Show $\int_{0}^{\epsilon}\rho(x)^{-2}dx= +\infty, \hspace{4mm} \forall \epsilon >0$ let  $\rho(x)=\sqrt{x}, \hspace{4mm} \forall x \in \mathbb{R}$
Show :
$$
 \int_{0}^{\epsilon}\rho(x)^{-2}dx= +\infty, \hspace{4mm} \forall \epsilon >0.
$$
My attempt:
\begin{align*}
\int_{0}^{\epsilon}\rho(x)^{-2}dx &= \int_{0}^{\epsilon}{1\over x}\,dx\\
&=[ \ln(x) ]_{0}^{\epsilon} \hspace{2mm} \forall \epsilon >0 \\
&=\ln(\epsilon)-\ln(0), \hspace{2mm} \forall \epsilon >0\\
&=\ln(\epsilon), \hspace{2mm} \forall \epsilon >0
\end{align*}
ln(x) :Natural logarithm
so i'm stuck help please 

$$\int_{0}^{\epsilon}{1\over x}dx = \lim_{y \rightarrow 0^{+}}\int_{y}^{\epsilon}{1\over x}dx = \lim_{y \rightarrow 0^{+}} \ (\log \epsilon -\log y) = + \infty$$
