I have this implicit equation $F(x, y) = x^2y+e^{x+y} = 0 $. Now this defines a function $y=f(x)$ everywhere except for $x=0$. I need to compute the limit for $x \rightarrow 0^+$.
I know that $y = -\dfrac{e^{x+y}}{x^2} < 0$ and that $y' = - \dfrac{2xy + e^{x+y}}{x^2+ e^{x+y}} $
But I cannot imagine how to compute the right handed limit in question. I know that $y(0^+)$ cannot be a constant else there would be a contradiction. But what if $y$ oscillates? How can I exclude this and be sure that the limit is $-\infty$? Thanks for tips.
If you rewrite the derivative using $y=-e^{x+y}/x^2$, you get $$ y'=\frac{(2-x)e^{x+y}}{x(x^2+e^{x+y})} $$ which is positive in a right neighborhood of $0$. Thus your function is increasing in such a neighborhood and therefore the limit exists, either finite or $-\infty$.
Now show that a finite limit leads to a contradiction.