Finding the Limit of (1-cos(a*x))/x^2: How to Solve

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In summary, the conversation discusses finding the limit of a function (1 - cos(a*x))/x^2 as x approaches 0. The conversation includes suggestions for solving the problem, such as using the Maclaurin series and L'Hopital's rule. Ultimately, the conversation suggests using known limits and algebraic properties to solve the problem, and also mentions using a graph to look for a pattern.
  • #1
Caldus
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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.
 
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  • #2
[tex]\lim _{x \rightarrow 0} \frac{1 - \cos ax}{x^2} = \frac{1 - \cos 0}{0^2} = \infty ^+[/tex]
I might be wrong though, but I think the limit is (positive) infinity and that the function has an asymptote at [tex]x = 0[/tex].
 
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  • #3
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 [tex]1/2*a^2x^2+O(x^4)[/tex] 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.
 
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  • #4
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
I figured it was too simple to be true. :wink:
 
  • #6
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
Hmm, [tex]\lim _{x \rightarrow 0} \frac{1 - \cos ax}{x^2}[/tex]

Using the maclaurin series:

[tex]\cos ax \approx 1 - \frac{(ax)^2}{2}[/tex]

Try that
 
  • #8
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
You could always graph it on a calculator and point.

cookiemonster
 
  • #10
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
Anyone know? Thanks.
 
  • #12
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.

cookiemonster
 
  • #13
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
cookiemonster said:
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.

cookiemonster

now that's very rigorous
 
  • #15
Caldus said:
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?
Then would you mind explaining what areas you have covered around this sort of area?
 

1. How do I find the limit of (1-cos(a*x))/x^2?

To find the limit of this expression, we can use the limit definition of a derivative. This involves taking the derivative of the expression with respect to x and then finding the limit as x approaches 0. This will give us the value of the limit.

2. What is the importance of finding this limit?

Finding the limit of a function is important as it helps us understand the behavior of the function as the input value approaches a certain point. This can help us make predictions about the function and its behavior.

3. Can I use any value for a and x?

Yes, you can use any value for a and x as long as x is not equal to 0. The limit will be the same regardless of the values chosen for a and x, as long as x is not 0.

4. Is there a shortcut method for finding this limit?

Yes, there is a shortcut method called L'Hôpital's rule which states that the limit of a quotient of two functions is equal to the limit of the derivative of the numerator divided by the derivative of the denominator. This can be used to find the limit of (1-cos(a*x))/x^2 as x approaches 0.

5. Are there any other applications for finding this limit?

Finding the limit of this expression can be useful in various fields such as physics, engineering, and economics. It can help us analyze the behavior of functions and make important decisions based on the information obtained from the limit.

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