# Calculus Limits

1. Mar 25, 2004

### Caldus

I am not sure what the answer to this is or how to go about getting an answer for this:

Find:

lim (1 - cos(a*x))/x^2
x->0

My teacher said the answer would have the term 'a' in it.

How do I go about getting the answer to a limit problem like this? Thank you.

2. Mar 25, 2004

### Chen

$$\lim _{x \rightarrow 0} \frac{1 - \cos ax}{x^2} = \frac{1 - \cos 0}{0^2} = \infty ^+$$
I might be wrong though, but I think the limit is (positive) infinity and that the function has an asymptote at $$x = 0$$.

Last edited: Mar 25, 2004
3. Mar 25, 2004

### kuengb

Sorry to say that Chen, but you are wrong. To solve this, look at the power series of cos(ax). Like this you will have $$1/2*a^2x^2+O(x^4)$$ on top of the line and another square term below. So the limit is...

You can also use L'Hôpital (twice), if you know it.

Last edited: Mar 25, 2004
4. Mar 25, 2004

### Muzza

It's trivial with l'Hopital's rule, but your teacher may or may not want you to use that ;)

lim (x -> 0) (1 - cos(a*x)) / x^2 =
lim (x -> 0) a * sin(ax) / (2x) =
lim (x -> 0) aa * cos(ax) / 2 =
a^2 * 1/2 =
a^2/2

5. Mar 25, 2004

### Chen

I figured it was too simple to be true.

6. Mar 25, 2004

### Caldus

I'm not supposed to know about L'Hopital's rule yet. How could I come up with that forumla? I have to show some kind of evidence. I don't know how to come up with that formula. Thanks.

7. Mar 25, 2004

### Zurtex

Hmm, $$\lim _{x \rightarrow 0} \frac{1 - \cos ax}{x^2}$$

Using the maclaurin series:

$$\cos ax \approx 1 - \frac{(ax)^2}{2}$$

Try that

8. Mar 25, 2004

### Caldus

I do not think I am supposed to know that either. Is there an easy way to prove that the limit is (ax^2)/2?

9. Mar 25, 2004

You could always graph it on a calculator and point.

10. Mar 25, 2004

### Caldus

Yeah I know I did that for when a = 1, 1.5, and 2. But I don't know how to find the forumla for finding the limit if a was anything.

11. Mar 25, 2004

### Caldus

Anyone know? Thanks.

12. Mar 25, 2004

Er, I guess you could always try delta-epsilon proofs, but I don't think you want to go there.

Why not just plot it for a bunch of values of a and look for a pattern? If you have 20 values of a of varying value supporting you, it's pretty solid for your purposes.

13. Mar 26, 2004

### arildno

Suppose you are allowed to know the limits: lim(x->0)(sin(x)/x)=1,lim(x->0)(cos(x)=1),
im(x->0)(a(x)*b(x))=lim(x->0)a(x))*(lim(x->0)b(x),
lim(x->0)(a(x)/b(x))=(lim(x->0)a(x))/(lim(x->0)b(x))
Then you may do without MacLaurin series/L'Hopital's Rule in your proof

14. Mar 26, 2004

### MathematicalPhysicist

now that's very rigorous

15. Mar 26, 2004

### Zurtex

Then would you mind explaining what areas you have covered around this sort of area?