELP

Lim 1-cos 3x / cos(squared) 5x - 1
x->0

4. $\displaystyle \frac{{1 - \cos 3x}} {{\cos ^2 5x - 1}} = - \frac{9} {{25}} \cdot \frac{{1 - \cos 3x}} {{9x^2 }}\left( {\frac{{5x}} {{\sin 5x}}} \right)^2 .$

The answer should be obvious from there.

5. ## still need help

that doesn't make it obvious. could you explain a little farther. Thanks