# Solving a trig limit

• Sep 26th 2007, 08:53 AM
Gamerdude
Solving a trig limit
I've been trying to work out this limit problem for the past few days. I got a few friends to look at it, and they haven't been able to solve it either. I think I'm just overlooking something...but I'm almost positive I need to apply the cosine double angle identity...I just get stuck after that. Here's the problem:

lim cos(2x) - cos(x)
x->0 ----------------
x

I just can't figure out how to evaluate it. Any help would be appreciated.
• Sep 26th 2007, 09:01 AM
red_dog
$\displaystyle \displaystyle\lim_{x\to 0}\frac{\cos 2x-\cos x}{x}=\lim_{x\to 0}\frac{-2\sin\frac{x}{2}\sin\frac{3x}{2}}{x}=$
$\displaystyle \displaystyle =-\lim_{x\to 0}\left(\frac{\sin\frac{x}{2}}{\frac{x}{2}}\right) \cdot\lim_{x\to 0}\sin\frac{3x}{2}=-1\cdot 0=0$
• Sep 26th 2007, 09:07 AM
Gamerdude
Such a fast response...wow, thanks a bunch. What identity did you use? I just want to make sure I understand what you did so I can do it later myself if I have to.
• Sep 26th 2007, 09:09 AM
Jhevon
here's an alternate method:

$\displaystyle \lim_{x \to 0} \frac {\cos 2x - \cos x}x = \lim_{x \to 0} \frac {2 \cos^2 x - 1 - \cos x }x$

$\displaystyle = \lim_{x \to 0} \frac {(2 \cos x + 1)(\cos x - 1)}x$

$\displaystyle = \lim_{x \to 0} (2 \cos x + 1) \cdot \lim_{x \to 0} \frac {\cos x - 1}x$

$\displaystyle = 3 \cdot 0$

$\displaystyle = 0$
• Sep 26th 2007, 09:15 AM
Gamerdude
Quote:

Originally Posted by Jhevon
here's an alternate method:

$\displaystyle \lim_{x \to 0} \frac {\cos 2x - \cos x}x = \lim_{x \to 0} \frac {2 \cos^2 x - 1 - \cos x }x$

$\displaystyle = \lim_{x \to 0} \frac {(2 \cos x + 1)(\cos x - 1)}x$

$\displaystyle = \lim_{x \to 0} (2 \cos x + 1) \cdot \lim_{x \to 0} \frac {\cos x - 1}x$

$\displaystyle = 3 \cdot 0$

$\displaystyle = 0$

Hmm, that's a method I was trying out...but when I got to the part when you split up the two functions, I kept the x as the denominator for both parts. I thought this was required...but in your example, I guess this isn't necessary?
• Sep 26th 2007, 09:17 AM
Jhevon
Quote:

Originally Posted by Gamerdude
Hmm, that's a method I was trying out...but when I got to the part when you split up the two functions, I kept the x as the denominator for both parts. I thought this was required...but in your example, I guess this isn't necessary?

you cannot do that. remember, $\displaystyle \frac 1x \cdot \frac 1x = \frac 1{x^2}$. so if you distributed the x among both fractions, you would be changing the value
• Sep 26th 2007, 09:19 AM
Gamerdude
Quote:

Originally Posted by Jhevon
you cannot do that. remember, $\displaystyle \frac 1x \cdot \frac 1x = \frac 1{x^2}$. so if you distributed the x among both fractions, you would be changing the value

Ah, gotcha. That makes sense. I get mixed up on the simplest things sometimes...it's embarrassing. Thank you for your help, both of you. I greatly appreciate it. I thanked both of you for your helpful responses.