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

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The limit of the expression (2/x^2) - (1/(1-cos(x))) as x approaches 0 is determined to be -1/6 through multiple applications of L'Hôpital's Rule. Despite initial confusion due to graphical calculator outputs suggesting a limit of zero, the correct approach involves algebraic manipulation and series expansion of cos(x). Specifically, using the Taylor series expansion for cos(x) allows for simplification to an indeterminate form that can be resolved through L'Hôpital's Rule applied four times. This method confirms the limit as -1/6.

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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.
 
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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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