# Definition:One-Sided Limit of Real Function

## Definition

A one-sided limit of real function is a right-hand limit of real function or a left-hand limit of real function.

### Limit from Right

Let $\Bbb I = \openint a b$ be an open real interval.

Let $f: \Bbb I \to \R$ be a real function.

Let $L \in \R$.

Suppose that:

$\forall \epsilon \in \R_{>0}: \exists \delta \in \R_{>0}: \forall x \in \Bbb I: a < x < a + \delta \implies \size {\map f x - L} < \epsilon$

where $\R_{>0}$ denotes the set of strictly positive real numbers.

That is, for every real strictly positive $\epsilon$ there exists a real strictly positive $\delta$ such that every real number in the domain of $f$, greater than $a$ but within $\delta$ of $a$, has an image within $\epsilon$ of $L$.

Then $\map f x$ is said to tend to the limit $L$ as $x$ tends to $a$ from the right, and we write:

$\map f x \to L$ as $x \to a^+$

or

$\ds \lim_{x \mathop \to a^+} \map f x = L$

This is voiced

the limit of $\map f x$ as $x$ tends to $a$ from the right

and such an $L$ is called:

a limit from the right.

### Limit from Left

Let $\openint a b$ be an open real interval.

Let $f: \openint a b \to \R$ be a real function.

Let $L \in \R$.

Suppose that:

$\forall \epsilon \in \R_{>0}: \exists \delta \in \R_{>0}: \forall x \in \R: b - \delta < x < b \implies \size {\map f x - L} < \epsilon$

where $\R_{>0}$ denotes the set of strictly positive real numbers.

That is, for every real strictly positive $\epsilon$ there exists a real strictly positive $\delta$ such that every real number in the domain of $f$, less than $b$ but within $\delta$ of $b$, has an image within $\epsilon$ of $L$.

Then $\map f x$ is said to tend to the limit $L$ as $x$ tends to $b$ from the left, and we write:

$\map f x \to L$ as $x \to b^-$

or

$\ds \lim_{x \mathop \to b^-} \map f x = L$

This is voiced:

the limit of $\map f x$ as $x$ tends to $b$ from the left

and such an $L$ is called:

a limit from the left.

## Examples

### Example: $\dfrac 1 {1 + e^{1 / x} }$

Let $f: \R \to \R$ be the real function defined as:

$\map f x = \dfrac 1 {1 + e^{1 / x} }$

Then:

 $\ds \lim_{x \mathop \to 0^+} \map f x$ $=$ $\ds 0$ $\ds \lim_{x \mathop \to 0^-} \map f x$ $=$ $\ds 1$