One-sided limit

Short description: Limit of a function approaching a value point from values below or above the value point

The function $f(x)=x^2+\mathrm{sign}(x),$ where $\mathrm{sign}(x)$ denotes the sign function, has a left limit of $-1,$ a right limit of $+1,$ and a function value of $0$ at the point $x=0.$

In calculus, a one-sided limit refers to either one of the two limits of a function $f(x)$ of a real variable $x$ as $x$ approaches a specified point either from the left or from the right.

The limit as $x$ decreases in value approaching $a$ ($x$ approaches $a$ "from the right" or "from above") can be denoted:

$$\underset{x\to a^{+}}{\lim }f(x)\text{ or }\underset{x\downarrow a}{\lim }f(x)\text{ or }\underset{x\searrow a}{\lim }f(x)\text{ or }f(x+)$$

The limit as $x$ increases in value approaching $a$ ($x$ approaches $a$ "from the left" or "from below") can be denoted:

$$\underset{x\to a^{-}}{\lim }f(x)\text{ or }\underset{x\uparrow a}{\lim }f(x)\text{ or }\underset{x\nearrow a}{\lim }f(x)\text{ or }f(x-)$$

If the limit of $f(x)$ as $x$ approaches $a$ exists then the limits from the left and from the right both exist and are equal. In some cases in which the limit $$\underset{x\to a}{\lim }f(x)$$ does not exist, the two one-sided limits nonetheless exist. Consequently, the limit as $x$ approaches $a$ is sometimes called a "two-sided limit".

It is possible for exactly one of the two one-sided limits to exist (while the other does not exist). It is also possible for neither of the two one-sided limits to exist.

Formal definition

Definition

If $I$ represents some interval that is contained in the domain of $f$ and if $a$ is a point in $I$ then the right-sided limit as $x$ approaches $a$ can be rigorously defined as the value $R$ that satisfies: $$\text{for all }\epsilon >0\text{ there exists some }\delta >0\text{ such that for all }x\in I,\text{ if }0<x-a<\delta \text{ then }|f(x)-R|<\epsilon ,$$ and the left-sided limit as $x$ approaches $a$ can be rigorously defined as the value $L$ that satisfies: $$\text{for all }\epsilon >0\text{ there exists some }\delta >0\text{ such that for all }x\in I,\text{ if }0<a-x<\delta \text{ then }|f(x)-L|<\epsilon .$$

We can represent the same thing more symbolically, as follows.

Let $I$ represent an interval, where $I\subseteq \mathrm{d}\mathrm{o}\mathrm{m}\mathrm{a}\mathrm{i}\mathrm{n}(f)$, and $a\in I$.

$$\underset{x\to a^{+}}{\lim }f(x)=R⟺(\mathrm{\forall }\epsilon \in {\mathbb{ℝ}}_{+},\mathrm{\exists }\delta \in {\mathbb{ℝ}}_{+},\mathrm{\forall }x\in I,(0<x-a<\delta ⟶|f(x)-R|<\epsilon ))$$

$$\underset{x\to a^{-}}{\lim }f(x)=L⟺(\mathrm{\forall }\epsilon \in {\mathbb{ℝ}}_{+},\mathrm{\exists }\delta \in {\mathbb{ℝ}}_{+},\mathrm{\forall }x\in I,(0<a-x<\delta ⟶|f(x)-L|<\epsilon ))$$

Intuition

In comparison to the formal definition for the limit of a function at a point, the one-sided limit (as the name would suggest) only deals with input values to one side of the approached input value.

For reference, the formal definition for the limit of a function at a point is as follows:

$$\underset{x\to a}{\lim }f(x)=L⟺\mathrm{\forall }\epsilon \in {\mathbb{ℝ}}_{+},\mathrm{\exists }\delta \in {\mathbb{ℝ}}_{+},\mathrm{\forall }x\in I,0<|x-a|<\delta ⟹|f(x)-L|<\epsilon$$

To define a one-sided limit, we must modify this inequality. Note that the absolute distance between $x$ and $a$ is $$|x-a|=|(-1)(-x+a)|=|(-1)(a-x)|=|(-1)||a-x|=|a-x|$$.

For the limit from the right, we want $x$ to be to the right of $a$, which means that $a<x$, so $x-a$ is positive. From above, $x-a$ is the distance between $x$ and $a$. We want to bound this distance by our value of $\delta$, giving the inequality $x-a<\delta$. Putting together the inequalities $0<x-a$ and $x-a<\delta$ and using the transitivity property of inequalities, we have the compound inequality $0<x-a<\delta$.

Similarly, for the limit from the left, we want $x$ to be to the left of $a$, which means that $x<a$. In this case, it is $a-x$ that is positive and represents the distance between $x$ and $a$. Again, we want to bound this distance by our value of $\delta$, leading to the compound inequality $0<a-x<\delta$.

Now, when our value of $x$ is in its desired interval, we expect that the value of $f(x)$ is also within its desired interval. The distance between $f(x)$ and $L$, the limiting value of the left sided limit, is $|f(x)-L|$. Similarly, the distance between $f(x)$ and $R$, the limiting value of the right sided limit, is $|f(x)-R|$. In both cases, we want to bound this distance by $\epsilon$, so we get the following: $|f(x)-L|<\epsilon$ for the left sided limit, and $|f(x)-R|<\epsilon$ for the right sided limit.

Examples

Example 1: The limits from the left and from the right of $g(x):=-\frac{1}{x}$ as $x$ approaches $a:=0$ are $$\underset{x\to 0^{-}}{\lim }-1/x=+\mathrm{\infty }\text{ and }\underset{x\to 0^{+}}{\lim }-1/x=-\mathrm{\infty }$$ The reason why $\underset{x\to 0^{-}}{\lim }-1/x=+\mathrm{\infty }$ is because $x$ is always negative (since $x\to 0^{-}$ means that $x\to 0$ with all values of $x$ satisfying $x<0$), which implies that $-1/x$ is always positive so that $\underset{x\to 0^{-}}{\lim }-1/x$ diverges to $+\mathrm{\infty }$ (and not to $-\mathrm{\infty }$) as $x$ approaches $0$ from the left. Similarly, $\underset{x\to 0^{+}}{\lim }-1/x=-\mathrm{\infty }$ since all values of $x$ satisfy $x>0$ (said differently, $x$ is always positive) as $x$ approaches $0$ from the right, which implies that $-1/x$ is always negative so that $\underset{x\to 0^{+}}{\lim }-1/x$ diverges to $-\mathrm{\infty }.$

Plot of the function $1/(1+2^{-1/x}).$

Example 2: One example of a function with different one-sided limits is $f(x)=\frac{1}{1+2^{-1/x}},$ (cf. picture) where the limit from the left is $\underset{x\to 0^{-}}{\lim }f(x)=0$ and the limit from the right is $\underset{x\to 0^{+}}{\lim }f(x)=1.$ To calculate these limits, first show that $$\underset{x\to 0^{-}}{\lim }{2}^{-1/x}=\mathrm{\infty }\text{ and }\underset{x\to 0^{+}}{\lim }{2}^{-1/x}=0$$ (which is true because $\underset{x\to 0^{-}}{\lim }-1/x=+\mathrm{\infty }\text{ and }\underset{x\to 0^{+}}{\lim }-1/x=-\mathrm{\infty }$) so that consequently, $$\underset{x\to 0^{+}}{\lim }\frac{1}{1+2^{-1/x}}=\frac{1}{1+\underset{x\to 0^{+}}{\lim }{2}^{-1/x}}=\frac{1}{1+0}=1$$ whereas $\underset{x\to 0^{-}}{\lim }\frac{1}{1+2^{-1/x}}=0$ because the denominator diverges to infinity; that is, because $\underset{x\to 0^{-}}{\lim }1+2^{-1/x}=\mathrm{\infty }.$ Since $\underset{x\to 0^{-}}{\lim }f(x)\ne \underset{x\to 0^{+}}{\lim }f(x),$ the limit $\underset{x\to 0}{\lim }f(x)$ does not exist.

Relation to topological definition of limit

The one-sided limit to a point $p$ corresponds to the general definition of limit, with the domain of the function restricted to one side, by either allowing that the function domain is a subset of the topological space, or by considering a one-sided subspace, including $p.$ Alternatively, one may consider the domain with a half-open interval topology.

Abel's theorem

A noteworthy theorem treating one-sided limits of certain power series at the boundaries of their intervals of convergence is Abel's theorem.

See also

• Projectively extended real line
• Semi-differentiability
• Limit superior and limit inferior