Continuity of a function 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?
 A: Here is the definition of continuity in terms of  the epsilon-delta definition: $f$ is continuous at $a$ if and only if for any $\epsilon>0$, there exists $\delta>0$ such that if $|x-a|<\delta$, then $|f(x)-f(a)|<\epsilon$. 
Now we have $f(x)=\displaystyle\frac{x^2}{1-x^2}$. Then for any $a\in(0,1)$, we have (as you have calculated)
$$\tag{1}\left|\frac{x^2}{1-x^2}-\frac{a^2}{1-a^2}\right|=\left|\frac{(x+a)(x-a)}{(1-x^2)(1-a^2)}\right|=\frac{|x+a|\cdot|x-a|}{|(1-x^2)(1-a^2)|}\leq \frac{2|x-a|}{[1-(\frac{1+a}{2})^2](1-a^2)}$$
if $x\in(\displaystyle\frac{a}{2},\frac{1+a}{2})$.
Therefore, for any $\epsilon>0$, there exists $\delta=\min\{\displaystyle\frac{\epsilon}{2}[1-(\frac{1+a}{2})^2](1-a^2),\frac{a}{2},\frac{1-a}{2}\}>0$ such that if $|x-a|<\delta$, then 
$$-\delta<x-a,\mbox{ or equivalently }, x>a-\delta>a-\frac{a}{2}=\frac{a}{2}$$
and 
$$x-a<\delta,\mbox{ or equivalently }, x<a+\delta<a+\frac{1-a}{2}=\frac{1+a}{2}.$$
That is 
$$\tag{2} x\in(\frac{a}{2},\frac{1+a}{2}).$$
Hence, using $(1)$ and $(2)$, we have
$$|f(x)-f(a)|=\left|\frac{x^2}{1-x^2}-\frac{a^2}{1-a^2}\right|<\frac{2\delta}{[1-(\frac{1+a}{2})^2](1-a^2)}\leq\epsilon.$$ 
