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I am trying to solve the following problem:

Let $f(x) = x^2$ and let $\epsilon > 0$ be given.

Find a $\delta$ so that $|x - 2| \leq \delta \implies |f(x) - 4| \leq \epsilon$.

I've rewritten the $\epsilon \text{-} \delta$ condition thusly,

\begin{align*} |f(x) - 4| = |x^2 - 4| = |(x-2)(x+2)| = |x - 2||x + 2| &\leq \epsilon\\ |x - 2| &\leq \frac{\epsilon}{|x+2|}. \end{align*}

Therefore, choose $\delta \leq \frac{\epsilon}{|x+2|}$ (which is positive since $|x+2|$ is positive).

Then we have \begin{align*} |x-2| \leq \delta \leq \frac{\epsilon}{|x+2|} \implies |f(x) - 4| \leq \epsilon. \end{align*}

Is this right? Are you allowed to define $\delta$ in terms of $x$?

Davis Berlind
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Yes, you are allowed to use $x$ when choosing $\delta$ but that would be done when you were doing the limit at an unknown point $x$. An example would be proving the continuity of $f(x)=x^2$ at any point $x \in \Bbb R$

When you are proving continuity at a given point it is necessary to set a hard upper limit for $\delta$. If you say "Let us guarantee that $\delta \lt \frac 12$", for example, then $x+2$ is less than $\frac 52$ so your $\delta \lt \frac {\epsilon}{|x+2|}$ can have $\le \frac {2\epsilon}5$ added and you have a limit for $\delta$ that does not depend on $x$.

If you haven't seen it already there is the notion of uniform continuity were you are not allowed to vary $\delta(\epsilon)$ as a function of $x$. This is a stronger notion, which is required for a number of theorems.

Ross Millikan
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  • That's what I thought at first too. But the OP is not asking about the $x$ at which we are taking the limit. The OP is asking about the $\color{red}{x}$ from the "... there exists $\delta>0$ such that $|\color{red}{x}-a|<\delta \implies |f(\color{red}{x})-f(a)|<\varepsilon$" part. Since this $\color{red}{x}$ is effectively quantified after $\delta$, $\delta$ can't possibly depend on this $\color{red}{x}$. So the answer to the question that the OP actually asked has to be an unequivocal '"no". – zipirovich Sep 21 '18 at 04:04
  • @zipirovich: Thinking about it, I agree with you. I have updated the answer – Ross Millikan Sep 21 '18 at 04:43
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No you have to get rid of that x in $\frac{\epsilon}{|x+2|}$ but you are on the right track . You can assume that 0< x <4 by agreeing in advance to look for a positive $\delta$ which is less than 2 . Now the denominator |x+2|=x+2<6 so your $\frac{\epsilon}{|x+2|}$ >$\frac{\epsilon}{6}$ so id you take $\delta$ = min of$\frac{\epsilon}{6}$ and 2 you will have $\delta$ < $\frac{\epsilon}{|x+2|}$ as you desire and your argument goes thru .Note that you geta formula for $\delta$ which just depends on $\epsilon$ and 2 . Good luck ,this method works on alot of this type problem .

user439545
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Short answer: NO! This $\delta$ absolutely can NOT depend on this $x$.

Short explanation, without even going into all the details of $\varepsilon$-$\delta$ proofs. The definition of the statement $\lim\limits_{x\to a}f(x)=L$ is that

For any $\varepsilon>0$ there exists $\delta>0$ such that for any $x$ satisfying $0<|x-a|<\delta$ we have that $|f(x)-f(a)|<\varepsilon$.

Note that the $x$ that you're asking about depends on $\delta$. With what you did, we have an $x$ that depends on $\delta$ that depends on this $x$. That's a vicious circle, and logically that doesn't make sense.

zipirovich
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