Finding the best possible $\delta$ for a continuous function. I am trying to understand the following problem... I understand half of it, but I get confused with something. First of all, I was wondering if there is a relation between $\delta$ and $\epsilon$ other than the one stated in the question (I think it was $\delta < 1/\epsilon$ or something like that, but I can't find it in my notes). Thanks a lot in advance!
Question: Consider $f:\mathbb{R}\to\mathbb{R}$ given by $f(x)=x^2$. We know that it is continuous at $a$ for all $a\in\mathbb{R}$. So, for every $a$, for every $\epsilon>0$ there is a $\delta>0$ such that $|x-a|<\delta$ implies $|f(x)-f(a)|<\epsilon$. For $a>1$ and $\epsilon=1$ find the best possible $\delta$. Is this best possible $\delta$ independent of $a>1$.
Solution: Our $\delta$ will be the best one such that $x\in(a-\delta,a+\delta)$ implies that $x^2\in(a^2-1,a^2+1)$, that is, $x\in((a^2-1)^{1/2},(a^2+1)^{1/2})$. Clearly very small $\delta$'s are fine, until either $a-\delta$ hits $(a^2-1)^{1/2}$ or $a+\delta$ hits $(a^2+1)^{1/2}$. So we have to work out $a-(a^2-1)^{1/2}$ and $(a^2+1)^{1/2}-a$ and our best $\delta$ will be whichever is the smaller of these two numbers.
So I understand everything up to here, but I get lost now. How do you get these 2 values?
The first is $\frac{1}{a+(a^2-1)^{1/2}}$ and the second is $\frac{1}{(a^2+1)^{1/2}+a}$, of which $\frac{1}{(a^2+1)^{1/2}+a}$ is the smaller.
And I also don't see how one works and the other won't work :/
So any $\delta\le\frac{1}{(a^2+1)^{1/2}+a}$ will work, and any $\delta\ge\frac{1}{(a^2+1)^{1/2}+a}$ won't work.
Thus we clearly cannot find a $\delta$ which works uniformly for all $a>1$. The best $\delta$ goes to $0$ when $a$ gets larger and larger.
 A: You have:
$\displaystyle a-(a^2+1)^{\frac{1}{2}}=\frac{a+(a^2+1)^{\frac{1}{2}}}{a+(a^2+1)^{\frac{1}{2}}}(a-(a^2+1)^{\frac{1}{2}})=\frac{a^2-a^2-1}{a+(a^2+1)^{\frac{1}{2}}}=\frac{-1}{a+(a^2+1)^{\frac{1}{2}}}$
It's by $(x+y)(x-y)=x^2-y^2$.
A: It is a well-known technique using the difference of two squares identity to show, say, that $$\frac{1}{a+(a^{2}-1)^{1/2}}=\frac{(a-(a^{2}-1)^{1/2})}{(a-(a^{2}-1)^{1/2})(a+(a^{2}-1)^{1/2})}=\frac{a-(a^{2}-1)^{1/2}}{a^{2}-(a^{2}-1)}=a-(a^{2}-1)^{1/2}$$
Can you see now why one of your quantities is bigger than the other?  
Now, our best $\delta$ should obviously be in $[a-(a^{2}-1)^{1/2},(a^{2}+1)^{1/2}-a]$, since that was how we found those two quantities in the first place. Notice that if we choose a $\delta$, and it works for some $x_{0}\in(x-\delta,x+\delta)$, then it also works for all of $(x_{0},x+\delta)$.  
Now what if we chose $\delta=(a^{2}+1)^{1/2}-a$?. Then when $x=a+\delta$, $x^{2}=a^{2}+1$ as required. But when $x=a-\delta$, $x^{2}=(2a-(a^{2}+1)^{1/2})^{2}>a^{2}-1$, so we are 'missing' some of our interval.
