How is $\frac{\ln r^2}{r^{-2}}$ equivalent to $\frac{r^{-1}}{-2r^{-3}}$

Question

I am trying to find the limit as $(x,y)$ approaches $(0,0)$ of

$$(x^2 + y^2)\ln(x^2 + y^2)$$

Using polar coordinates, I understand that taking $x = r\cos\theta$ and $y=r\sin\theta$, we get

$$r^2\ln(r^2)$$

Which can be written as $$\frac{\ln(r^2)}{r^{-2}}$$

However, I do not understand why this tutorial says that is equivalent to $$\frac{r^{-1}}{-2r^{-3}}$$

which is equivalent to

$$\frac{-1}{2r^2}$$

(shouldn't it be $\frac{-r^2}{2}$?)

Can someone please explain how $\dfrac{\ln r^2}{r^{-2}}$ is equivalent to $\dfrac{r^{-1}}{-2r^{-3}}$

Answer 1

L'hopital rule:

$$\displaystyle \lim_{r\to0} \dfrac{\ln(r^2)}{r^{-2}} = \lim_{r\to0} \dfrac{[\ln(r^2)]'}{[r^{-2}]'} = \lim_{r\to0} \dfrac{2r^{-1}}{-2r^{-3}}$$