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

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 $\frac{0}{0}$...
We can try using de l'Hospital Rule deriving top and bottom and then apply the limit.
We get:
${\lim}_{x \to 0} \left(\frac{2 x}{\cos \left(x\right) - 1}\right) = \frac{0}{0}$
again:
${\lim}_{x \to 0} \left(\frac{2}{- \sin \left(x\right) - 0}\right) =$
in this case depending on the side you chose to approach zero you get different situations.
The two lateral limits are different so for $x \to 0$ the limit does not exists:
On the other hand the lateral limits give you
${\lim}_{x \to {0}^{+}} \left(\frac{2}{- \sin \left(x\right) - 0}\right) = \frac{2}{\text{a negative very small number}} = - \infty$
${\lim}_{x \to {0}^{-}} \left(\frac{2}{- \sin \left(x\right) - 0}\right) = \frac{2}{\text{a positive very small number}} = + \infty$

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