# Real Analysis: Limits

## Limits

**Motivation**

We would like to be able to describe what happens to the values of a function as the values in its domain get closer and closer to a particular point. For example, if $f$ is a real function, we might be curious as to what happens to $f(x)$ as $x$ approaches $0,$ or perhaps $2$ or $-7.$ Now, if $f$ is actually defined at the value in question, we may well ask why we can't just plug that value in and get the answer. This is a compelling argument, and one that requires a good counterargument.

The first concern is that while $f$ may be defined at the value in question, its value might not coincidence with how we might expect it to act given its immediate neighborhood. Consider the following function:

$$f(x) = \left\{ \begin{array}{cc} (x-2)^2 + 1 & x < 0\\2 & x = 0\\ x - 1 & x > 0 \end{array} \right.$$

As $x$ gets closer and closer to $2,$ $f$ gets closer and closer to $1.$ However, $f(2) = 2.$ Thus we would like to be able to associate the value $2$ with $f(2)$ somehow.

The second concern is that $f$ may not be defined at the value in question in the first place. For example if $f(x) = x + b$ and $f$ is defined over $\mathbb{R} - \{0\},$ we might like a way to say that $f$ gets *really* close to $b$ as $x$ gets *really* close to $0,$ even though $f$ is not defined at $0.$ This second concern is in fact quite important, and it is used to define the *derivative* of a function, which describes the slope of the line tangent to a function at a given point. Thus, we'd like to turn this notion of "value of a function as the values in its domain get really close to some value in particular" into the formal definition of a limit.

**Limits**

The **limit** of a function $f$ from a metric space $(X, d_X)$ into a metric space $(Y, d_Y)$ at a point $c$ equals $L$ if $c$ is a limit point of $X$ and if for every $\varepsilon > 0$ there exists a $\delta > 0$ such that $d_Y(f(a), L) < \varepsilon$ whenever $0 < d_X(c, x) < \delta.$ If this condition is met, then we express this mathematically as

$$\lim_\limits{x \rightarrow c} f(x) = L,$$

which is read aloud as "the limit of $f$ as $x$ approaches $c$ is $L.$" This formal definition of a limit is known as the "$\varepsilon-\delta$ definition" after the Greek letters used in its formulation.

**Details on the Definition of a Limit**

There are two details to note about the definition. One is that the definition requires that $0 < d_X(x, c).$ This requirement precludes us from considering what happens at $c$ itself, as $d(c, c) = 0$ no matter the metric. It could be the case that $f(c) = L,$ that $f(c) \neq L$, or that $c \notin X$ at all, but this has no bearing on the value or existence of the limit. Only those values in $X$ that are sufficiently close to $c$ and their images under $f$ in $Y$ are considered.

The other detail to note is that $c$ must be a limit point of $X.$ This ensures that $X$ has infinitely many points arbitrarily close to $c$ such that we may always ask for a closer one. Stated another way, for every $\delta > 0$, it follows that $B_{\delta}(c) - \{c\} \neq \varnothing.$ This fact precludes the existence of limits at isolated points of $X$, which precludes the existence of limits for functions of the standard metric spaces of $\mathbb{N}$ or $\mathbb{Z}.$

**Continuity Limit Theorem**

The continuity theorem is a powerful theorem about limits. It states that if $f$ is continuous at a limit point $c$ in its domain, then $\lim\limits_{x \rightarrow c} f(x) = f(c).$ This theorem sidesteps the need to evaluate limits by using the epsilon-delta definition for continuous functions, allowing us to simply evaluate the function directly at the limit point instead. The many implications for this theorem for real functions are covered in the next section.

**Composition Limit Theorem**

The composition limit theorem is another powerful theorem about limits that states the conditions under which the composition of two functions has a readily calculable limit. Consider two functions, $g : (X, d_X) \rightarrow (Y_1, d_Y)$ and $f : (Y_2, d_Y) \rightarrow (Z, d_Z),$ where $Y_1,Y_2 \subseteq Y.$ The composition limit theorem states that if $\lim\limits_{x \rightarrow c} g(x) = L,$ $L$ is a limit point of $\text{Dom}(f \circ g),$ and $f$ is continuous at $L,$ then $\lim\limits_{x \rightarrow c} f(g(x)) = f\left(\lim\limits_{x \rightarrow c}g(x)\right).$ Like the continuity limit theorem, the composition limit theorem allows us to directly evaluate limits of composite functions without needing to resort to the epsilon delta definition.

**Open Set Definition of a Limit**

The above definition for a limit is the standard definition for limits in metric spaces. Despite its generality, it is readily amenable to concrete computations, as all you need are an actual function and an actual metric and you can then set to calculating whether the function has a limit at some point of interest. However, there is an even more general definition of a limit that is made in terms of open sets: A function $f$ from $(X', d_X)$ to $(Y, d_Y),$ where $(X', d_X)$ is a subspace of $(X, d_X),$ has a limit $L$ as $x$ approaches $c$ if $c$ is a limit point of $X'$ and for every open set $S \subseteq Y$ containing $L,$ there exists an open set $T \subseteq X$ such that $x \in T$ and $f((T - \{x\})\cap X') \subseteq S.$ This definition is more abstract, but it immediately generalizes from metric spaces to topological spaces. While this generality comes at the cost of applicability, it still implies the standard definition.

## Problems

Limits are unique: Show that if $\lim\limits_{x \rightarrow c} f(x) = a$ and $\lim\limits_{x \rightarrow c} f(x) = b,$ then $a = b.$

Assume $\lim\limits_{x \rightarrow c} f(x) = a$ and $\lim\limits_{x \rightarrow c} f(x) = b.$ Then by definition of limit, it follows that for every $\varepsilon_1 > 0$ there is a $\delta_1 > 0$ such that $d(f(x), a) < \varepsilon_1$ whenever $d(x, c) < \delta_1,$ and for every $\varepsilon_2 > 0$ there is a $\delta_2 > 0$ such that $d(f(x), b) < \varepsilon_2$ whenever $d(x, c) < \delta_2.$ Set $\frac{1}{2}\varepsilon = \varepsilon_1 = \varepsilon_2$ and pick $\delta = \text{min}(\delta_1, \delta_2).$ Then $d(f(x), a) < \frac{1}{2}\varepsilon$ and $d(f(x), b) < \frac{1}{2}\varepsilon$ whenever $d(x, c) < \delta.$ It follows that $d(f(x), a) + d(f(x), b) < \varepsilon.$ By the triangle inequality, it follows that $d(a, b) < \varepsilon,$ and therefore $a = b.$

Consider the metric spaces $(X, d_X)$ and $(Y, d_Y),$ as well as a function $f : X \rightarrow Y.$ Can a limit exist if either or both of the metric spaces are equipped with the discrete metric? If so, under what conditions? If not, why not?

If $d_X$ is the discrete metric on $X$, then no limits exist on $f$ because discrete metrics preclude the possibility of limit points.

If $d_Y$ is the discrete metric, then $\lim\limits_{x \rightarrow c} f(x) = L$ if and only if $c$ is a limit point of $X$ and $f(x) = L$ for all values of $x$ within some open ball $B_{\delta}(c).$

First, assume $\lim\limits_{x \rightarrow c} f(x) = L.$ Then $x$ is a limit point of $X$ and for any $\varepsilon > 0$ there exists a $\delta > 0$ such that $d_Y(f(x), L) < \varepsilon$ whenever $d_X(x, c) < \delta.$ Pick $\varepsilon = \frac{1}{2}.$ Since $d_Y$ is the discrete metric, it must be that $f(x) = L$ for all $x$ when $d_X(x, c) < \delta,$ as then $d_Y(f(x), L) = 0 < \varepsilon.$ It cannot be that $f(x) \neq L$ in this case, as otherwise $d_Y(f(x), L) = 1 > \varepsilon.$

Conversely, assume $c$ is a limit point of $X$ and $f(x) = L$ whenever $x \in B_{\delta}(c) - \{c\}$ for some $\delta > 0.$ Pick $\varepsilon > 0.$ It follows $d_Y(f(x), L) = d_Y(L, L) = 0 < \varepsilon$ whenever $d_X(x, c) < \delta.$ Therefore $\lim\limits_{x \rightarrow c} f(x) = L.$

Let $f$ be a function from $(X, d_X)$ to $(Y, d_Y)$ where $f(x) = a,$ i.e. where $f$ is a constant function. Show that $\lim\limits_{x \rightarrow c} f(x) = a$ for all limit points $x \in X.$

Let $c$ be a limit point of $X$, and pick $\varepsilon > 0$. We would like to find a $\delta$ such that $d_Y(f(x), a) < \varepsilon$ whenever $d_X(x, c) < \delta.$ Pick any $\delta > 0,$ such as $\delta = 1.$ Then for all $x$ where $d_X(x, c) < \delta,$ it follows that $d_Y(f(x), a) = d_Y(a, a) = 0 < \varepsilon,$ as $f(x) = a$ by definition. Therefore $\lim\limits_{x \rightarrow c} f(x) = a.$

Continuity Limit Theorem: Show that if $c$ is a limit point of $X,$ then a function $f : (X, d_X) \rightarrow (Y, d_Y)$ is continuous at $c$ if and only if $\lim\limits_{x \rightarrow c} f(x) = f(c).$

Assume $c$ is a limit point of $X$ and $f$ is continuous at $c.$ Then $c \in X$ and for every $\varepsilon > 0$ there exists a $\delta > 0$ such that $d_Y(f(x), f(c)) < \varepsilon$ whenever $d_X(x, c) < \delta.$ It follows that $\lim\limits_{x \rightarrow c} f(x) = f(c).$

Conversely, assume $\lim\limits_{x \rightarrow c} f(x) = f(c).$ By definition of limit, $c$ is a limit point of $X,$ and for every $\varepsilon > 0$ there exists a $\delta > 0$ such that $d_Y(f(x), f(c)) < \varepsilon$ whenever $0 < d_X(x, c) < \delta.$ Since $c \in X,$ this condition also holds when $x = c.$ Therefore $f$ is continuous at $c.$

Composition Limit Theorem: Consider the functions $g : (X, d_X) \rightarrow (Y_1, d_Y)$ and $f : (Y_2, d_Y) \rightarrow (Z, d_Z)$ where $Y_1, Y_2 \subseteq Y.$ Show that if $\lim\limits_{x \rightarrow c} g(x) = L,$ $c$ is a limit point of $\text{Dom}(f \circ g),$ and $f$ is continuous at $L,$ then $\lim\limits_{x \rightarrow c} f(g(x)) = f\left(\lim\limits_{x \rightarrow c} g(x)\right).$

We would like to show that for any given $\varepsilon > 0$ there exists a $\delta > 0$ such that $d_Z( f(g(x)), f(L)) < \varepsilon$ whenever $d_X(x, c) < \delta.$

Since $f$ is continuous at $L,$ for every $\varepsilon > 0$ there exists a $\delta_f > 0$ such that $d_Z(f(y), f(L)) < \varepsilon$ whenever $d_Y(y, L) < \delta_f.$ Likewise, since $\lim\limits_{x \rightarrow c} g(x) = L,$ for every $\varepsilon_g > 0$ there exists a $\delta > 0$ such that $d_Y(g(x), L) < \varepsilon_g$ whenever $0 < d_X(x, c) < \delta.$ Pick $\varepsilon_g = \delta_f.$ Then $d_Z(f(g(x)), f(L)) < \varepsilon$ whenever $d_Y(g(x), L) < \varepsilon_g,$ which in turn is true whenever $0 < d_X(x, c) < \delta.$ Since $c$ is a limit point of $\text{Dom}(f \circ g)$, it follows that $\lim\limits_{x \rightarrow c} f(g(x)) = f(L) = f\left(\lim\limits_{x \rightarrow c} g(x) \right).$

Counterexample: Show that it is not enough for $L$ to be a limit point of $Y_2$ in the composition limit theorem for the theorem to hold.

Consider $f : [0, \infty) \rightarrow \mathbb{R}$ where $f(x) = \sqrt{x}$ and $g : (-\infty, 0] \rightarrow \mathbb{R}$ where $g(x) = \sqrt{-x}.$ Note that $\lim\limits_{x \rightarrow c} g(x) = 0$ and that $0$ is a limit point of $\text{Dom}(f).$ However, $\text{Dom}(f \circ g) = \{0\}.$ Since $\text{Dom}(f \circ g)$ is finite, it follows that $0$ is an isolated point and therefore not a limit point of $\text{Dom}(f \circ g).$ As a result, $\lim\limits_{x \rightarrow c} f(g(x))$ is undefined, even though $f\left(\lim\limits_{x \rightarrow 0}g(x)\right) = f(0) = 0$ is defined.

Counterexample: In the composition limit theorem, show that swapping the condition that $f$ be continuous at $L$ for the condition that it have a limit at $L$ breaks the theorem.

Consider the function $f : (0, \infty) \rightarrow \mathbb{R}$ where $f(x) = \sqrt{x}.$ We can see that $0$ is a limit point of $\text{Dom}(f),$ and that $\lim\limits_{x \rightarrow 0} f(x) = 0.$ However, consider the function $g : \mathbb{R} \rightarrow \mathbb{R}$ where $g(x) = x.$ Then $\lim\limits_{x \rightarrow 0} f(g(x)) = \lim\limits_{x \rightarrow 0} \sqrt{x} = 0,$ but $f(\lim\limits_{x \rightarrow 0}g(x)) = f(0),$ which is undefined.

Show that the open set definition of a limit implies the standard epsilon-delta definition of a limit.

Let $f$ be a function from a metric space $(X', d_X)$ to $(Y, d_Y)$ where $(X', d_X)$ is a subspace of $(X, d_X).$ Further, assume $c$ is a limit point of $X'$ and that for every open set $S \subseteq Y$ containing $L$ there exists an open set $T \subseteq X$ such that $c \in T$ and $f(T - \{c\}) \subseteq S.$

We would like to prove that for every $\varepsilon > 0$ there exists a $\delta > 0$ such that $d(f(x), L) < \varepsilon$ whenever $d(x, c) < \delta.$

First, set $\varepsilon > 0.$ Then $B_{\varepsilon}(L)$ is an open set in $Y$ that contains $L.$ This implies that there exists some open set $T \subseteq X$ such that $c \in T$ and $f(T - \{c\}) \subseteq B_{\varepsilon}(L).$ Since $T$ is an open set containing $c,$ there is some $\delta > 0$ such that $B_{\delta}(c) \subseteq T.$ Then whenever $0 < d_X(x, c) < \delta,$ it follows that $x \in B_{\delta}(c) - \{c\} \subseteq T - \{c\}.$ From this it follows that $f(x) \in f(B_{\delta}(c) - \{c\}) \subseteq f(T - \{c\}) \subseteq B_{\varepsilon}(L).$ This in turn implies that $d_Y(f(x), L) < \varepsilon.$