The equation $ax^3-x^2+1=0$ has three real solutions if $a^2<4/27$. The three solutions are plotted below against parameter $a$
Is there some intuitive reason why instead of being two continuous functions, both the yellow branch and green branch are discontinuous at $a=0$? This leads to the consequence that $x^2-1=0$ can't be solved by solving $ax^3-x^2+1=0$ and bring $a\to 0$.
P.S. The solutions are shown below, if anyone's interested.
$$ \cases{x_1 = \frac{1}{3} \left(\frac{\sqrt[3]{-27 a^2+3 \sqrt{3} \sqrt{27 a^4-4 a^2}+2}}{\sqrt[3]{2} a}+\frac{\sqrt[3]{2}}{a \sqrt[3]{-27 a^2+3 \sqrt{3} \sqrt{27 a^4-4 a^2}+2}}+\frac{1}{a}\right)\\ x_2=-\frac{\left(1-i \sqrt{3}\right) \sqrt[3]{-27 a^2+3 \sqrt{3} \sqrt{27 a^4-4 a^2}+2}}{6 \sqrt[3]{2} a}-\frac{1+i \sqrt{3}}{3\ 2^{2/3} a \sqrt[3]{-27 a^2+3 \sqrt{3} \sqrt{27 a^4-4 a^2}+2}}+\frac{1}{3 a}\\ x_3=-\frac{\left(1+i \sqrt{3}\right) \sqrt[3]{-27 a^2+3 \sqrt{3} \sqrt{27 a^4-4 a^2}+2}}{6 \sqrt[3]{2} a}-\frac{1-i \sqrt{3}}{3\ 2^{2/3} a \sqrt[3]{-27 a^2+3 \sqrt{3} \sqrt{27 a^4-4 a^2}+2}}+\frac{1}{3 a}} $$
One obvious thing is that the three functions given are all odd functions: the contents of the cube root is an even function, so if we write it as $b = \frac{1}{2}(2-27a^2-3\sqrt{3}\sqrt{27a^4-4a^2})$, we have three functions of the form $$ \frac{1}{3a} \left(1+\omega b^{-1/3}+\omega^{-1} b^{1/3} \right), $$ where $\omega$ is one of the cube roots of $1$ (namely $1$ and $(-1\pm i\sqrt{3})/2$), and all three of these are the product of an odd and even function and so must be odd.
If you replace $b$ by $\frac{1}{2}(2-27a^2-3\sqrt{3}a\sqrt{27a^2-4})$ (i.e. pull an $a$ out of the square root, so choosing a branch so that $\sqrt{27a^4-4a^2}$ is instead odd), this is no longer the case, and you'll find that both finite solutions are now continuous at zero: