Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
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
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
@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
up vote 0 down vote accepted

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

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.