How do you find the limit of #(x^(2)/(sinx-x))# as x approaches 0?

1 Answer
Gió
Jun 22, 2016

I think that the limit does not exists but you can find the lateral ones!

Explanation:

If you do it directly you get the form #0/0#...
We can try using de l'Hospital Rule deriving top and bottom and then apply the limit.
We get:
#lim_(x->0)((2x)/(cos(x)-1))=0/0#
again:
#lim_(x->0)((2)/(-sin(x)-0))=#
in this case depending on the side you chose to approach zero you get different situations.
The two lateral limits are different so for #x->0# the limit does not exists:
On the other hand the lateral limits give you
#lim_(x->0^+)((2)/(-sin(x)-0))=2/("a negative very small number")=-oo#
#lim_(x->0^-)((2)/(-sin(x)-0))=2/("a positive very small number")=+oo#

graph{x^2/(sin(x)-x) [-10, 10, -5, 5]}

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