One-sided limit
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In calculus, a one-sided limit refers to either one of the two limits of a function f ( x ) {\displaystyle f(x)} of a real variable x {\displaystyle x} as x {\displaystyle x} approaches a specified point either from the left or from the right.
The limit, as x {\displaystyle x} decreases in value approaching a {\displaystyle a} (x {\displaystyle x} approaches a {\displaystyle a} "from the right" or "from above"), is denoted:
lim x → a + f ( x ) or lim x ↓ a f ( x ) or lim x ↘ a f ( x ) or f ( a + ) . {\displaystyle \lim _{x\to a^{+}}f(x)\quad {\text{ or }}\quad \lim _{x\,\downarrow \,a}\,f(x)\quad {\text{ or }}\quad \lim _{x\searrow a}\,f(x)\quad {\text{ or }}\quad f(a+).}
The limit, as x {\displaystyle x} increases in value approaching a {\displaystyle a} (x {\displaystyle x} approaches a {\displaystyle a} "from the left" or "from below"), is denoted:
lim x → a − f ( x ) or lim x ↑ a f ( x ) or lim x ↗ a f ( x ) or f ( a − ) . {\displaystyle \lim _{x\to a^{-}}f(x)\quad {\text{ or }}\quad \lim _{x\,\uparrow \,a}\,f(x)\quad {\text{ or }}\quad \lim _{x\nearrow a}\,f(x)\quad {\text{ or }}\quad f(a-).}
If the limits from the left and right both exist and are equal, then the limit of f ( x ) {\displaystyle f(x)} as x {\displaystyle x} approaches a {\displaystyle a} exists. Conversely, if the limit of f ( x ) {\displaystyle f(x)} as x {\displaystyle x} approaches a {\displaystyle a} exists, then the limits from left and right both exist and are equal. Consequently, the limit as x {\displaystyle x} approaches a {\displaystyle a} is sometimes called a "two-sided limit".[citation needed] It is denoted: lim x → a f ( x ) . {\displaystyle \lim _{x\to a}f(x).}
In some cases in which the two-sided limit does not exist, the two individual one-sided limits nonetheless exist and they are then necessarily unequal.
It is possible for only one of the two one-sided limits to exist. It is also possible for neither of the two one-sided limits to exist.
Formal definition
Definition
If I {\displaystyle I} represents some interval that is contained in the domain of a function f {\displaystyle f} and if a {\displaystyle a} is a point in I {\displaystyle I}, then the right-sided limit as x {\displaystyle x} approaches a {\displaystyle a} can be rigorously defined as the value R {\displaystyle R} that satisfies:
for all ε > 0 {\displaystyle \varepsilon >0} there exists some δ > 0 {\displaystyle \delta >0} such that for all x ∈ I {\displaystyle x\in I}, if 0 < x − a < δ {\displaystyle 0<x-a<\delta } then | f ( x ) − R | < ε {\displaystyle |f(x)-R|<\varepsilon },
and the left-sided limit as x {\displaystyle x} approaches a {\displaystyle a} can be rigorously defined as the value L {\displaystyle L} that satisfies:
for all ε > 0 {\displaystyle \varepsilon >0} there exists some δ > 0 {\displaystyle \delta >0} such that for all x ∈ I {\displaystyle x\in I}, if 0 < a − x < δ {\displaystyle 0<a-x<\delta } then | f ( x ) − L | < ε {\displaystyle |f(x)-L|<\varepsilon }.
These definitions can be represented more symbolically as follows: Let I {\displaystyle I} represent an interval, where I ⊆ d o m a i n ( f ) {\displaystyle I\subseteq \mathrm {domain} (f)} and a ∈ I {\displaystyle a\in I}, then lim x → a + f ( x ) = R ⟺ ∀ ε ∈ R + , ∃ δ ∈ R + , ∀ x ∈ I , 0 < x − a < δ ⟶ | f ( x ) − R | < ε , lim x → a − f ( x ) = L ⟺ ∀ ε ∈ R + , ∃ δ ∈ R + , ∀ x ∈ I , 0 < a − x < δ ⟶ | f ( x ) − L | < ε . {\displaystyle {\begin{aligned}\lim _{x\to a^{+}}f(x)=R&\iff \forall \varepsilon \in \mathbb {R} _{+},\exists \delta \in \mathbb {R} _{+},\forall x\in I,0<x-a<\delta \longrightarrow |f(x)-R|<\varepsilon ,\\\lim _{x\to a^{-}}f(x)=L&\iff \forall \varepsilon \in \mathbb {R} _{+},\exists \delta \in \mathbb {R} _{+},\forall x\in I,0<a-x<\delta \longrightarrow |f(x)-L|<\varepsilon .\end{aligned}}}
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:
lim x → a f ( x ) = L ⟺ ∀ ε ∈ R + , ∃ δ ∈ R + , ∀ x ∈ I , 0 < | x − a | < δ ⟹ | f ( x ) − L | < ε . {\displaystyle \lim _{x\to a}f(x)=L~~~\iff ~~~\forall \varepsilon \in \mathbb {R} _{+},\exists \delta \in \mathbb {R} _{+},\forall x\in I,0<|x-a|<\delta \implies |f(x)-L|<\varepsilon .}
To define a one-sided limit, we must modify this inequality. Note that the absolute distance between x {\displaystyle x} and a {\displaystyle a} is
| x − a | = | ( − 1 ) ( − x + a ) | = | ( − 1 ) ( a − x ) | = | ( − 1 ) | | a − x | = | a − x | . {\displaystyle |x-a|=|(-1)(-x+a)|=|(-1)(a-x)|=|(-1)||a-x|=|a-x|.}
For the limit from the right, we want x {\displaystyle x} to be to the right of a {\displaystyle a}, which means that a < x {\displaystyle a<x}, so x − a {\displaystyle x-a} is positive. From above, x − a {\displaystyle x-a} is the distance between x {\displaystyle x} and a {\displaystyle a}. We want to bound this distance by our value of δ {\displaystyle \delta }, giving the inequality x − a < δ {\displaystyle x-a<\delta }. Putting together the inequalities 0 < x − a {\displaystyle 0<x-a} and x − a < δ {\displaystyle x-a<\delta } and using the transitivity property of inequalities, we have the compound inequality 0 < x − a < δ {\displaystyle 0<x-a<\delta }.
Similarly, for the limit from the left, we want x {\displaystyle x} to be to the left of a {\displaystyle a}, which means that x < a {\displaystyle x<a}. In this case, it is a − x {\displaystyle a-x} that is positive and represents the distance between x {\displaystyle x} and a {\displaystyle a}. Again, we want to bound this distance by our value of δ {\displaystyle \delta }, leading to the compound inequality 0 < a − x < δ {\displaystyle 0<a-x<\delta }.
Now, when our value of x {\displaystyle x} is in its desired interval, we expect that the value of f ( x ) {\displaystyle f(x)} is also within its desired interval. The distance between f ( x ) {\displaystyle f(x)} and L {\displaystyle L}, the limiting value of the left sided limit, is | f ( x ) − L | {\displaystyle |f(x)-L|}. Similarly, the distance between f ( x ) {\displaystyle f(x)} and R {\displaystyle R}, the limiting value of the right sided limit, is | f ( x ) − R | {\displaystyle |f(x)-R|}. In both cases, we want to bound this distance by ε {\displaystyle \varepsilon }, so we get the following: | f ( x ) − L | < ε {\displaystyle |f(x)-L|<\varepsilon } for the left sided limit, and | f ( x ) − R | < ε {\displaystyle |f(x)-R|<\varepsilon } for the right sided limit.
Examples
Example 1. The limits from the left and from the right of g ( x ) := − 1 x {\textstyle g(x):=-{\frac {1}{x}}} as x {\displaystyle x} approaches a := 0 {\displaystyle a:=0} are, respectively lim x → 0 − − 1 x = + ∞ and lim x → 0 + − 1 / x = − ∞ . {\displaystyle \lim _{x\to 0^{-}}-{\frac {1}{x}}=+\infty \qquad {\text{ and }}\qquad \lim _{x\to 0^{+}}{-1/x}=-\infty .} The reason why lim x → 0 − − 1 x = + ∞ {\textstyle \lim _{x\to 0^{-}}-{\frac {1}{x}}=+\infty } is because x {\displaystyle x} is always negative (since x → 0 − {\displaystyle x\to 0^{-}} means that x → 0 {\displaystyle x\to 0} with all values of x {\displaystyle x} satisfying x < 0 {\displaystyle x<0}), which implies that − 1 / x {\displaystyle -1/x} is always positive so that lim x → 0 − − 1 x {\textstyle \lim _{x\to 0^{-}}-{\frac {1}{x}}} diverges to + ∞ {\displaystyle +\infty } (and not to − ∞ {\displaystyle -\infty }) as x {\displaystyle x} approaches 0 {\displaystyle 0} from the left. Similarly, lim x → 0 + − 1 x = − ∞ {\textstyle \lim _{x\to 0^{+}}-{\frac {1}{x}}=-\infty } since all values of x {\displaystyle x} satisfy x > 0 {\displaystyle x>0} (said differently, x {\displaystyle x} is always positive) as x {\displaystyle x} approaches 0 {\displaystyle 0} from the right, which implies that − 1 / x {\displaystyle -1/x} is always negative so that lim x → 0 + − 1 x {\textstyle \lim _{x\to 0^{+}}-{\frac {1}{x}}} diverges to − ∞ . {\displaystyle -\infty .}
Example 2. One example of a function with different one-sided limits is f ( x ) = 1 1 + 2 − 1 / x {\textstyle f(x)={\frac {1}{1+2^{-1/x}}}}, where the limit from the left is lim x → 0 − f ( x ) = 0 {\displaystyle \lim _{x\to 0^{-}}f(x)=0} and the limit from the right is lim x → 0 + f ( x ) = 1. {\displaystyle \lim _{x\to 0^{+}}f(x)=1.} To calculate these limits, first show that lim x → 0 − 2 − 1 / x = ∞ and lim x → 0 + 2 − 1 / x = 0 , {\displaystyle \lim _{x\to 0^{-}}2^{-1/x}=\infty \qquad {\text{ and }}\qquad \lim _{x\to 0^{+}}2^{-1/x}=0,} which is true because lim x → 0 − − 1 / x = + ∞ {\textstyle \lim _{x\to 0^{-}}{-1/x}=+\infty } and lim x → 0 + − 1 / x = − ∞ {\textstyle \lim _{x\to 0^{+}}{-1/x}=-\infty } so that consequently, lim x → 0 + 1 1 + 2 − 1 / x = 1 1 + lim x → 0 + 2 − 1 / x = 1 1 + 0 = 1 {\displaystyle \lim _{x\to 0^{+}}{\frac {1}{1+2^{-1/x}}}={\frac {1}{1+\displaystyle \lim _{x\to 0^{+}}2^{-1/x}}}={\frac {1}{1+0}}=1} whereas lim x → 0 − 1 1 + 2 − 1 / x = 0 {\textstyle \lim _{x\to 0^{-}}{\frac {1}{1+2^{-1/x}}}=0} because the denominator diverges to infinity; that is, because lim x → 0 − 1 + 2 − 1 / x = ∞ {\displaystyle \lim _{x\to 0^{-}}1+2^{-1/x}=\infty }. Since lim x → 0 − f ( x ) ≠ lim x → 0 + f ( x ) {\displaystyle \lim _{x\to 0^{-}}f(x)\neq \lim _{x\to 0^{+}}f(x)}, the limit lim x → 0 f ( x ) {\displaystyle \lim _{x\to 0}f(x)} does not exist.
Relation to topological definition of limit
The one-sided limit to a point p {\displaystyle 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 . {\displaystyle p.}[verification needed] Alternatively, one may consider the domain with a half-open interval topology.[citation needed]
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.[citation needed]