Edit: I Think the equation came out wrong in latex. Here is the updated one.

34. $\displaystyle lim_{x\to\infty}\frac{sin^2x - 2x}{x^2sin^2x}$

I have done it out, and for the numerator I get $\displaystyle 2cos^2x - 2sin^2x -2$ and I just know it wants to be a pythagorean and cancel out leaving me with 0. I think I am messing up somewhere because I took it to the second derivation and it just gets funny after that.

Any help would be appreciated.