Limit

Definition

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:

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

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

• For every $\epsilon >0$
• 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 $|f(x)-L|<\epsilon$ (explicitly, $f(x)\in (L-\epsilon ,L+\epsilon )$).

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

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

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

• For every $\epsilon >0$
• 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 $|f(x)-L|<\epsilon$ (explicitly, $f(x)\in (L-\epsilon ,L+\epsilon )$.

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

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

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

• For every $\epsilon >0$
• 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 $|f(x)-L|<\epsilon$ (explicitly, $f(x)\in (L-\epsilon ,L+\epsilon )$.

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

$\!lim_{x\to c}f(x)=L$

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 $\epsilon >0$, or equivalently, chooses the target interval $(L-\epsilon ,L+\epsilon )$.
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 $|f(x)-L|<\epsilon$ (i.e., $f(x)\in (L-\epsilon ,L+\epsilon )$), the prover wins. If $|f(x)-L|\geq \epsilon$, the skeptic wins.

We say that the limit statement

$\!lim_{x\to c}f(x)=L$

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 $\epsilon$ chosen by the skeptic. Thus, it is an expression of $\delta$ as a function of $\epsilon$.

We say that the limit statement

$\!lim_{x\to c}f(x)=L$

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