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consider following definition of Right Hand Limit,Left Hand Limit and limit point

RHL

Given a limit point c of set A and a function $f: A \rightarrow \mathbb{R}$ we say that $\lim_{x \rightarrow c^{+}}$ if $\forall \epsilon >0 \, \exists \delta>0 \, x \in A \,:\,0<x-c<\delta \implies |f(x) - L| < \epsilon$

LHL

Given a limit point c of set A and a function $f: A \rightarrow \mathbb{R}$ we say that $\lim_{x \rightarrow c^{-}}$ if $\forall \epsilon >0 \, \exists \delta>0 \, x \in A \,:\,0<c-x<\delta \implies |f(x) - L| < \epsilon$

limit point

a point c is called limit point of set A if $\forall \delta >0 \exists (x\neq c) \in A \,\text{s.t}\, x \in (c-\delta , c+\delta)$

$x$ for each delta has just to lie in $(c-\delta,c+\delta)$ according to above definition.Which is enough for uniqueness of limits(general,not one-sided).But I don't think it is enough for uniqueness of one-sided limits.

Uniqueness of Both RHL and LHL cannot be asserted(one can prove that at least one of them has to be unique if exists ). Cause we need at least two $x's$ for each $\delta$ (one to left ,one to right) that are less than $\delta$ distance away from c, so that one-sided limit definition is not vacuously satisfied.But one fundamental thing is that we want limits to be unique if they exist.

EDIT: I am confused about whether if c is a limit point then there exists two sequences $(x_n>c) \rightarrow c$ and $(x_n < c) \rightarrow c$

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    $\begingroup$ Your definitions are almost OK. But note that $c$ being a limit point of $A$ does not necessarily imply that for every $\delta>0$ you have some $x\in A$ with $0<x-c<\delta$. All it implies is that there is some $x\in A$ with $0<|x-a|<\delta $. For one sided limits it is best to assume that the function is defined in some interval of type $(c, d) $ for $\lim_{x\to c^{+}} f(x) $ and it is defined in some interval of type $(d, c) $ for $\lim_{x\to c^{-}} f(x) $. $\endgroup$ – Paramanand Singh Aug 8 at 3:28
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    $\begingroup$ Uniqueness of limit is consequence of the definition of limit. It has nothing to do with definition of limit point. If you can prove that usual two sides limits are unique then you can note that the same proof can be adapted with minor changes to deal with one sided limits. $\endgroup$ – Paramanand Singh Aug 8 at 3:30
  • $\begingroup$ @ParamanandSingh Thanks! finally someone understood what I am confused about .My point is that these definitions are NOT okay because you cannot guarantee uniqueness of limits. These definitions are given here mathonline.wikidot.com/left-hand-and-right-hand-limits. Could you please write an answer so that I will accept $\endgroup$ – viru Aug 8 at 12:24
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We first start with the usual definition of limit.

Let $c$ be a real number / point. A neighborhood of $c$ is an open interval $I$ containing $c$. If $I$ is a neighborhood of $c$ then the set $I\setminus \{c\} $ is said to be a deleted neighborhood of $c$.

Let $A$ be a non empty subset of $\mathbb {R} $ and let $f:A\to\mathbb {R} $ be a function. Further let $c$ be a point such that there exists some deleted neighborhood of $c$ contained in $A$.

A number $L$ is said to be the limit of $f$ at $c$ (written symbolically as $L=\lim\limits_{x\to c} f(x) $) if the following condition holds: for every $\epsilon>0$ there exist a $\delta>0$ corresponding to it such that $|f(x) - L|<\epsilon$ whenever $x\in A$ and $0<|x-c|<\delta$.

Such a number $L$ may or may not exist, but if it exists then the above definition ensures that it must be unique (see proof later).

The above definition can be generalized a bit to include domains which are not necessarily neighborhoods. And the question here provides such a general definition using concept of limit points.

The same generalization however does not apply to one sided limits because of lack of corresponding concepts like left / right limit points. The idea of one sided limits is restricted to introductory courses of calculus where typical domains are intervals/neighborhoods. In that context here is a definition of left hand limit.

Let $A$ be a non-empty subset of $\mathbb {R} $ and let $f:A\to\mathbb {R} $ be a function. Further let $c$ be a point such that there exists a number $h>0$ with $(c-h, c) \subseteq A$.

A number $L$ is said to be the left hand limit of $f$ at $c$ (written symbolically as $L=\lim\limits_{x\to c^{-}} f(x) $) if the following condition holds: for every $\epsilon>0$ there exists a $\delta>0$ corresponding to it such that $|f(x) - L|<\epsilon $ whenever $x\in A$ and $0<c-x<\delta$.

The key prerequisite in the notion of limit is that if $c$ is the point under consideration then the function must be defined at points arbitrarily near to $c$. For one sided limits the function must be defined at points arbitrarily near to and on the corresponding side of $c$.

The uniqueness of a limit is based on the fact that values of the function are near the limit $L$. If $L_1,L_2$ are distinct then you can not ensure values $f(x) $ to be arbitrarily near to both $L_1,L_2$ at the same time. If you go too close to $L_1$ then you have to necessarily move away from $L_2$.

This is more technically expressed as follows:

Theorem (with easy proof) : Let $a, b$ be two real numbers with $a\neq b$. Then there exists a neighborhood $I$ of $a$ and a neighborhood $J$ of $b$ such that $I\cap J=\emptyset$.

This key fact is the basis of the general notion of a Hausdorff space.

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  • $\begingroup$ I was trying to prove that number of discontinuities of monotonically increasing function is finite or countably infinite.There I was required to use these one-sided limits.Thank you so much for explaining it more clearly :) $\endgroup$ – viru Aug 9 at 5:44
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If the right limit of a function at a given point does exist, then it is unique.

Similarly if the left limit of a function at a given point does exist, it is unique .

If both right limit and left limit exist and they are equal, then the limit of the function does exist and it is equal to the common value of the left limit and the right limit.

If the limit of a function at a given point does exist, it is unique.

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  • $\begingroup$ my question is basically about whether RHL and LHL are properly defined for limit point.I think I didn't correctly ask what I am confused about $\endgroup$ – viru Aug 7 at 5:15
  • $\begingroup$ Your definitions for RHL and LHL are correct. You should note that limit point of a set is different from the limit of a function at a point. $\endgroup$ – Mohammad Riazi-Kermani Aug 7 at 10:13
  • $\begingroup$ could you prove that RHL and LHL are unique if it exists $\endgroup$ – viru Aug 8 at 2:09
  • $\begingroup$ Assume that you have two different RHL, $l_1\ne l_2$ and let $\epsilon = |l_1-l_2|/2$ try to get a contradiction. $\endgroup$ – Mohammad Riazi-Kermani Aug 8 at 10:28

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