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

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Homework Help Overview

The problem involves evaluating the limit as x approaches 0 for the expression (1 - cos(4x)) / (9x^2). This falls under the subject area of limits in calculus, specifically dealing with trigonometric functions.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the form of the limit, noting it as 0/0, and suggest using L'Hopital's rule or power series expansion. There is also mention of using trigonometric identities to manipulate the expression.

Discussion Status

Some participants have offered alternative methods for approaching the limit, while others express uncertainty about the techniques available to them based on their current coursework. There is an acknowledgment of confusion regarding the appropriate limit laws to apply.

Contextual Notes

One participant notes that they have not yet covered L'Hopital's rule or power series in their studies, which may limit their approach to the problem. Additionally, there is a reference to a potential misunderstanding of the limit laws applicable to this situation.

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Homework Statement


Evaluate the limits that exist:

lim x->0 of [tex]\frac{1-cos4x}{9x^{2}}[/tex]

Homework Equations


lim x->0 [tex]\frac{1-cosax}{ax}[/tex] = 0


The Attempt at a Solution


so far I've got this
[tex]\frac{4x}{9x^{2}}\frac{1-cos4x}{4x}[/tex]
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?
 
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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.
 
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.
 
L4N0 said:
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...
 

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