# Limits Definition and relaxation of the point of accumulation condition

I am doing a self study reading Thomson-Bruckner-Bruckner book on analysis, found here:

http://classicalrealanalysis.info/documents/TBB-AllChapters-Landscape.pdf

The chapter on continuous functions defines Limit of a function as follows,

Definition 5.1: (Limit) Let $f:E \mapsto \mathbb{R}$ be a function with domain E and suppose that $x_o$ is a point of accumulation of $E$. Then we write $\lim_{x \rightarrow x_o} f(x) = L$ if for every $\epsilon > 0$ there is a $\delta > 0$ so that $|f(x) - L| < \epsilon$ whenever $x$ is a point of $E$ differing from $x_o$ and satisfying $|x - x_o| < \delta$.

The point being made is that $x_o$ is an accumulation point of $E$. In order to illustrate this point, the author has the following exercise problem,

Show that $\lim_{x \rightarrow \ -2} \sqrt{x} = L$ is true for any $L$, if the definition of limits excludes that $x_o$ has to be a point of accumulation of the set $E$. My reasoning was that this ought to be vacuously true since $E \bigcap (-2-\delta, -2+ \delta) = \emptyset$ for any $\delta < 2$. However, I would like to make the case using the $\epsilon - \delta$ argument. I am not sure how to proceed.

• I have to doubt quality of the exposition if the author really states that the equality is true for any $L$, since this implies that any two real numbers are equal.
– MPW
Commented Feb 21, 2017 at 16:08
• @MPW, I had by mistake left out the fact that the definition of limit point be changed to exclude the fact that $x_o$ is a point of accumulation of the domain of definition of $f$. I have made the change. Thanks for pointing the error. Commented Feb 21, 2017 at 16:15

## 1 Answer

You were actually almost done. As soon as you know the implication gets vacuously true if $0 < \delta < 2$, for every $\varepsilon > 0$ you've got a choice of $\delta$ (say $\delta := 1$).

Alright, this problem is in fact a little logical issue. Recall that you are to find at least one $\delta > 0$ allowing the condition pertaining to $\delta$. You ended up with finding infinitely many such $\delta$, namely anyone in the open interval $]0,2[$. So for whatever $\varepsilon > 0$, you've got infinitely many choices of such $\delta$, namely those in $]0,2[$. So simply giving them $\delta := 1$, say, suffices.

• @"Walking Blues" From the definition I reason for each $\epsilon > 0$ there has to be some $\delta(\epsilon) > 0$ which in our case has a region for which the statement is vacuously true no matter what the epsilon value is? And that is sufficient? I am still a bit lost, if you could elaborate, I would appreciate it. Commented Feb 21, 2017 at 17:24
• @RameshKadambi, You might want to check the revised answer :).
– Yes
Commented Feb 21, 2017 at 17:31
• @RameshKadambi, found that the answer's got some typos (it should be $]0, 2[$ instead of $]0,2]$), which are now corrected.
– Yes
Commented Feb 22, 2017 at 5:24
• @"Walking Blues" Thank you. It was extremely helpful. I understood the solution nonetheless. Commented Feb 22, 2017 at 18:11