Limit

Definition for finite limit for function of one variable

Two-sided limit

Suppose $f$ is a function of one variable and $c\in \mathbb{R}$ is a point such that $f$ is defined to the immediate left and immediate right of $c$ (note that $f$ may or may not be defined at $c$). In other words, there exists some value $t>0$ such that $f$ is defined on $(c-t,c)\cup (c,c+t)$.

For a given value $L\in \mathbb{R}$, we say that:

$\underset{x\to c}{lim}f(x)=L$

if the following holds (the single sentence is broken down into multiple points to make it clearer):

• For every $ϵ>0$
• there exists $\delta >0$ such that
• for all $x\in \mathbb{R}$ satisfying $0<|x-c|<\delta$ (explicitly, $x\in (c-\delta ,c)\cup (c,c+\delta )$),
• we have $|f(x)-L|<ϵ$ (explicitly, $f(x)\in (L-ϵ,L+ϵ)$).

The limit (also called the two-sided limit) $\underset{x\to c}{lim}f(x)$ is defined as a value $L\in \mathbb{R}$ such that $\underset{x\to c}{lim}f(x)=L$. By the uniqueness theorem for limits, there is at most one value of $L\in \mathbb{R}$ for which $\underset{x\to c}{lim}f(x)=L$. Hence, it makes sense to talk of the limit when it exists.

Left hand limit

Suppose $f$ is a function of one variable and $c\in \mathbb{R}$ is a point such that $f$ is defined to the immediate left of $c$ (note that $f$ may or may not be defined at $c$). In other words, there exists some value $t>0$ such that $f$ is defined on $(c-t,c)$.

For a given value $L\in \mathbb{R}$, we say that:

$\underset{x\to c^{-}}{lim}f(x)=L$

if the following holds (the single sentence is broken down into multiple points to make it clearer):

• For every $ϵ>0$
• there exists $\delta >0$ such that
• for all $x\in \mathbb{R}$ satisfying $0<c-x<\delta$ (explicitly, $x\in (c-\delta ,c)$),
• we have $|f(x)-L|<ϵ$ (explicitly, $f(x)\in (L-ϵ,L+ϵ)$.

The left hand limit (acronym LHL) $\underset{x\to c^{-}}{lim}f(x)$ is defined as a value $L\in \mathbb{R}$ such that $\underset{x\to c^{-}}{lim}f(x)=L$. By the uniqueness theorem for limits (one-sided version), there is at most one value of $L\in \mathbb{R}$ for which $\underset{x\to c^{-}}{lim}f(x)=L$. Hence, it makes sense to talk of the left hand limit when it exists.

Right hand limit

Suppose $f$ is a function of one variable and $c\in \mathbb{R}$ is a point such that $f$ is defined to the immediate right of $c$ (note that $f$ may or may not be defined at $c$). In other words, there exists some value $t>0$ such that $f$ is defined on $(c,c+t)$.

For a given value $L\in \mathbb{R}$, we say that:

$\underset{x\to c^{+}}{lim}f(x)=L$

if the following holds (the single sentence is broken down into multiple points to make it clearer):

• For every $ϵ>0$
• there exists $\delta >0$ such that
• for all $x\in \mathbb{R}$ satisfying $0<x-c<\delta$ (explicitly, $x\in (c,c+\delta )$),
• we have $|f(x)-L|<ϵ$ (explicitly, $f(x)\in (L-ϵ,L+ϵ)$.

The right hand limit (acronym RHL) $\underset{x\to c^{+}}{lim}f(x)$ is defined as a value $L\in \mathbb{R}$ such that $\underset{x\to c^{+}}{lim}f(x)=L$. By the uniqueness theorem for limits (one-sided version), there is at most one value of $L\in \mathbb{R}$ for which $\underset{x\to c^{+}}{lim}f(x)=L$. Hence, it makes sense to talk of the right hand limit when it exists.

Relation between the limit notions

The two-sided limit exists if and only if (both the left hand limit and right hand limit exist and they are equal to each other).

Definition of finite limit for function of one variable in terms of a game

The formal definitions of limit, as well as of one-sided limit, can be reframed in terms of a game. This is a special instance of an approach that turns any statement with existential and universal quantifiers into a game.

Two-sided limit

Consider the limit statement, with specified numerical values of $c$ and $L$ and a specified function $f$:

$\underset{x\to c}{lim}f(x)=L$

The game is between two players, a Prover whose goal is to prove that the limit statement is true, and a Skeptic (also called a Verifier or sometimes a Disprover) whose goal is to show that the statement is false. The game has three moves:

1. First, the skeptic chooses $ϵ>0$, or equivalently, chooses the target interval $(L-ϵ,L+ϵ)$.
2. Then, the prover chooses $\delta >0$, or equivalently, chooses the interval $(c-\delta ,c+\delta )\setminus \{c\}$.
3. Then, the skeptic chooses a value $x$ satisfying $0<|x-c|<\delta$, or equivalently, $x\in (c-\delta ,c+\delta )\setminus \{c\}$, which is the same as $(c-\delta ,c)\cup (c,c+\delta )$.

Now, if $|f(x)-L|<ϵ$ (i.e., $f(x)\in (L-ϵ,L+ϵ)$), the prover wins. If $|f(x)-L|\ge ϵ$, the skeptic wins.

We say that the limit statement

$\underset{x\to c}{lim}f(x)=L$

is true if the prover has a winning strategy for this game. The winning strategy for the prover basically constitutes a strategy to choose an appropriate $\delta$ in terms of the $ϵ$ chosen by the skeptic. Thus, it is an expression of $\delta$ as a function of $ϵ$.

We say that the limit statement

$\underset{x\to c}{lim}f(x)=L$

is false if the skeptic has a winning strategy for this game. the winning strategy for the skeptic involves a choice of $ϵ$, and a strategy that chooses a value of $x$ (constrained in the specified interval) based on the prover's choice of $\delta$.

Conceptual definition and various cases

Formulation of conceptual definition

Below is the conceptual definition of limit. Suppose $f$ is a function defined in a neighborhood of the point $c$, except possibly at the point $c$ itself. We say that:

$\underset{x\to c}{lim}f(x)=L$

if:

• For every choice of neighborhood of $L$ (where the term neighborhood is suitably defined)
• there exists a choice of neighborhood of $c$ (where the term neighborhood is suitably defined) such that
• for all $x\ne c$ that are in the chosen neighborhood of $c$
• $f(x)$ is in the chosen neighborhood of $L$.

Functions of one variable case

The following definitions of neighborhood are good enough to define limits.

• For points in the interior of the domain, for functions of one variable: We can take an open interval centered at the point. For a point $c$, such an open interval is of the form $(c-t,c+t),t>0$. Note that if we exclude the point $c$ itself, we get $(c-t,c)\cup (c,c+t)$.
• For the point $+\mathrm{\infty }$, for functions of one variable: We take intervals of the form $(a,\mathrm{\infty })$, where $a\in \mathbb{R}$.
• For the point $-\mathrm{\infty }$, for functions of one variable: We can take interval of the form $(-\mathrm{\infty },a)$, where $a\in \mathbb{R}$.

We can now list the nine cases of limits, combining finite and infinite possibilities:

Case	Definition
$\underset{x\to c}{lim}f(x)=L$	For every $ϵ>0$, there exists $\delta >0$ such that for all $x$ satisfying $0<|x-c|<\delta$ (i.e., $x\in (c-\delta ,c)\cup (c,c+\delta )$), we have $|f(x)-L|<ϵ$ (i.e., $f(x)\in (L-ϵ,L+ϵ)$).
$\underset{x\to c}{lim}f(x)=-\mathrm{\infty }$	For every $a\in \mathbb{R}$, there exists $\delta >0$ such that for all $x$ satisfying $0<|x-c|<\delta$ (i.e., $x\in (c-\delta ,c)\cup (c,c+\delta )$), we have $f(x)<a$ (i.e., $f(x)\in (-\mathrm{\infty },a)$).
$\underset{x\to c}{lim}f(x)=\mathrm{\infty }$	For every $a\in \mathbb{R}$, there exists $\delta >0$ such that for all $x$ satisfying $0<|x-c|<\delta$ (i.e., $x\in (c-\delta ,c)\cup (c,c+\delta )$), we have $f(x)>a$ (i.e., $f(x)\in (a,\mathrm{\infty })$).
$\underset{x\to -\mathrm{\infty }}{lim}f(x)=L$	For every $ϵ>0$, there exists $a\in \mathbb{R}$ such that for all $x$ satisfying $x<a$ (i.e., $x\in (-\mathrm{\infty },a)$), we have $|f(x)-L|<ϵ$ (i.e., $f(x)\in (L-ϵ,L+ϵ)$).
$\underset{x\to -\mathrm{\infty }}{lim}f(x)=-\mathrm{\infty }$	For every $b\in \mathbb{R}$, there exists $a\in \mathbb{R}$ such that for all $x$ satisfying $x<a$ (i.e., $x\in (-\mathrm{\infty },a)$), we have $f(x)<b$ (i.e., $f(x)\in (-\mathrm{\infty },b)$).
$\underset{x\to -\mathrm{\infty }}{lim}f(x)=\mathrm{\infty }$	For every $b\in \mathbb{R}$, there exists $a\in \mathbb{R}$ such that for all $x$ satisfying $x<a$ (i.e., $x\in (-\mathrm{\infty },a)$), we have $f(x)>b$ (i.e., $f(x)\in (b,\mathrm{\infty })$).
$\underset{x\to \mathrm{\infty }}{lim}f(x)=L$	For every $ϵ>0$, there exists $a\in \mathbb{R}$ such that for all $x$ satisfying $x>a$ (i.e., $x\in (a,\mathrm{\infty })$), we have $|f(x)-L|<ϵ$ (i.e., $f(x)\in (L-ϵ,L+ϵ)$).
$\underset{x\to \mathrm{\infty }}{lim}f(x)=-\mathrm{\infty }$	For every $b\in \mathbb{R}$, there exists $a\in \mathbb{R}$ such that for all $x$ satisfying $x>a$ (i.e., $x\in (a,\mathrm{\infty })$), we have $f(x)<b$ (i.e., $f(x)\in (-\mathrm{\infty },b)$).
$\underset{x\to \mathrm{\infty }}{lim}f(x)=\mathrm{\infty }$	For every $b\in \mathbb{R}$, there exists $a\in \mathbb{R}$ such that for all $x$ satisfying $x>a$ (i.e., $x\in (a,\mathrm{\infty })$), we have $f(x)>b$ (i.e., $f(x)\in (b,\mathrm{\infty })$).