Question

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

Answer 1

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]}