Solve Cosine IND Limit Without L'Hospital

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    Cosine Limit
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Homework Help Overview

The problem involves evaluating the limit \(\lim_{x \to{0}}{\frac{1-\cos(1-\cos x)}{3x^4}}\) without using L'Hospital's rule. The subject area pertains to limits and trigonometric functions.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the possibility of using Taylor series expansions, though some express unfamiliarity with this method. Others inquire about applying L'Hospital's rule and express frustration with repeated indeterminate forms.

Discussion Status

The discussion is ongoing, with participants exploring different approaches to the limit. Some guidance has been offered regarding the use of Taylor series and L'Hospital's rule, but there is no consensus on a clear path forward. Multiple interpretations of the problem are being considered.

Contextual Notes

Some participants mention constraints related to their current knowledge of Taylor series, indicating that they may be limited to using L'Hospital's rule. There is a noted difficulty in managing repeated indeterminate forms during the evaluation process.

Hernaner28
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It's the following one:

\displaystyle\lim_{x \to{0}}{\frac{1-\cos(1-\cos x)}{3x^4}}

In case we have to apply L'Hospital, appart from it, how could I solve this without it?
Thanks!
 
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You could use the Taylor series expansion of cos(x) around x=0.
 
Mmm I haven't learned Taylor series expansions yet. So anyway, could you tell me how to apply L'Hospital here? There are a lot of steps! I keep getting indeterminations. I cannot figure it out yet...
Thanks for the reply :)
 
Hernaner28 said:
Mmm I haven't learned Taylor series expansions yet. So anyway, could you tell me how to apply L'Hospital here? There are a lot of steps! I keep getting indeterminations. I cannot figure it out yet...
Thanks for the reply :)

If you don't have Taylor series yet, then you'll probably want to stick with l'Hopital. But the idea is to use cos(x)=1-x^2/2!+ terms of higher order in x. It does make things easier.
 
Sorry but I didn't understand the idea.. could you explain the first steps of the resolution? I keep getting IND 0/0 -- I know I've got to apply l'Hopitale every time I get the indtermination but there're just too many.. it never ends.
 
Hernaner28 said:
Sorry but I didn't understand the idea.. could you explain the first steps of the resolution? I keep getting IND 0/0 -- I know I've got to apply l'Hopitale every time I get the indtermination but there're just too many.. it never ends.

l'Hopital will end at the fourth derivative. It has to. Then the denominator becomes a constant. It is a little hard to keep track of the numerator, I will admit.
 
Yo just plug that mofo numerator equation into WolframAlpha:
http://www.wolframalpha.com/input/?i=fourth+derivative+of+1-cos(1-cosx)

I agree, it's a nasty numerator, but you can just plug in x = 0 now. Looking at it real quick, and it's looks like the numerator at 0 equals 3, so the limit is 3.

Edit: Oh wait, the limit wouldn't be 3, it would 3/(3*4*3*2*1) = 1/24
 

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