# Simple trig limit with cos

## Homework Statement

Evaluate the limits that exist:

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

## Homework Equations

lim x->0 $$\frac{1-cosax}{ax}$$ = 0

## 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?

Dick
Homework Helper
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:
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.

Dick
Homework Helper
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.

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

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

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...