# Homework Help: Finding a Trig Limit by hand, no L'Hopitals!

1. Mar 28, 2012

### gorlitsa

1. The problem statement, all variables and given/known data

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

2. Relevant equations

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

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

3. 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'Hopitals 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?

2. Mar 28, 2012

### kru_

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

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

3. Mar 28, 2012

### 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!