I have a test tonight and need help with one example. Find the limit of (sin(5x)cos(3x))/sin(2x) as x approaches 0. I'm pretty sure the answer is 2.5, but I used a calculator to get that and we aren't allowed to use one on this part of the test. Thanks in advance!

Well if I remember correctly sin(5x)=5*cos5x which the limit of = 5 in the same manner the limit of cos(3x)=1 you end up with 5*1/2=2.5 There are identities and rules that apply to trig functions that can make your life easier. put them on your cheat sheet.

This is the type of stuff that reminds me that while I had a blast and wouldn't trade those years for anything, I am in fact, glad to be done with college. I've got nothin' for ya. Good luck on your test.

I don't think that's correct, at least after my glance of trig identities online. sin(nx) = 2(cos(x)*sin((n-1)x)-sin(n-2)x Which would mean: sin(5x) =2(cos(5)*sin((5-1)x))-sin(5-2)x =2cos(5)*sin(4x)-sin(3x) Honestly, looking at the problem you gave I would initially say infinity since the sine on the bottom would go to zero as x goes to zero. I don't think that is right considering what your calculator gave. It's been too long since I've done calculus and I don't remember all the tricks anymore or I would help you out more. *edit* Ok so I forgot, both top and bottom turn to zero. That means you should use L'Hopital's Rule. You need to take the derivative of both the top and bottom, then try the limit again.