# Trig limit

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

Related Calculus and Beyond Homework Help News on Phys.org
Defennder
Homework Helper
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:
malawi_glenn
Homework Helper
I get -1/6 too, by graphing with calculator and evaluting it.

Gib Z
Homework Helper
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

malawi_glenn