I was trying to do an exercise: proving that $\frac{x^2}{1-x^2}$ is continuous on $(0,1)$. I did it but I want to be sure that it's right, could you tell me if my argument is wrong?
$\frac{x^2}{1-x^2}-\frac{a^2}{1-a^2}=\frac{(x+a)(x-a)}{(1-x^2)(1-a^2)}$, now $x+a\leq 1+a$. $1-x^2=1-x^2+a^2-a^2=1-a^2-(x^2-a^2)=1-a^2-(x-a)(x+a)\geq 1-a^2-(x-a)a\geq$ $1-a^2+\delta a$. So $\frac{(x+a)(x-a)}{(1-x^2)(1-a^2)}\leq \frac{(1+a)\delta}{(1-a^2+\delta a)(1-a^2)}\leq\varepsilon$ and so we can just take $\delta\leq\frac{(1-a^2)^2}{1+a-a\varepsilon}$. Is that right?