Limit

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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 $(c-t,c+t)\setminus \{c\}=(c-t,c)\cup (c,c+t)$.

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 $\epsilon >0$ (the symbol $\epsilon$ is a Greek lowercase letter pronounced "epsilon")
• there exists $\delta >0$ such that (the symbol $\delta$ is a Greek lowercase letter pronounced "delta")
• for all $x\in \mathbb{R}$ satisfying $0<|x-c|<\delta$ (explicitly, $x\in (c-\delta ,c)\cup (c,c+\delta )=(c-\delta ,c+\delta )\setminus \{c\}$),
• we have $|f(x)-L|<\epsilon$ (explicitly, $f(x)\in (L-\epsilon ,L+\epsilon )$).

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.

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Note: Although the definition customarily uses the letters $ϵ$ and $\delta$, any other letters can be used. The reason for sticking to a standard letter choice is that it reduces cognitive overload.

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 $\epsilon >0$
• 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) $\underset{x\to c^{-}}{lim}f(x)$ 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 $\epsilon >0$
• 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) $\underset{x\to c^{+}}{lim}f(x)$ 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.

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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 $f$:

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

Note that there is one trivial sense in which the above statement can be false, or rather, meaningless, namely, that $f$ is not defined on the immediate left or immediate right of $c$. In that case, the limit statement above is false, but moreover, it is meaningless to even consider the notion of limit.

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. Otherwise, the skeptic wins (see the subtlety about the domain of definition issue below the picture).

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

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 $\epsilon$, and a strategy that chooses a value of $x$ (constrained in the specified interval) based on the prover's choice of $\delta$.

Slight subtlety regarding domain of definition: The domain of definition issue leads to a couple of minor subtleties:

• A priori, it is possible that the $x$ chosen by the skeptic is outside the domain of $f$, so it does not make sense to evaluate $f(x)$. In the definition given above, this would lead to the game being won by the skeptic. In particular, if $f$ is not defined on the immediate left or right of $c$, the skeptic can always win by picking $x$ outside the domain.
• It may make sense to restrict discussion to the cases where $f$ is defined on the immediate left or right of $c$. Explicitly, we assume that $f$ is defined on the immediate left and immediate right, i.e., there exists $t>0$ such that $f$ is defined on the interval $(c-t,c+t)\setminus \{c\}$. In this case, it does not matter what rule we set regarding the case that the skeptic picks $x$ outside the domain. To simplify matters, we could alter the rules in any one of the following ways, and the meaning of limit would remain the same as in the original definition:
  • We could require (as part of the game rules) that the prover pick $\delta$ such that $(c-\delta ,c+\delta )\setminus \{c\}\subseteq \mathrm{d}\mathrm{o}\mathrm{m}f$. This pre-empts the problem of picking $x$-values outside the domain.
  • We could require (as part of the game rules) that the skeptic pick $x$ in the domain, i.e., pick $x$ with $0<|x-c|<\delta$ and $x\in \mathrm{d}\mathrm{o}\mathrm{m}f$.
  • We could alter the rule so that if the skeptic picks $x$ outside the domain, the prover wins (instead of the skeptic winning).

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Non-existence of limit

The statement $\underset{x\to c}{lim}f(x)$ does not exist could mean one of two things:

1. $f$ is not defined around $c$, i.e., there is no $t>0$ for which $f$ is defined on $(c-t,c+t)\setminus \{c\}$. In this case, it does not even make sense to try taking a limit.
2. $f$ is defined around $c$, around $c$, i.e., there is $t>0$ for which $f$ is defined on $(c-t,c+t)\setminus \{c\}$. So, it does make sense to try taking a limit. However, the limit still does not exist.

The formulation of the latter case is as follows:

For every

$L\in \mathbb{R}$

, there exists

$\epsilon >0$

such that for every

$\delta >0$

, there exists

$x$

satisfying

$0<|x-c|<\delta$

and such that

$|f(x)-L|\ge \epsilon$

.

We can think of this in terms of a slight modification of the limit game, where, in our modification, there is an extra initial move by the prover to propose a value $L$ for the limit. The limit does not exist if the skeptic has a winning strategy for this modified game.

An example of a function that does not have a limit at a specific point is the sine of reciprocal function. Explicitly, the limit:

$\underset{x\to 0}{lim}\sin (\frac{1}{x})$

does not exist. The skeptic's winning strategy is as follows: regardless of the $L$ chosen by the prover, pick a fixed $\epsilon <1$ (independent of $L$, so $\epsilon$ can be decided in advance of the game -- note that the skeptic could even pick $\epsilon =1$ and the strategy would still work). After the prover has chosen a value $\delta$, find a value $x\in (0-\delta ,0+\delta )\setminus \{0\}$ such that the $\sin (1/x)$ function value lies outside $(L-\epsilon ,L+\epsilon )$. This is possible because the interval $(L-\epsilon ,L+\epsilon )$ has width $2\epsilon$, hence cannot cover the entire interval $[-1,1]$, which has width 2. However, the range of the $\sin (1/x)$ function on $(0-\delta ,0+\delta )\setminus \{0\}$ is all of $[-1,1]$.

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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 $L$.

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

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Real-valued functions of multiple variables case

We consider the multiple input variables as a vector input variable, as the definition is easier to frame from this perspective.

The correct notion of neighborhood is as follows: for a point $\overline{c}$, we define the neighborhood parametrized by a positive real number $r$ as the open ball of radius $r$ centered at $\overline{c}$, i.e., the set of all points $\overline{x}$ such that the distance from $\overline{x}$ to $\overline{c}$ is less than $r$. This distance is the same as the norm of the difference vector $\overline{x}-\overline{c}$. The norm is sometimes denoted $|\overline{x}-\overline{c}|$. This open ball is sometimes denoted $B_r(\overline{c})$.

Suppose $f$ is a real-valued (i.e., scalar) function of a vector variable $\overline{x}$. Suppose $\overline{c}$ is a point such that $f$ is defined "around" $\overline{c}$, except possibly at $\overline{c}$. In other words, there is an open ball centered at $\overline{c}$ such that $f$ is defined everywhere on that open ball, except possibly at $\overline{c}$.

With these preliminaries out of the way, we can define the notion of limit. We say that:

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

if the following holds:

• For every $\epsilon >0$
• there exists $\delta >0$ such that
• for all $\overline{x}$ satisfying $0<|\overline{x}-\overline{c}|<\delta$ (i.e., $\overline{x}$ is in a ball of radius $\delta$ centered at $\overline{c}$ but not the point $\overline{c}$ itself -- note that the $|\cdot |$ notation is for the norm, or length, of a vector)
• we have $|f(\overline{x})-L|<\epsilon$. Note that $f(\overline{x})$ and $L$ are both scalars, so the $|\cdot |$ here is the usual absolute value function.

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