Talk:One-sided limit

I removed the following example:

• We have

$\lim _{x\downarrow 0}\arctan(x)={+\pi \over 2},$

whereas

$\lim _{x\uparrow 0}\arctan(x)={-\pi \over 2}.$

As far as I know, the arctangent is continuous everywhere on the real line, and this limit is zero. I would like to replace the example with one that does work, but I don't know any from trigonometry. It is true that lim x-> infty arctan x is pi/2, but one sided limits aren't very enlightening at infinity. If someone would like to explain what was supposed to be there, I'd like to have it back. -lethe ^talk 05:11, 18 November 2005 (UTC)

I've put it back with the needed correction. It now says the following:

• We have

$\lim _{x\downarrow 0}\arctan(1/x)={+\pi \over 2},$

whereas

$\lim _{x\uparrow 0}\arctan(1/x)={-\pi \over 2}.$

Michael Hardy 23:48, 18 November 2005 (UTC)

Now that you've said what it was supposed to be, it should have been obvious. Anyway, thanks. -lethe ^talk 00:10, 19 November 2005 (UTC)

I found it a bit strange on Michael's behalf to remove an elementary example with a picture, in favor of an example which duplicates an existing one, that is containing 1/x with x->0.

By no means do I plan it as a revenge, but after Lethe put my example back, I removed Michael's second example. Now we have two very different and somewhat representative examples, as things should be. Oleg Alexandrov (talk) 00:38, 19 November 2005 (UTC)

microhitlers

The following sentance does not seem appropriate to me but i'm not familiar with this subject so I don't want to rush in and make corrections

"The f(x±) notation is singularly and uniformly awful (600 microHitlers)"

Plugwash (talk) 19:45, 9 February 2009 (UTC)

Definitions

You say:

The right-sided limit can be rigorously defined as:

$\forall \varepsilon >0\;\exists \delta >0\;\forall x\in I\;(0<x-a<\delta \Rightarrow |f(x)-L|<\varepsilon )$

Similarly, the left-sided limit can be rigorously defined as:

$\forall \varepsilon >0\;\exists \delta >0\;\forall x\in I\;(0<a-x<\delta \Rightarrow |f(x)-L|<\varepsilon )$

Where $I$ represents some interval that is within the domain of $f$

This appears to be nonsense. Presumably it should read something like:

A real number $L$ or one of the extended real numbers $\infty$, $-\infty$ can be rigorously defined as a right-sided limit of $f$at $a\in L(D\cap (a,\infty ))$ if respectively:

$\forall \varepsilon \in \mathbb {R} ^{+}\;\exists \delta \in \mathbb {R} ^{+}\;\forall x\in D\;(0<x-a<\delta \Rightarrow |f(x)-L|<\varepsilon )$

$\forall y\in \mathbb {R} \;\exists \delta \in \mathbb {R} ^{+}\;\forall x\in D\;(0<x-a<\delta \Rightarrow f(x)>y)$

or

$\forall y\in \mathbb {R} \;\exists \delta \in \mathbb {R} ^{+}\;\forall x\in D\;(0<x-a<\delta \Rightarrow f(x)<y)$

Similarly, a real number $L$ or one of the extended real numbers $\infty$, $-\infty$ can be rigorously defined as a left-sided limit of $f$at $a\in L(D\cap (-\infty ,a))$ if respectively:

$\forall \varepsilon \in \mathbb {R} ^{+}\;\exists \delta \in \mathbb {R} ^{+}\;\forall x\in D\;(0<a-x<\delta \Rightarrow |f(x)-L|<\varepsilon )$

$\forall y\in \mathbb {R} \;\exists \delta \in \mathbb {R} ^{+}\;\forall x\in D\;(0<a-x<\delta \Rightarrow f(x)>y)$

or

$\forall y\in \mathbb {R} \;\exists \delta \in \mathbb {R} ^{+}\;\forall x\in D\;(0<a-x<\delta \Rightarrow f(x)<y)$

Where $D$ is the domain of $f$, $L(S)$ denotes the set of limit points of a set $S$ and $\mathbb {R} ^{+}$ denotes the set of positive real numbers.

The set of left(right)-sided limits of $f$ at $a$ contain at most one element which, if it exists, is the left(right)-sided limit and is denoted as above. If either set is empty the corresponding limit is not defined. For $a\notin L(D\cap (a,\infty ))$ the right limit is not defined, and for $a\notin L(D\cap (-\infty ,a))$ the left limit is not defined.

As it stands all real numbers would be left(right)-sided limits of all functions at all points (one need only choose the empty interval for $I$, when the clause $\forall x\in I\ldots$ becomes vacuously true). We would have, for example,

$\lim _{x\to y^{+}}(0)=1\quad \quad {\text{for all }}y$.

Also if

$f(x)=0\quad x\in \mathbb {Q}$

$f(x)=1\quad x\notin \mathbb {Q}$

then we could say that $f(x)$ is continuous from the left and right throughout the real line.

Without the restriction of $a$ to $L(D\cap (a,\infty ))$ or $L(D\cap (-\infty ,a))$ if $D$ contained points isolated from below or above then all reals would also be left and/or right limits at these points. Martin Rattigan (talk) 02:04, 28 April 2015 (UTC)

Existence

You say:

The two one-sided limits exist and are equal if the limit of $f(x)$ as $x$ approaches $a$ exists.

This would not be necessarily true according to the definition I suggested in the previous section, e.g.:

$\lim _{x\to 1}Arcsin(x)={\frac {\pi }{2}}$,

but I have specified that

$\lim _{x\to 1^{+}}Arcsin(x)$

is undefined because $1\notin L(dom(Arcsin)\cap (1,\infty ))$.

On what grounds do you assert this? — Preceding unsigned comment added by Martin Rattigan (talk • contribs) 05:13, 28 April 2015 (UTC)

"rigorously defined" needs a link to syntax definition

I love that there is a rigorous definition tabled. I suggest that for the lay reader interested in learning more about that there is no easy or obvious way to do that. It's a classic case of "what to search for". Does one search "rigorous definitions in math" for example and wade through a lot of results in the hope of finding a page that describes how to read this? I think not. The beauty of the web, and of wikipedia is that we can link directly from the words "rigorously defined" or via a parenthesized or "see also" link point readers to a Wikipedia page that describes the syntax of this rigorous definition. As I am such a lay reader with no clue where to turn, I can't even add it, just drop a note here saying how nice it would be: if assumed knowledge were low and links to learning more are embedded. --120.29.241.112 (talk) 03:12, 15 June 2018 (UTC)

In probability theory?

"In probability theory it is common to use the short notation: $f(x-)$ for the left limit and $f(x+)$ for the right limit." — Really? I always believed this is common in math analysis (and all theories that use it, including probability). Boris Tsirelson (talk) 04:38, 24 September 2019 (UTC)