Simple Trig Limit: Evaluate lim x->0 of (1-cos4x)/9x^2

[L4N0]

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]

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.

 

[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.

 

[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.

 

[L4N0]

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

 

[dynamicsolo]

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...