Let y = f(x) be the solution of the differential equation. The $\underset{\mathrm{x}\to 0^{+}}{lim} \mathrm{f}\left(\mathrm{x}\right)$ is equal to

$\begin{array}{l}(\mathrm{x}+1)\mathrm{ydx}={\mathrm{x}}^2\mathrm{dy}\\ \left(\frac{\mathrm{x}+1}{{\mathrm{x}}^2}\right)\mathrm{dx}=\frac{1}{\mathrm{y}}\mathrm{dy}\\ \left(\frac{1}{\mathrm{x}}+\frac{1}{{\mathrm{x}}^2}\right)\mathrm{dx}=\frac{1}{\mathrm{y}}\mathrm{dy}\\ \log \mathrm{x}-\frac{1}{\mathrm{x}}=\log \mathrm{y}+\log \mathrm{C}\end{array}$
$\begin{array}{l}\log \mathrm{x}-\frac{1}{\mathrm{x}}=\log \left(\mathrm{cy}\right)\\ \log \left(\frac{\mathrm{x}}{\mathrm{cy}}\right)=\frac{1}{\mathrm{x}}\\ \frac{\mathrm{x}}{\mathrm{cy}}={\mathrm{e}}^{1/\mathrm{x}}\Rightarrow \mathrm{cy}={\mathrm{xe}}^{-1/\mathrm{x}}\\ \mathrm{y}=\mathrm{cx}\cdot {\mathrm{e}}^{-1/\mathrm{x}}\\ \mathrm{e}=\mathrm{c}\cdot {\mathrm{e}}^{-1}\\ \mathrm{y}={\mathrm{xe}}^2\cdot {\mathrm{e}}^{-1/\mathrm{x}}\\ ={\mathrm{xe}}^{2-\frac{1}{\mathrm{x}}}\\ ={\mathrm{e}}^2\cdot \frac{\mathrm{x}}{{\mathrm{e}}^{1/\mathrm{x}}}\\ \underset{\mathrm{x}\to 0}{lim} f\left(\mathrm{x}\right)={\mathrm{e}}^2\cdot \left(\frac{0}{\mathrm{\infty }}\right)=0\end{array}$