Finding Limit of (2/x^2)-(1/(1-cos(x))) as x Approaches 0

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Homework Help Overview

The discussion revolves around finding the limit of the expression (2/x^2) - (1/(1-cos(x))) as x approaches 0. Participants are exploring the behavior of this limit and the methods to evaluate it, particularly focusing on the application of L'Hôpital's rule and series expansion.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants mention attempts to apply L'Hôpital's rule multiple times, with some consistently arriving at -1/6. Others question the validity of this result based on graphical evaluations that suggest the limit may be zero.

Discussion Status

The discussion is ongoing, with various interpretations of the limit being explored. Some participants have suggested using series expansion to simplify the expression, while others express confusion regarding discrepancies between analytical and graphical results.

Contextual Notes

There are indications of potential confusion stemming from the behavior of the function near x = 0, as well as the challenges posed by repeated applications of L'Hôpital's rule. Participants are also grappling with the implications of their findings when substituting values close to zero.

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

i have tried to use l'hopital's rule, but i keep on getting -1/6 however from graphing the function on my graphics calculator i know that it is equal to zero

any help is appreciated
 
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I graphed it on graphmatica, and it looks like -1/6 is correct.

EDIT: Looks like you have to apply L-Hospital rule a hell lot of times before you can get the limit.
 
Last edited:
I get -1/6 too, by graphing with calculator and evaluting it.
 
It's quite simple when we take \cos x = 1 - \frac{x^2}{2} + \frac{x^4}{24} + O(x^6).
 
yeh after you do algebraic manipulation to get it in the indeterminate form of

2-2cos(x)-x^2
-----------------
(x^2)-(x^2)cos(x)

you have to apply l'hospital's rule 4 times to get a non indeterminate form which is -1/6
the only problem is when i substitute in say 0.00001 or -0.00001 into the original equation, i get zero, i don't understand why
 
crappy calculator, cancellation etc.
 
Try using the series truncation I gave, after a little algebraic manipulation the answer comes almost immediately, its quite easy =]
 

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