Trig limit

sean/mac · May 26, 2008

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

Defennder · May 26, 2008

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.

malawi_glenn · May 26, 2008

I get -1/6 too, by graphing with calculator and evaluting it.

Gib Z · May 26, 2008

It's quite simple when we take [tex]\cos x = 1 - \frac{x^2}{2} + \frac{x^4}{24} + O(x^6)[/tex].

sean/mac · May 26, 2008

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

malawi_glenn · May 26, 2008

crappy calculator, cancellation etc.

Gib Z · May 27, 2008

Try using the series truncation I gave, after a little algebraic manipulation the answer comes almost immediately, its quite easy =]

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Trig limit

sean/mac

Defennder

malawi_glenn

Gib Z

sean/mac

malawi_glenn

Gib Z

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