Show that the below equation has exactly two local solutions.

Question

Show that the equation $y=x^2+o(x^2)$ as $x\to0$ for given $y>0$ and for $(x,y)$ near $(0,0)$ has exactly two solutions given by $x= \pm \sqrt{y} + o(\sqrt{y})$ as $y\to 0$.

Answer 1

By the implicit function theorem, the above equation can be solved for $x^2$ uniquely as $$ x^2 = y + g(y) $$ where $g(y)=o(\vert y \vert)$. Using the Taylor series, $$ \sqrt{1+u} = 1 + o(1), \qquad u\to 0, $$ we obtain $$ x=\pm(y+g(y))^{1/2} = \pm \sqrt{y} \sqrt{1 + \frac{g(y)}{y}} = \pm\sqrt{y}(1+o(1)) = \pm \sqrt{y} + o(\sqrt{y}), \quad y \to 0+ $$