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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?

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On first glance, you're forgetting to take the absolute value. –  Alex Becker Jan 11 '12 at 1:32
    
Also $x-a$ can be positive or negative, so $1-a^2 - (x-a)(x+a)$ cannot be directly compared to $1-a^2 + \delta a$ like you did. // Are you specifically asked to use the epsilon-delta definition of continuity? This problem is simpler using the standard properties of continuous functions. –  Srivatsan Jan 11 '12 at 1:35
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The denominator $1-x^2$ is never zero in $(0,1)$ and so the function is continuous because it's the quotient of two continuous functions. –  lhf Jan 11 '12 at 1:46
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@Srivatsan Yeah, I'm asked to do it with the epsilon-delta definition –  John Jan 11 '12 at 2:10
    
@Srivatsan: I don't understand, if $|x-a|<\delta$ then $-\delta< x-a<\delta$, so $x-a>-\delta$, right? But now that I think about it, if I put the absolute values I have a problem...could you help me to solve this problem, please? –  John Jan 11 '12 at 2:50

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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 $$|f(x)-f(a)|=\left|\frac{x^2}{1-x^2}-\frac{a^2}{1-a^2}\right|<\frac{2\delta}{[1-(\frac{1+a}{2})^2](1-a^2)}\leq\epsilon.$$

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why $\frac{|x+a||x-a|}{|(1-x^2)(1-a^2)|}\leq2|x-a|$? –  John Jan 11 '12 at 3:21
    
Oh yes, that's a mistake. Originally I thought $\frac{1}{1-x^2}\leq 1$. See my edited answer. –  Paul Jan 11 '12 at 4:44

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