<Home>Teaching><MATH 131A (Winter 2021)>Week 8 Discussion Notes

Week 8 Discussion Notes

Homework Comments
- 9.3
Continuity
- Introduction
True or False

Homework Comments

9.3

Here is the problem statement:

Suppose $\displaystyle\lim_{n\to\infty} a_n = a$ , $\displaystyle\lim_{n\to\infty} b_n = b$ , and $s_n = \dfrac{a_n^3+4a_n}{b_n^2+1}$ . Prove $\displaystyle\lim_{n\to\infty} s_n = \frac{a^3 + 4a}{b^2 + a}$ carefully, using the limit theorems.

I emphasized the word "carefully" because it's the reason why I was fairly strict with grading. Many of you had something like this as the first line:

\lim_{n\to\infty} s_n = \frac{\lim_{n\to\infty} \p{a_n^3 + 4a_n}}{\lim_{n\to\infty} \p{b_n^2 + 1}} \quad \p{\text{Theorem 9.6}}

Theorem 9.6 says the following:

If $\set{s_n}_n$ converges to $s$ and $\set{t_n}_n$ converges to $t$ with $t \neq 0$ , then $\set{\frac{t_n}{s_n}}_n$ converges to $\frac{t}{s}$ .

Remember that you need to check assumptions before using a fact. The assumptions for Theorem 9.6 are that $\displaystyle\lim_{n\to\infty} s_n$ and $\displaystyle\lim_{n\to\infty} t_n$ exist. In other words, if you wrote the line above before checking that the limits of the numerator and denominator exist, you used Theorem 9.6 before checking all assumptions. So while you get the "correct" answer in the end, the reasoning is wrong, which is why I took off a lot of points.

To prove this properly, Theorem 9.6 turns out to be the last limit law that you use in this problem. A correct solution would look like the following:

By Theorem 9.4,

\lim_{n\to\infty} a_n^3 = \p{\lim_{n\to\infty} a_n}^3 = a^3 \quad\text{and}\quad \lim_{n\to\infty} b_n^2 = \p{\lim_{n\to\infty} b_n}^2 = b^2,

and by Theorem 9.2, $\displaystyle\lim_{n\to\infty} 4a_n = 4a$ . Thus, by Theorem 9.3,

\begin{aligned} \lim_{n\to\infty} \p{a_n^3 + 4a_n} = \p{\lim_{n\to\infty} a_n^3} + \p{\lim_{n\to\infty} 4a_n} &= a^3 + 4a, \quad\text{and} \\ \lim_{n\to\infty} \p{b_n^2 + 1} = \p{\lim_{n\to\infty} b_n^2} + \p{\lim_{n\to\infty} 1} &= b^2 + 1. \end{aligned}

Since $b^2 + 1 \geq 1$ , we know that $b^2 + 1 \neq 0$ , so now we can apply Theorem 9.6 to get

\lim_{n\to\infty} s_n = \frac{\lim_{n\to\infty} \p{a_n^3 + 4a_n}}{\lim_{n\to\infty} \p{b_n^2 + 1}} = \frac{a^3 + 4a}{b^2 + 1}.

Continuity

Introduction

Definition

Suppose $\func{f}{D}{\R}$ is a function. Then $f$ is continuous at $x_0 \in D$ if one of the following holds:

If $\set{x_n}_n$ is a sequence in $D$ converging to $x_0$ , then $\set{f\p{x_n}}_n$ converges to $f\p{x_0}$ .
For any $\epsilon > 0$ , there exists $\delta > 0$ such that whenever $y \in D$ satisfies $\abs{x_0 - y} < \delta$ , then $\abs{f\p{x_0} - f\p{y}} < \epsilon$ .

Remark.

From lecture, you already know that (i) and (ii) are equivalent definitions, which means we can use whichever one's more convenient for us. As a rule of thumb, (i) is easier to use to prove something is discontinuous, and (ii) is easier to use to prove something is continuous.

Example 1.

(17.10(b))

Prove that

f\p{x} = \begin{cases} \sin{\frac{1}{x}} & \text{if } x > 0, \\ 0 & \text{if } x = 0 \end{cases}

is not continuous at $x_0 = 0$ .

Solution.

Following the remark above, we're going to want to use (i). To prove discontinuity, we just need to find a sequence $\set{x_n}_n$ in $\pco{0, \infty}$ such that $\set{x_n}_n$ converges to $0$ , but $\set{f\p{x_n}}_n$ does not converge to $f\p{0} = 0$ . Since $\sin{\theta} = 1$ at $\frac{\pi}{2} + 2\pi n$ for all $n \in \N$ , we can set

\frac{1}{x_n} = \frac{\pi}{2} + 2\pi n \implies x_n = \frac{1}{\frac{\pi}{2} + 2\pi n}.

Then $x_n \to 0$ as $n \to \infty$ , but $f\p{x_n} = 1$ for all $n \in \N$ , so $f$ is not continuous at $x_0 = 0$ .

Example 2.

(17.4)

Prove that $\sqrt{x}$ is continuous on $\pco{0, \infty}$ .

Solution.

We're going to use definition (ii) here, which means that we want to show an inequality of the form

\abs{\sqrt{x_0} - \sqrt{y}} < \epsilon,

so let's take a look at the left-hand side:

\abs{\sqrt{x_0} - \sqrt{y}} \cdot \abs{\frac{\sqrt{x_0} + \sqrt{y}}{\sqrt{x_0} + \sqrt{y}}} = \frac{\abs{x_0 - y}}{\sqrt{x_0} + \sqrt{y}} \leq \frac{\abs{x_0 - y}}{\sqrt{x_0}}.

This inequality is okay except when $x_0 = 0$ , which tells me that I want to deal with this case separately.

Let $\epsilon > 0$ . If $x_0 = 0$ , then if $\abs{y} < \delta$ ,

\abs{\sqrt{y}} = \sqrt{y} < \sqrt{\delta}.

So, $\delta = \epsilon^2$ gives $\abs{\sqrt{y}} < \epsilon$ , so $\sqrt{x}$ is continuous at $x_0 = 0$ .

If $x_0 \neq 0$ , then we can use the inequality above: if $\abs{x_0 - y} < \delta$ , then

\abs{\sqrt{x_0} - \sqrt{y}} \leq \frac{\abs{x_0 - y}}{\sqrt{x_0}} < \frac{\delta}{\sqrt{x_0}}.

Thus, $\delta = \epsilon\sqrt{x_0}$ works: $\abs{\sqrt{x_0} - \sqrt{y}} < \epsilon\frac{\sqrt{x_0}}{\sqrt{x_0}} = \epsilon$ . Thus, $\sqrt{x}$ is continuous on all of $\pco{0, \infty}$ .

Another consequence of the sequence definition of continuity is that facts about sequences translate directly into facts about continuity.

Theorem

Suppose $\func{f}{D}{\R}$ and $\func{g}{D}{\R}$ are continuous functions. Then:

$f + g$ is continuous.
$fg$ is continuous.
$\frac{f}{g}$ is continuous at all points $x_0 \in D$ such that $g\p{x_0} \neq 0$ .

Proof.

(sketch)

Continuity at a point $x_0 \in D$ basically boils down to sequences. For $f + g$ , if $\set{x_n}_n$ is a sequence in $D$ that converges to $x_0$ , then we need to show that $\set{\p{f + g}\p{x_n}}_n$ converges to $\p{f + g}\p{x_0}$ , but this boils down to the limit laws for sequences. $\square$

This theorem tells us that sequences help us with proving continuity. The converse is also true: we can use continuity to help us calculate limits.

Example 3.

Calculate $\displaystyle\lim_{n\to\infty} e^{1/n}$ .

Solution.

$e^x$ is continuous (which is a fact we'll just take for granted), and $\frac{1}{n}$ converges to $0$ . Thus, by the sequence definition of continuity,

\lim_{n\to\infty} e^{1/n} = e^{\lim_{n\to\infty} \p{1/n}} = e^0 = 1.

To end this section, here's an important connection between density and continuity:

Proposition

Let $\func{f}{\R}{\R}$ and $\func{g}{\R}{\R}$ be continuous functions such that $f\p{q} = g\p{q}$ for all $q \in \Q$ . Then $f = g$ everywhere.

Remark.

The idea behind this fact is that a continuous function will "fill in gaps" through limits. For example, let's say I give you all the information about a function $f\p{x}$ except at a single point:

Because of continuity, there's only one way you can fill in the gap. In other words, information about a continuous function except at one point is the same thing as information about the continuous function everywhere.

The proposition says something a bit stronger: information about a continuous function at the rational numbers is the same as information about the function on the whole real line.

Proof.

To prove the statement, we just need to use density of $\Q$ . Let $x_0 \in \R$ , so by density of $\Q$ , there exists a sequence of rational numbers $\set{q_n}_n$ which converge to $x_0$ . Since we know that $f$ and $g$ agree at every rational, we have $f\p{q_n} = g\p{q_n}$ for all $n$ . Then by the sequence definition of continuity,

f\p{x_0} = \lim_{n\to\infty} f\p{q_n} = \lim_{n\to\infty} g\p{q_n} = g\p{x_0}.

Since $x_0$ was arbitrary, this means $f = g$ everywhere on $\R$ , which was what we wanted to show. $\square$

True or False

Example 4.

True or false: $\func{f}{\R \setminus \set{0}}{\R}$ defined by $f\p{x} = \frac{1}{x}$ is continuous.

Solution.

True. Continuity of a function only depends on points in the domain. $x_0 = 0$ is a problematic point for $f$ , but $0$ is outside the domain of $f$ , so we don't care what happens there.

Example 5.

True or false: If $f + g$ is continuous, then $f$ or $g$ is must be continuous at at least one point.

Solution.

False. For example, let

f\p{x} = \begin{cases} 1 & \text{if } x \in \Q, \\ 0 & \text{otherwise}. \end{cases}

Then if $g = -f$ , we get $f + g = 0$ which is continuous everywhere, but $f$ and $g$ are both discontinuous everywhere. This is because $\Q$ and $\R \setminus \Q$ are both dense in $\R$ , so given any $x_0 \in \R$ , there exist a sequence $\set{q_n}_n$ of rational numbers and a sequence $\set{r_n}_n$ of irrational numbers which both converge to $x_0$ . But

f\p{q_n} = 1 \quad\text{and}\quad f\p{r_n} = 0 \quad\text{for all } n \in \N,

so by the sequence definition of continuity, $f$ is discontinuous everywhere.

Example 6.

True or false: If $f + g$ is continuous and $f$ is continuous, then $g$ is continuous.

Solution.

True. You can write

g = \underbrace{\p{f + g}}_{\text{continuous}} + \underbrace{\p{-f}}_{\text{continuous}},

so $g$ is a sum of continuous functions, hence continuous.

Week 8 Discussion Notes

Table of Contents

Homework Comments

9.3

Continuity

Introduction

Definition

Remark.

Example 1.

Solution.

Example 2.

Solution.

Theorem

Proof.

Example 3.

Solution.

Proposition

Remark.

Proof.

True or False

Example 4.

Solution.

Example 5.

Solution.

Example 6.

Solution.