Can delta be defined in terms of x in a epsilon-delta proof?

Question

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 $x$?

Answer 1

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.

Answer 2

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 .

Answer 3

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.