Given,

$$y(x + 1)dx - {x^2}dy = 0$$

$$ \Rightarrow \left( {{{x + 1} \over {{x^2}}}} \right)dx = {{dy} \over y}$$

$$ \Rightarrow {1 \over x}dx + {{dx} \over {{x^2}}} = {{dy} \over y}$$

Integrating both sides, we get

$$\int {{{dx} \over x} + \int {{{dx} \over {{x^2}}} = \int {{{dy} \over y}} } } $$

$$ \Rightarrow \ln |x| - {1 \over x} = \ln |y| + C$$ ..... (1)

Given $$y(1) = e$$

$$\therefore$$ $$x = 1$$ and $$y = e$$

Putting value of x and y in equation (1), we get

$$\ln |1| - {1 \over 1} = \ln |e| + C$$

$$ \Rightarrow 0 - 1 = 1 + C$$

$$ \Rightarrow C = - 2$$

$$\therefore$$ Equation (1) becomes,

$$\ln |x| = - {1 \over x} = \ln |y| - 2$$

$$ \Rightarrow \ln |y| = \ln |x| - {1 \over x} + 2$$

$$ \Rightarrow y = {e^{\ln |x|}}\,.\,{e^{2 - {1 \over x}}}$$

$$ \Rightarrow y = x\,.\,{e^{2 - {1 \over x}}}$$

Now,

$$\mathop {\lim }\limits_{x \to {0^ + }} y$$

$$ = \mathop {\lim }\limits_{x \to {0^ + }} \left( {x\,.\,{e^{2 - {1 \over x}}}} \right)$$

$$ = 0\,.\,{e^{ - \alpha }}$$

$$ = 0$$