Let y = f ( x ) be the solution of the differential equation y ( x...

Given,

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

$⇒ \dfrac{x + 1}{x2}dx = \dfrac{dy}{y}$

$⇒ \left(\dfrac{1}{x} + \dfrac{1}{x2}\right)dx = \dfrac{dy}{y}$

Now integrating both side we get,

$\log_{e}x - \dfrac{1}{x} = \log_{e}y + c$

Now on using~$y\left(1\right) = e$~we get,& $c = - 2$

So, the equation of curve becomes~$\log_{e}x - \dfrac{1}{x} = \log_{e}y - 2$

$⇒ y = e^{\ln x} - \dfrac{1}{x} + 2$

Hence,~$\lim_{x \to {0}^{+}}e^{\ln x - 1} - \dfrac{1}{x} + 2= e^{- \infty}= 0$.