Functional limits

and a little continuity

This material is covered in sections 4.1 and 4.2 of our text, as well as in this YouTube video.

Definition

Let's jump right in with the definition of limit:

Def 4.2.1 Functional Limit Let $f:A\to\mathbb R$, and let $c$ be a limit point of the domain $A$. We say that $\displaystyle \lim_{x\to c}f(x) = L$ provided that, for all $\varepsilon > 0$, there exists $\delta > 0$ such that whenever $$0 < |x − c| < \delta \text{ and } x\in A,$$ it follows that $|f(x) − L| < \varepsilon$.

Comments

  • This definition and its application captures the fundamental idea of epsilonics that is at the forefront of our student learning objectives outlined in our transition plan.
  • We say $\displaystyle \lim_{x\to c} f(x) = L$ if we can make $f(x)$ as close as we like to $L$ by taking $x$ to be as close as necessary to $c$, but not equal to $c$.
  • In the formal definition, $\varepsilon$ measures "as close as we'd like $f(x)$ to be to $L$" and $\delta$ measures "as close as $x$ needs to be to $c$".

Examples

Let's take a look at a couple of specific examples of $\varepsilon$-$\delta$ proofs of statements of the form $$\lim_{x\to c} f(x) = L.$$

Typically, these proofs have the form:

Let $\varepsilon > 0$ and choose $\delta>0$ such that yada yada. Then, $0<|x-c|<\delta$ implies $$|f(x)-L| < \cdots < \varepsilon.$$

Note that the "yada yada" is generally some sort of upper bound on $\delta$ and the $\cdots$ are computations using that upper bound that magically lead to $\varepsilon$.

Example 1

Claim: $\displaystyle \lim_{x\to2} (4x-1) = 7$.

Proof: Let $\varepsilon>0$ and choose $\delta > 0$ such that $\delta < \varepsilon/4$.

Then, $0<|x-2|<\delta$ implies

$$\begin{align} |4x - 1 - 7| &= |4x - 8| \\ &= 4|x-2| < 4 \times \varepsilon/4 = \varepsilon. \Box \end{align}$$

Example 2

Claim: $\displaystyle \lim_{x\to3} (x^2 - 2x - 1) = 2$.

Proof: Let $\varepsilon>0$ and choose $\delta > 0$ such that $\delta < \min(1,\varepsilon/5)$.

Then, $0<|x-3|<\delta$ implies that $|x+1|<5$, since $\delta<1$ so that $x$ could be close to $4$. In addition,

$$\begin{align} |x^2 - 2x - 1 - 2| &= |x^2 - 2x - 3| \\ &= |(x-3)(x+1)| < 5|x-3| \\ &< 5 \times \varepsilon/5 = \varepsilon. \Box \end{align}$$

Algebraic limit laws

Cor 4.2.4 Algebraic Limit Theorem for Functional Limits Let $f$ and $g$ be functions defined on a domain $A\subset\mathbb R$, and assume $\displaystyle \lim_{x\to c} f(x) = L$ and $\displaystyle \lim_{x\to c} g(x) = M$ for some limit point $c$ of $A$. Then,

  1. $\displaystyle \lim_{x\to c} kf(x) = kL$ for all $k\in\mathbb R$
  2. $\displaystyle \lim_{x\to c} (f(x)+g(x)) = L+M$
  3. $\displaystyle \lim_{x\to c} (f(x)g(x)) = LM$
  4. $\displaystyle \lim_{x\to c} (f(x)/g(x)) = L/M$, provided $M\neq0$.

Comments

  • Essentially, the algebraic limit laws say that limits interact nicely with respect to algebraic operations.
  • There's a similar theorem for sequential limits. In fact, Cor 4.2.4 is stated as a corollary because the texts proves it as consequence of the sequential limit laws.
  • A simple consequence of the limit laws is that if $f$ is a polynomial or rational function and $c$ is in the domain of $f$, then $$\lim_{x\to c}f(x) = f(c).$$ A function with this property is called continuous.

Proofs

The textbook goes into the relationship between functional and sequential limits in great detail and, in particular, derives the functional limit laws as consequences of the sequential limit laws. That's all good stuff to know and you might want to have a look.

Both sets of limit laws follow directly from the definition, which offers another look at epsilonics. So, let's take a look at how one of those might go.

As we examine this proof, recall that the theorem explicitly assumes that $\displaystyle \lim_{x\to c} f(x) = L$ and $\displaystyle \lim_{x\to c} g(x) = M$.

The sum rule

Claim: $\displaystyle \lim_{x\to c} (f(x)+g(x)) = L+M$.

Proof: Let $\varepsilon>0$ and choose $\delta>0$ such that $0<|x-c|<\delta$ implies both $$|f(x)-L| < \frac{\varepsilon}{2} \: \text{ and } \: |g(x)-M| < \frac{\varepsilon}{2}.$$ Then, for $0<|x-c|<\delta$, we have $$\begin{align}|(f(x)+g(x)) - (L+M)| &= |(f(x) - L) + (g(x) - M)| \\ &\leq |f(x) - L| + |g(x)-M| < \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon.\end{align}$$

A couple of crazy examples

Our text features strange examples at the beginning of each chapter, generally to help us understand why we need to think about this stuff carefully. This examples in this chapter are called Dirichlet's function and Thomae's function.

Relationship with continuity

Both of these examples are strange with respect to limits and also with respect to continuity.

Recall that, a simple definition of continuity is $$\lim_{x\to c}f(x) = f(c).$$

We'll discuss that in more detail next time.

Dirichlet's function

Dirichlet's function is defined by $$f(x) = \begin{cases}1 & x\in\mathbb Q \\ 0 & x\not\in\mathbb Q.\end{cases}$$ Note that on any open interval, $f$ assumes values $0$ and $1$ and only those values.

In particlar, given any $c\in\mathbb R$, the maximum value of $|f(x)-L|$ over any $\delta$ neighborhood of $c$ is always at least $1/2$. Thus, we can't make $|f(x)-L|<1/2$ for all $x$ in a neighborhood of $c$.

Dirichlet's picture

I guess that the graph of Dirichlet's function might look something like so:

Thomae's function

Thomae's function is even crazier. It's defined by $$g(x) = \begin{cases}\frac{1}{n} & \text{if } x = \frac{m}{n} \text{ in reduced terms} \\ 0 & \text{if } x\not\in\mathbb Q.\end{cases}$$

It has the very peculiar property of being continuous at each irrational number but discontinuous at each rational.

Thomae's picture

The graph of Thomae's function might look like so:

Thomae's limits

It turns out that $$\lim_{x\to c} g(x) = 0$$ for all $c\in\mathbb R$.

It's kinda fun to think about why that might be!