One-sided limit

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:

$$\lim_{x \to a^+}f(x) \quad \text{ or } \quad \lim_{x\,\downarrow\,a}\,f(x) \quad \text{ or } \quad \lim_{x \searrow a}\,f(x) \quad \text{ or } \quad f(x+)$$

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

$$\lim_{x \to a^-}f(x) \quad \text{ or } \quad \lim_{x\,\uparrow\,a}\, f(x) \quad \text{ or } \quad \lim_{x \nearrow a}\,f(x) \quad \text{ or } \quad f(x-)$$

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 $$\lim_{x \to a} f(x)$$ 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.

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: $$\text{for all } \varepsilon > 0\;\text{ there exists some } \delta > 0 \;\text{ such that for all } x \in I, \text{ if } \;0 < x - a < \delta \text{ then } |f(x) - R| < \varepsilon,$$ and the left-sided limit as $$x$$ approaches $$a$$ can be rigorously defined as the value $$L$$ that satisfies: $$\text{for all } \varepsilon > 0\;\text{ there exists some } \delta > 0 \;\text{ such that for all } x \in I, \text{ if } \;0 < a - x < \delta \text{ then } |f(x) - L| < \varepsilon.$$

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

Let $$I$$ represent an interval, where $$I \subseteq \mathrm{domain}(f)$$, and $$a \in I $$.



\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))

$$

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 \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 $$x$$ 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 $$\varepsilon$$, so we get the following: $$|f(x) - L| < \varepsilon$$ for the left sided limit, and $$|f(x) - R| < \varepsilon$$ 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 $$\lim_{x \to 0^-} {-1/x} = + \infty \qquad \text{ and } \qquad \lim_{x \to 0^+} {-1/x} = - \infty$$ The reason why $$\lim_{x \to 0^-} {-1/x} = + \infty$$ 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 $$\lim_{x \to 0^-} {-1/x}$$ diverges to $$+ \infty$$ (and not to $$- \infty$$) as $$x$$ approaches $$0$$ from the left. Similarly, $$\lim_{x \to 0^+} {-1/x} = - \infty$$ 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 $$\lim_{x \to 0^+} {-1/x}$$ diverges to $$- \infty.$$

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 $$\lim_{x \to 0^-} f(x) = 0$$ and the limit from the right is $$\lim_{x \to 0^+} f(x) = 1.$$ To calculate these limits, first show that $$\lim_{x \to 0^-} 2^{-1/x} = \infty \qquad \text{ and } \qquad \lim_{x \to 0^+} 2^{-1/x} = 0$$ (which is true because $$\lim_{x \to 0^-} {-1/x} = + \infty \text{ and } \lim_{x \to 0^+} {-1/x} = - \infty$$) so that consequently, $$\lim_{x \to 0^+} \frac{1}{1 + 2^{-1/x}} = \frac{1}{1 + \displaystyle\lim_{x \to 0^+} 2^{-1/x}} = \frac{1}{1 + 0} = 1$$ whereas $$\lim_{x \to 0^-} \frac{1}{1 + 2^{-1/x}} = 0$$ because the denominator diverges to infinity; that is, because $$\lim_{x \to 0^-} 1 + 2^{-1/x} = \infty.$$ Since $$\lim_{x \to 0^-} f(x) \neq \lim_{x \to 0^+} f(x),$$ the limit $$\lim_{x \to 0} f(x)$$ 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.