Finding a Trig Limit by hand, no L'Hopitals

[gorlitsa]

Homework Statement

$\stackrel{lim}{x\rightarrow 0}\frac{cos^2x-1}{2xsinx}$

Homework Equations

$\stackrel{lim}{x\rightarrow 0}\frac{1-cosx}{x}=0$

$\stackrel{lim}{x\rightarrow 0}\frac{sinx}{x}=1$

The Attempt at a Solution

I found this problem online (and can't remember where). It was in a limits section, so it can be solved without using l'hospital's rule. I gave it to my class today, and then we all got stuck. We know the answer is -.5 but can't get it algebraically. Any advice?

 

[kru_]

Remember that 1 = cos^2(x) + sin^2(x).

Now substitute and simplify and you can use one of your known limits.

 

[gorlitsa]

Thank you! I got thrown into this class half way through the year, and while I have re-learned the calc, my trig is horrible. Thank you!