$\epsilon$-method for directional limits.

Question

Good evening. I'm going to ask a question which might appear dumb or worse, but I consider myself both a student, a researcher and a continuous learner and there are times in which even the tiniest things have to be properly sift. Especially for a Mathematician!

Few years ago I was teaching about limits and most of all about directional limits. Now, I am a very strong fan and supporter of the Heine method with limits, for I have always wanted to sterilise the thinking of the students from the left-right dichotomy (and to prepare them to what will happen in $\mathbb{R}^n$). But at the times I just taught directional limits in the usual way (Cauchy criterion and so on).

Now, here is the point: when dealing with directional limits at finite points, especially when calculating them at the boundary of a function's domain ($\mathrm{e.g.}\quad \Omega: x\in (-\infty, -2)\cup(-2, 7)\cup(7, +\infty)$) I used to explain them both via a graphic way but also via the "so called" $\epsilon-$ method.

I have to say that I never found this method in any book I have read, but I am rather sure it is not something that strange or uknown. What is this method about? It's just a translation of what, say, $a^+$ or $a^-$ means, in terms of infinitesimals. In this way we write

$$\lim_{x\to 2^+} f(x) \equiv \lim_{\substack{x\to 2 + \epsilon \\\\ \epsilon > 0 \\\\ \epsilon \to 0}} f(x) $$

One example

$$f(x) = \exp\left(\dfrac{3x+3}{x^2-3x}\right)$$

The domain reads $\Omega: x\in(-\infty, 0) \cup (0, 3) \cup (3, +\infty)$. Say we want to calculate the limit at $3$, with no graphic help and no other method. Then for what concerns the exponent:

$$\lim_{x\to 3^+} \dfrac{3x+3}{x^2-3x} \equiv \lim_{\substack{x\to 3 + \epsilon \\\\ \epsilon > 0 \\\\ \epsilon \to 0}} \dfrac{3x+3}{x^2-3x}$$

And now we directly substitute, and evaluate the limit at the end:

$$\lim_{\substack{x\to 3 + \epsilon \\\\ \epsilon > 0 \\\\ \epsilon \to 0}} \frac{9 + 3\epsilon + 3}{9 + \epsilon^2 + 6\epsilon - 9 - 3\epsilon} = \lim_{\substack{x\to 3 + \epsilon \\\\ \epsilon > 0 \\\\ \epsilon \to 0}} \frac{12 + 3\epsilon}{\epsilon^2 + 3\epsilon} \longrightarrow +\infty$$

Whence $$\lim_{x\to 3^+} \exp\left(\frac{3x+3}{x^2-3x}\right) \to e^{+\infty} \to +\infty $$

Where it was easy to see the behaviour, since the numerator goes like $12$ and the denominator is a positive infinitesimal quantity.

In the same way, we can obtain

$$\lim_{x\to 3^-} \exp\left(\frac{3x+3}{x^2-3x}\right) \equiv \lim_{\substack{x\to 3 - \epsilon \\\\ \epsilon > 0 \\\\ \epsilon \to 0}}\exp\left(\frac{3x+3}{x^2-3x}\right) = \lim_{\substack{x\to 3 - \epsilon \\\\ \epsilon > 0 \\\\ \epsilon \to 0}} \exp\left( \frac{12-3\epsilon}{\epsilon^2 - 3\epsilon}\right)$$

The numerator goes like $12$ again. Here we see that since $\epsilon >0$ but infinitesimal, then $\epsilon^2 < \epsilon$ and $\epsilon^2 < 3\epsilon$. Thence $\epsilon^2 - 3\epsilon$ is a negative infinitesimal quantity (or we can neglect $\epsilon^2$ and remain just with $-3\epsilon$) and this term goes to $-\infty$ namely:

$$\sim \lim_{\substack{x\to 3 - \epsilon \\\\ \epsilon > 0 \\\\ \epsilon \to 0}} \exp\left( \frac{12}{- 3\epsilon}\right) \to e^{-\infty} \to 0$$

In a very simple and similar way one can deal with limits at $0$.

The positive side of this, is that we indeed work with infintesimals, which are always positive ($\epsilon > 0,\ \epsilon \to 0$) so there is no sign ambiguity.

Now, finally, the question : is there some reason why this method couldn't or shouldn't be used? As I said before, I never found this in any book I have ever read, and this made me think that it might be not a valid or suitable method for directional limits. So I appeal to you who can give me more information, kindly excusing possible ill-notations that I may have used in this question.

Thank you so much!

Answer 1

What you making here i think is basically seeing that, if sequence $x_n$ is convergent to $k$ then there is convergent to 0 sequence $\varepsilon_n$ such that $x_n=\varepsilon_n+k$.

So if we consider $\lim_{x\to k}f(x)$ then from heine definition, we consider limit $\lim_{n\to\infty} f(x_n)$ for any sequence $x_n\to k$ (for any such sequence we have limit to this tk be convergent). Let set $x_n=y_n+k$, as $x_n$ is any sequence convergent to $k$ then $y_n$ too will be any sequence convergent to 0 in here so $\lim_{x\to k}f(x)$ will be equal to $\lim_{y\to 0}f(y+c)$.

You can too consider limits via nonstandard analysis. Here we make field of hyperreal numbers, such numbers basically will be real numbers extended with infinie numbers, infinitesimals, and real numbers + infenitesimals (if $\varepsilon$ is positive infenitesimal then for every real $x$ and any positive real number $r$, $x+\varepsilon<r$).

Here the definition of limit is much closer to what you want to achieve, namely:

$\lim_{x\to k}f(x)=g\Longleftrightarrow \text{ for any $z\approx k$, $f(z)\approx g$}$

Whete $a\approx b$ means that $b-a$ is infenitesimal (so there is infenitesimal $\varepsilon$ [positive or negative] such that $a+\varepsilon=b$). So in here considering a limit is basically consideration wheter for any infenitesimal $k+\varepsilon$ statement $f(k+\varepsilon)\approx g$ holds. Simmilary one side limit will be same with consideration of only positive/negative infenitesimals (right side limit will be consideration wheter $f(k+\varepsilon)\approx g$ for any infenitesimal $\varepsilon>0$, simmilary $<0$ in case of left side limit).

I think it's valuebale in this point to mention that infenitesimals and infinite numbers can be, let's say, complicated. Like we for instance we can consider infenitesimal to be rational or not, so in case of let's say function $f(x)=\begin{cases} x²,x\in\mathbb{Q}\\ -x²,x²\in \mathbb{R}-\mathbb{Q}\end{cases}$, the limit $\lim_{h\to 0}\frac{f(5+h)-f(5)}{h}$ doesn't exist. Because:

1. if infenitesimal $\varepsilon\in \mathbb{Q}^*$, then $\frac{f(5+\varepsilon)-f(5)}{\varepsilon}=\frac{(5+\varepsilon)²-5²}{\varepsilon}=10+\varepsilon \approx 10$
2. if infenitesimal $\varepsilon\in\mathbb{R}^*-\mathbb{Q}^*$ then $\frac{f(5+\varepsilon)-f(5)}{\varepsilon}=\frac{-(5+\varepsilon)²-5}{\varepsilon}=\frac{-30}{\varepsilon}-10-\varepsilon$

.Now $\frac{-30}{\varepsilon}$ will be either infinite or minus infinite number (it's dependent wheter infenitesimal is positive or negative), so definitely it's distinct than $10$ as in case $1.$ so the limit doesn't exist because our condition is that it's infinietely close to same real number (in case of divergent sequence we want it to always be infinite number for argument infinietely close to something) no can matter what infenitesimal we choose.