# Real Analysis: Limits

## Holes, Jumps, and Removable Discontinuities

**Motivation**

Sometimes we come across a function that has certain values we might want to either change or fill in. For example, if a function is defined everywhere but a single point, is it possible to "fill in" the value of the function at that point in a reasonable way? Or what if the function is defined at a particular point, but it somehow doesn't fit with what we'd like it to be? Can we find a reasonable replacement for it? The answers to these questions relies on the existence of limit values at the points in questions.

This section focuses on categorizing several kinds of holes and jumps in the domains and images real functions. While some of these concepts could have been defined previously, a proper discussion requires the language of open sets, continuity, and limits. A full categorization of holes and jumps must include whether a function is defined at a point, the existence of one-sided and two-sided limits at that point, the continuity of the function at that point, and the continuity of the function around that point.

**Holes and Removable Holes**

Let $f$ be a function from $(X', d_X)$ to $(Y, d_Y)$ where $(X', d_X)$ is a subspace of $(X, d_X).$ $f$ has a **hole** at $c \in X$ if $c$ is a limit point of $X'$ but $f$ is not defined at $c.$ If $\lim\limits_{x \rightarrow c} f(x)$ exists, then $c$ is a **removable hole**. Note that the holes of a function are a subset of the points outside it domain, but not necessarily equal to it.

Holes can be further categorize for real functions. If $f$ is a real function that has a hole at $c,$ and the left limit of $f$ at $c$ exists, then $f$ has a **left removable hole** at $c.$ Likewise, if the right limit of $f$ is defined at $c,$ then $f$ has a **right removable hole** at $c.$ Note that $c$ may be both left and right removable without being removable!

**Hole and Point Discontinuities**

If $f$ is continuous in a neighborhood around a hole at a limit point $c$ of its domain, then $c$ is a **hole discontinuity**. If instead $f$ is continuous in a neighborhood around $c$ and $f$ is defined at $c$ but discontinuous at $c,$ then $c$ is a **point discontinuity**.

If $c$ is a hole or a point discontinuity and the limit of $f$ at $c$ exists, then $c$ is a **removable discontinuity**. If $c$ is a hole or a point discontinuity and both the left and right limits of $f$ as $x$ approaches $c$ exist but are not equal, then $c$ is a **jump discontinuity**. Jump discontinuities may be further categorized into **point jump discontinuities** and **hole jump discontinuities** depending on whether $c$ is in the domain of $f$ or not, respectively.

**Algebraically Removable Holes**

Let $f$ be a function of the form $f(x) = \frac{g(x)}{g(x)}h(x).$ If $f$ is defined in a limit point $c$ of its domain, $g$ is nonzero in a neighborhood around $c,$ and $\lim\limits_{x \rightarrow c}h(x)$ is defined, then $c$ is an **algebraically removable hole**. This name is due to the fact that the $\frac{g(x)}{g(x)}$ term equals $1$ in all places where it is defined and nonzero, and thus $f$ is equal to $h$ at all points except those where $g$ is undefined or equal to $0.$ A hole at $c$ due to either of these conditions for $g$ can thus be removed by algebraically removing the $\frac{g(x)}{g(x)}$ term to leave only the $h(x)$ term.

**Limit Law for Algebraically Removable Holes**

The limit law for algebraically removable holes states that if $f$ is a real function of the form $f(x) = \frac{g(x)}{g(x)}h(x)$ with an algebraically removable hole at $c,$ then $\lim\limits_{x \rightarrow c} f(x) = \lim\limits_{x \rightarrow c} h(x).$ Since in many applications $h$ is a continuous function, this theorem is often called the limit law for removable discontinuities.

This limit law is very useful for efficiently evaluating limits of rational functions and functions like them. For example, consider the function $f(x) = \frac{x^2}{x}.$ This function is identical to $h(x) = x$ at all points except $x = 0,$ where it is undefined. We can break $f$ down into two terms to see this: $f(x) = \frac{x}{x} \cdot x.$ The $\frac{x}{x}$ term equals $1$ whenever $x \neq 0$ and is undefined when $x = 0.$ The limit law for algebraically removable holes lets us ignore this term. Since the remaining $x$ term is continuous, we can use the limit law for continuous functions to directly evaluate the limit to conclude that $\lim\limits_{x \rightarrow 0} \frac{x^2}{x} = 0.$

## Problems

Limit Law for Algebraically Removable Discontinuities: Show that if $f$ is a real function of the form $f(x) = \frac{g(x)}{g(x)}h(x),$ $c$ is a limit point of $\text{Dom}(f),$ $g$ is nonzero in an interval around $c,$ and $\lim\limits_{x \rightarrow c} h(x)$ is defined, then $\lim\limits_{x \rightarrow c} f(x) = \lim\limits_{x \rightarrow c}h(x).$

Let $f$ be a real function of the form $f(x) = \frac{g(x)}{g(x)}h(x),$ $c$ be a limit point of $\text{Dom}(f),$ $g$ be nonzero in an interval $(a, b)$ containing $c,$ and $\lim\limits_{x \rightarrow c}h(x)$ be defined.

First, use the algebraic limit theorem to break the limit for $f$ into two parts:

$ \lim\limits_{x \rightarrow c}f(x) = \lim\limits_{x \rightarrow c} \dfrac{g(x)}{g(x)}h(x) \\ \lim\limits_{x \rightarrow c}f(x) = \left(\lim\limits_{x \rightarrow c} \dfrac{g(x)}{g(x)}\right) \left(\lim\limits_{x \rightarrow c} h(x) \right) \\ $

Since $\lim\limits_{x \rightarrow c} h(x),$ is given, we must determine whether $\lim\limits_{x \rightarrow c} \frac{g(x)}{g(x)}$ is defined and, if so, what its value is. First, note that because $c$ is a limit point of $\text{Dom}(f),$ and $\text{Dom}(f) \subseteq \text{Dom}(g),$ it follows that $c$ is a limit point of $\text{Dom}(g).$ Next, because $g$ is nonzero in the interval $(a, b)$ containing $c,$ we conclude that $\frac{g(x)}{g(x)} = 1$ in $(a, b) - \{c\}.$ Pick $\delta = \text{min}\{c - a, b - a\}.$ Then for any $\varepsilon > 0$ it follows that $\left|\frac{g(x)}{g(x)} - 1\right| = |1 - 1| = 0 < \varepsilon$ whenever $0 < x < \delta.$ Therefore $\lim\limits_{x \rightarrow c} \frac{g(x)}{g(x)} = 1,$ and so we conclude that $\lim\limits_{x \rightarrow c} f(x) = \lim\limits_{x \rightarrow c} h(x).$