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:

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

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

or

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

or

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

Where $$D$$ is the domain of $$f$$, $$L(S)$$ denotes the set of limit points of a set $$S$$ and $$\reals^+$$ 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)