# Simple trig limit with cos

1. Sep 30, 2007

### L4N0

1. The problem statement, all variables and given/known data
Evaluate the limits that exist:

lim x->0 of $$\frac{1-cos4x}{9x^{2}}$$

2. Relevant equations
lim x->0 $$\frac{1-cosax}{ax}$$ = 0

3. The attempt at a solution
so far I've got this
$$\frac{4x}{9x^{2}}\frac{1-cos4x}{4x}$$
The second part has a limit of 0 but I don't know what to do about the first fraction. To me it looks undefined but according to some java applet I found on the internet the limit should be 8/9. What's the right answer?

2. Sep 30, 2007

### Dick

The limit has the form 0/0. That suggests you use L'Hopital's rule. Alternatively, if you know the power series expansion of cos(x)=1-x^2/2!+x^4/3!-..., you could substitute x->4x and use that.

Last edited: Sep 30, 2007
3. Sep 30, 2007

### L4N0

Thank you. So far we are in Ch 2. so we haven't done L'Hopital's rule or power series. I guess saying it is undefined is enough for now.

4. Sep 30, 2007

### Dick

The trouble with that is that is IS defined. If you can't use those, then you have to use tricks. Try a double angle formula like cos(2x)=1-2*sin(x)^2. Use that to express cos(4x). Then use lim x->0 sin(ax)/(ax)=1. I'm guessing you do know that from the relevant equation you posted.

5. Sep 30, 2007

### L4N0

oops, I got confused a bit...
Anyways, I used your suggestion and solved it.
Thanks again.

6. Sep 30, 2007

### dynamicsolo

Very sneaky little problem: the limit law it looks the most obvious to use is the wrong one!. Keep these kinds of manipulations in mind on exam problems...