Tag: Continuity

Find a function continuous at a single point of an interval

by RoRi

August 31, 2015

Find a function continuous at a single point of an interval, but discontinuous at all other points of the interval.

Consider the interval $(-a,a)$ for any $a \in \mathbb{R}$ , and define

$f(x) = \begin{cases} x & \text{if } x \text{ is rational},\\ 0 & \text{if } x \text{ is irrational}. \end{cases}$

for all $x \in (-a,a)$ .

We claim $f$ is continuous at 0, but discontinuous at every other point of the interval.

Proof. First, we show $f$ is continuous at 0. To do this we must show that for all $\varepsilon > 0$ there exists a $\delta > 0$ such that $|f(x)| < \varepsilon$ whenever $0 < |x| < \delta$ . So, let $\varepsilon > 0$ be given. Then, choose $\delta = \varepsilon$ . Then we have,

$|x| < \delta \quad \implies \quad |x| < \varepsilon \quad \implies \quad |f(x)| < \varepsilon$

since $|f(x)| \leq |x|$ from the definition of $f(x)$ . Thus, $f$ is continuous at 0.

Now, we must show $f$ is discontinuous at every other point of the interval. Let $p \neq 0$ be any nonzero point of the interval.

We consider two cases:
Case #1: $p$ is rational. Then $f(p) = p$ . Since $p \neq 0$ by assumption, we consider $\varepsilon = |p| > 0$ . Since the irrationals are dense in the reals (see Exercise #9 of Section I.3.12 here) we know that for any $\delta > 0$ there is some irrational number $x$ such that $|p-x| < \delta$ . But then, $|f(p) - f(x)| = |p - 0| = |p| \not < \varepsilon$ . This means $f$ is not continuous at $p$ since we have found an $\varepsilon > 0$ such that there is no $\delta > 0$ we can choose such that $|f(p) - f(x)| < \varepsilon$ whenever $|p - x| < \delta$ .

Case #2: $p$ is irrational. Since $p$ is irrational we know $f(p) = 0$ . Consider $\varepsilon = |p| > 0$ . Since the rationals are dense in the reals (see Exercise #6 of Section I.3.12 here) we know that for any $\delta > 0$ there exists some rational number $r$ such that $|p| < |r| < |p| + \delta$ . But then, $|f(p) - f(r)| = |0-r| = |r| \not < \varepsilon$ . This means $f$ is not continuous at $p$ . (Again, because we have shown that there exists an $\varepsilon > 0$ such that for every $\delta > 0$ there is some number $r$ such that $|f(p) - f(r)| \not < \varepsilon$ with $|p-r| < \delta.) \qquad \blacksquare$

(Note: There are much nicer ways to do this using sequences, but Apostol doesn’t develop those techniques, so we have to work straight from the epsilon-delta definitions.)

Consider the continuity of x(-1)^[1/x]

by RoRi

August 31, 2015

Define:

$f(x) = x(-1)^{\left \lfloor \frac{1}{x} \right \rfloor} \qquad \text{for } x \neq 0,$

where $\lfloor \ \rfloor$ denotes the greatest integer function, or floor function. Sketch the graph of $f(x)$ for $-2 \leq x \leq -\frac{1}{5}$ and $\frac{1}{5} \leq x \leq 2$ . Evaluate

$\lim_{x \to 0^+} f(x) \qquad \text{and} \qquad \lim_{x \to 0^-} f(x).$

Is it possible to define $f(0)$ in a way that makes $f(x)$ continuous at 0.

First, we sketch the graph of $f$ on the requested intervals.

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As $x \to 0^+$ , $f(x) \to 0$ .
As $x \to 0^-$ , $f(x) \to 0$ .

If we define $f(0) = 0$ then $f$ is continuous at 0.

Consider the continuity of (-1)^[1/x]

by RoRi

August 31, 2015

Define:

$f(x) = (-1)^{\left \lfloor \frac{1}{x} \right \rfloor} \qquad \text{for } x \neq 0,$

where $\lfloor \ \rfloor$ denotes the greatest integer function, or floor function. Sketch the graph of $f(x)$ for $-2 \leq x \leq -\frac{1}{5}$ and $\frac{1}{5} \leq x \leq 2$ . Evaluate

$\lim_{x \to 0^+} f(x) \qquad \text{and} \qquad \lim_{x \to 0^-} f(x).$

Is it possible to define $f(0)$ in a way that makes $f(x)$ continuous at 0.

First, we sketch the graph of $f$ on the requested intervals.

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As $x \to 0^+$ , $f(x)$ alternates between $+1$ and $-1$ .
As $x \to 0^-$ , $f(x)$ alternates between $+1$ and $-1$ .

There is no way to define $f(0)$ to make $f$ continuous at 0 since $f(x)$ will take both values $+1$ and $-1$ no matter how small we choose our $\delta > 0$ . (So, if we were to try to define $f(0) = 1$ , then for $\varepsilon = \frac{1}{2} > 0$ there is no $\delta> 0$ such that $f(x) < \varepsilon$ whenever $|x| < \delta$ , and similarly if we try to define $f(0) = -1$ .)

Consider the limit and continuity of the floor function of (1/x)

by RoRi

August 30, 2015

Define:

$f(x) = \left\lfloor \frac{1}{x} \right\rfloor \qquad \text{for } x \neq 0,$

where $\lfloor \ \rfloor$ denotes the greatest integer function, or floor function. Sketch the graph of $f(x)$ for $-2 \leq x \leq -\frac{1}{5}$ and $\frac{1}{5} \leq x \leq 2$ . Evaluate

$\lim_{x \to 0^+} f(x) \qquad \text{and} \qquad \lim_{x \to 0^-} f(x).$

Is it possible to define $f(0)$ in a way that makes $f(x)$ continuous at 0.

First, we sketch the graph of $f$ on the requested intervals.

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As $x \to 0^+$ , $f(x)$ takes on arbitrarily large positive values.
As $x \to 0^-$ , $f(x)$ takes on arbitrarily large negative values.

There is no way to define $f(0)$ to make $f$ continuous at 0.

Prove that sin (1/x) has no limit as x approaches 0

by RoRi

August 29, 2015

Define the function:

$f(x) = \sin \frac{1}{x} \qquad \text{for } x \neq 0.$

Prove there is no $A \in \mathbb{R}$ such that

$\lim_{x \to 0} f(x) = A.$

Proof. We prove this by contradiction. Suppose there does exist some $A \in \mathbb{R}$ such that $\lim_{x \to 0} f(x) = A$ . From the definition of limit this means that for all $\varepsilon > 0$ there exists a $\delta > 0$ such that

$| f(x) - A | < \varepsilon \qquad \text{whenever} \qquad 0 < |x| < \delta.$

First, we claim that for any such $A$ we must have $|A| \leq 1$ . This must be the case since if $|A| > 1$ then $|A| - 1 > 0$ , so we may choose $\varepsilon$ such that $0 < \varepsilon < (|A|-1)$ . But then, since $|f(x)| \leq 1$ for all $x$ , we have

$|f(x) - A| = |A - f(x)| \geq |A| - |f(x)| \geq |A| - 1 > \varepsilon.$

(Where we used part (i) this exercise (Section I.4.9, Exercise #1) for the inequality $|A - f(x) \geq |A| - |f(x)|$ .) This contradicts our choice of $\varepsilon$ , so $|A|$ must be less than or equal to 1.
Next, suppose $|A| \leq 1$ . Then choose $\varepsilon = \frac{1}{2} > 0$ . To obtain our contradiction we must show that there is no $\delta > 0$ such that

$|f(x) - A| < \frac{1}{2}, \qquad \text{whenever} \qquad 0 < |x| < \delta.$

By the Archimedean property of the real numbers we know that for any $x \in \mathbb{R}$ , there exists a positive integer $n$ with $n \equiv 1 \pmod{4}$ such that $0 < \frac{2}{n \pi} < |x|$ . (The Archimedean property guarantees us an integer, but then the inequality will hold for any larger integer, so we can choose one with the property of being $1 \pmod{4}$ .) But,

$0 < \frac{2}{n \pi} < |x| \quad \implies \quad 0 < \frac{2}{(n+2) \pi} < |x|.$

Then, from the definition of $f$ and since $n \equiv 1 \pmod{4}$ and $n+2 \equiv 3 \pmod{4}$ we have,

$\begin{align*} f \left( \frac{2}{n \pi} \right) &= \sin \left( \frac{n \pi}{2} \right) = 1 \\ f \left( \frac{2}{(n+2) \pi} \right) &= \sin \left( \frac{(n+2)\pi}{2} \right) = -1. \end{align*}$

But then,

$\begin{align*} \left| f \left( \frac{n \pi}{2} \right) - A \right| < \frac{1}{2} && \implies && |1 - A| &< \frac{1}{2} \\ && \implies && \frac{1}{2} &< A \leq 1 &(\text{from above } A \leq 1). \end{align*}$

Furthermore,

$\left| f \left( \frac{(n+2) \pi}{2} \right) - A \right| = |-1-A| > \frac{1}{2}.$

This contradicts that if $\lim_{x \to 0} f(x) = A$ then for every $\varepsilon > 0$ we have $|f(x) - A| < \varepsilon$ for all $0 < |x| < \delta$ . In other words, we found an $\varepsilon$ greater than 0 (in particular, we found $\varepsilon = \frac{1}{2}$ ) such that no matter how small we choose $\delta$ there exists an $x$ such that $|x|$ is smaller than $\delta$ , but $|f(x) - A|$ is bigger than $\varepsilon$ . This exactly contradicts the definition of $\lim_{x \to 0} f(x) = A$ . Hence, there can be no such number $A \in \mathbb{R}. \qquad \blacksquare$

This proof shows that there is now way to define $f(0)$ so that $f$ is continuous at 0. We see this since for $f$ to be continuous at 0 we must have $\lim_{x \to 0} f(x) = f(0)$ . But if we try to define $f(0) = A$ for any $A \in \mathbb{R}$ , we will fail to achieve continuity since the proof shows $\lim_{x \to 0} f(x) \neq A$ for any $A$ we might want to choose.

Give an alternate proof of the continuity of the sine and cosine functions

by RoRi

August 29, 2015

Use the inequality
$| \sin x | < |x|, \qquad \text{for } 0 < |x| < \frac{\pi}{2},$

to prove that the sine function is continuous at 0.
Recall the trig identity,
$\cos 2x = 1 - 2 \sin^2 x.$

Use this and part (a) to prove that the cosine function is continuous at 0.
Use the formulas for sine and cosine of a sum to prove that the sine and cosine functions are continuous for all $x \in \mathbb{R}$ .

Proof. To show $\sin$ is continuous at 0, we must show that $\lim_{x \to 0} \sin x = \sin 0 = 0$ . We show this limit is zero directly from the epsilon-delta definition of the limit, i.e., given arbitrary positive $\varepsilon$ , we have $|\sin x| < \varepsilon$ whenever $|x| < \delta$ . Let $\varepsilon$ be an arbitrary number greater than 0, $\varepsilon > 0$ . Then, let $\delta = \varepsilon$ . Using the given inequality we have,
$| \sin x | < |x| < \delta = \varepsilon, \qquad \text{whenver } 0 < |x| < \delta.$

Thus, $\sin x$ is continuous at $0. \qquad \blacksquare$
Proof. First, using the given trig identity we have
$\cos (2x) = 1 - 2 \sin^2 x \quad \implies \quad \cos x = 1 - 2 \sin^2 \frac{x}{2}.$

Thus,

$\begin{align*} \lim_{x \to 0} \cos x &= \lim_{x \to 0} \left( 1 - 2 \sin^2 \frac{x}{2} \right) \\ &= 1 - 2 \cdot \lim_{x \to 0} \left( \sin \left( \frac{x}{2} \right) \sin \left( \frac{x}{2} \right) \right) \\ &= 1 = \cos 0. \end{align*}$

Thus, cosine is continuous at 0.
Proof. Finally, to show sine and cosine are continuous for all $x \in \mathbb{R}$ , we show that $\lim_{x \to h} \sin x = sin h$ , and $\lim_{x \to h} \cos x = cos h$ . First, we recall the formulas for the sine and cosine of a sum,
$\begin{align*} \sin (x+h) &= \sin x \cos h + \sin h \cos x \\ \cos (x+h) &= \cos x \cos h - \sin x \sin h. \end{align*}$

So, we compute the limits

$\begin{align*} \lim_{x \to h} \sin x &= \lim_{x \to 0} \sin (x+h) = \lim_{x \to 0} (\sin x \cos h + \sin h \cos x) = \sin h \\ \lim_{x \to h} \cos x &= \lim_{x \to 0} \cos (x+h) = \lim_{x \to 0} (\cos x \cos h - \sin x \sin ) = \cos h. \end{align*}$

Therefore, sine and cosine are continuous for all $x \in \mathbb{R}. \qquad \blacksquare$

Consider the continuity of (tan x)/x at the point x=0

by RoRi

August 28, 2015

Consider the function

$f(x) = \frac{\tan x}{x} \qquad x \neq 0.$

Determine what happens to $f(x)$ as $x \to 0$ . Define $f(0)$ so that $f$ will be continuous at $x = 0$ , if possible.

First, we sketch the graph of $f(x) = \frac{\tan x}{x}$ on the intervals $\left[ -\frac{1}{4} \pi, 0\right)$ and $\left( 0, \frac{1}{4} \pi \right]$ .

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To determine what happens to $f(x)$ as $x \to 0$ we can take the limit, (recalling that $\lim_{x \to 0} \frac{\sin x}{x} = 1$ ),

$\begin{align*} \lim_{x \to 0} \frac{\tan x}{x} &= \lim_{x \to 0} \frac{\sin x}{x \cos x} \\ & =\lim_{x \to 0} \left( \left( \frac{\sin x}{x} \right) \cos x \right) \\ & =\lim_{x \to 0} \cos x \\ &= 1. \end{align*}$

Since $f(x) = \frac{\tan x}{x}$ is not defined at $x = 0$ it is not continuous there. However, since the limit exists we could redefine $f$ by

$f(x) = \begin{cases} \frac{\tan x}{x} & \text{if } x \neq 0 \\ 1 & \text{if } x = 0. \end{cases}$

Then, this new $f$ (which has the same values as the original function everywhere the original function is defined, but has the additional feature of being defined at $x=0$ ) is continuous at $x=0$ since it is defined there and $\lim_{x \to 0} f(x) = f(0) = 1$ .

Note what we have done here. We had a function $f(x) = \frac{\tan x}{x}$ which was undefined at the point $x = 0$ . We created a new function (which we also called $f$ , which is a bit confusing) that took the same values as the original function for all $x \neq 0$ , and took the value 1 when $x = 0$ . In this way we “removed” the discontinuity of the original $f$ at the point $x = 0$ , but we should be aware that this is actually a new function (since it has a different domain than the original).

Find all points at which tangent and cotangent are continuous

by RoRi

August 28, 2015

Find the points $x \in \mathbb{R}$ at which $\tan x$ is continuous and at which $\cot x$ is continuous.

Since

$\tan x = \frac{\sin x}{\cos x}$

and $\sin x$ , $\cos x$ are continuous everywhere we know (by Theorem 3.2) that $\tan x$ is continuous everywhere $\cos x$ is not zero. We proved in this exercise (Section 2.8, #1 (b) of Apostol) that

$\cos x = 0 \quad \iff \quad x = \frac{\pi}{2} + n \pi, \qquad n \in \mathbb{Z}.$

Thus, $\tan x$ is continuous for

$\left\{ x \in \mathbb{R} \mid x \neq \frac{\pi}{2} + n \pi, \ n \in \mathbb{Z} \right\}$

Similarly,

$\cot x = \frac{\cos x}{\sin x}$

and by Theorem 3.2 we then have $\cot x$ is continuous everywhere that $\sin x$ is not zero. We proved in the same exercise (Section 2.8, #1(a) of Apostol) that

$\sin x = 0 \quad \iff \quad x = n \pi \qquad n \in \mathbb{Z}.$

Hence, $\cot x$ is continuous for

$\left\{ x \in \mathbb{R} \mid x \neq n \pi, n \in \mathbb{Z} \right \}.$

Find a constant such that a given piecewise function is continuous everywhere

by RoRi

August 28, 2015

Given constants $a,b,c \in \mathbb{R}$ define:

$f(x) = \begin{cases} 2 \cos x & \text{if } x \leq c, \\ ax^2 + b & \text{if } x > c. \end{cases}$

If $b,c$ are fixed find all values of $a$ such that $f$ is continuous at $x = c$ .

By the definition of continuity, we know that $f$ continuous at $x = c$ means that $f(c)$ is defined and $\lim_{x \to c} f(x) = f(c)$ . Since $2 \cos x$ and $ax^2 + b$ are defined for all $x \in \mathbb{R}$ , we know that $f$ is defined for every $c \in \mathbb{R}$ . Then, to show that it is continuous we must show

$\lim_{x \to c} f(x) = f(c).$

From the definition of $f$ we know

$f(c) = 2 \cos c.$

Then, taking the limit as $x$ approaches $c$ from the right (since $\lim_{x \to c^-} f(x) = f(c)$ since $2 \cos x$ is continuous and $f(x) = 2 \cos x$ as $x$ approaches $c$ from the left),

$\lim_{x \to c^+} f(x) = \lim_{x \to c^+} ax^2 + b = ac^2 + b.$

So, we must have

$ac^2 + b = 2 \cos c \quad \implies \quad a = \frac{2 \cos c - b}{c^2} \quad \text{if } c \neq 0.$

If $c = 0$ then,

$ac^2 + b = 2 \cos c \quad \implies \quad b= 2 \quad \text{and } a \text{ is arbitrary}.$

Find a constant such that a given piecewise function is continuous

by RoRi

August 28, 2015

For a constants $a,b,c \in \mathbb{R}$ , define:

$f(x) = \begin{cases} \sin x & \text{if } x \leq c, \\ ax + b & \text{if } x > c. \end{cases}$

If $b,c$ are fixed, find all values for $a$ such that $f$ is continuous at $x = c$ .

By the definition of continuity of a function at a point, we know that $f$ is continuous at $x = c$ means $f$ is defined at $c$ , and $\lim_{x \to c} f(x) = f(c)$ .

Since $\sin x$ and $ax + b$ are defined for all $x \in \mathbb{R}$ , we know that $f$ is defined at $x = c$ for all $c \in \mathbb{R}$ . So, we must then find values of $a$ such that

$\lim_{x \to c} f(x) = f(c).$

From the definition of $f$ , we know

$f(c) = \sin c.$

Then, we evaluate the limit as $x \to c$ through values greater than $c$ (since the limit as $x \to c$ through values less than $c$ is $f(c)$ since $\sin x$ is a continuous function, and for values less than $c$ , $f(x) = \sin x$ ),