Math 242: Calculus I

1.8 Continuity

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1.8.1 Definition of Continuity

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Definition

A function $f$ is continuous at a value $a$ in the domain of $f$ if and only if

$$ \displaystyle\lim_{x\rightarrow a} f(x) = f(a) $$

This means that the two-sided limit as $x=a$ exists, $a$ is in the domain of $f$ so $f(a)$ is in the codomain, and they are equal!

In a manner similar to finding the domain of a function, one does not ask oneself, "What are all of the numbers in the domain of $f$ where $f$ is continuous."

Instead one should always ask,

Where is a function $f$ discontinuous?

Example 1

Where is the function the following continuous?

$$ f(x) = \frac{x^2 - 5x + 6}{x-3} $$

Solution

Instead ask, "Where is $f$ discontinuous?"

We are looking for values for which

• the denominator is equal to zero

• we try to compute an even root of a negative number

• we try to compute the logarithm of a number that is not positive

• we try to compute the inverse sine of a number greater than 1 or less than -1

Only this first of this list applies to this function. The denominator is zero when $x=3$. Hence $3$ is not in the domain of this function. Therefore the function cannot be continuous as $x=3$.

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1.8.2 Functions Continuous on an Interval

Definition

If a function $f$ is continuous at every element of an interval, $x\in[a,b]$, then we say that $f$ is continuous on the interval $[a,b]$.

If a function is continuous at every point in the domain of $f$, the we simply say that $f$ is continuous.

Example 2

Show that the function $h(x) = \frac{x-2}{2x-8}$ is continuous on the interval $(-\infty, 4)$.

Solution

Function $h$ is discontinuous only at the value where $2x-8=0$.

$$ \begin{align*} 2x-8 &= 0 \\ 2x &= 8 \\ x &= 4 \end{align*} $$

This function $h$ is continuous at every other point on the number line. The domain of $h$ is $(-\infty,4)\cup(4,\infty)$.

The number $4$ is not in the domain of $h$, and it is also not in the interval $(-\infty,4)$, so $h$ is continuous on the interval $(-\infty,4)$.

Example 3

Show that the function $f(x) = \frac{1}{x-5}$ is discontinuous at $a=5$.

Solution

At $a=5$, the value of the denominator of $f$ is equal to zero. Thus $5$ is not in the domain of $f$. Therefore is cannot be continuous at $a=5$.

Example 4

Show that the Heaviside function is discontinuous at the value $a=0$.

Solution

We already showed that $\lim_{x\rightarrow 0+} H(x) = 1$, but $\lim_{x\rightarrow 0^-} H(x) = 0$. Since the two-sided limit of the Heaviside function does not exist at $x=0$, $H$ cannot be continuous at $a=0$.

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1.8.3 Properties of Continuous Functions

Theorem 1.8.1

If $f$ and $g$ are continuous at $x=a$ and $c\in\mathbb{R}$ is a real constant, then the following functions are also continuous

1. $f+g$
2. $f-g$
3. $cf$
4. $fg$
5. $\frac{f}{g}$, if $g(a)\neq 0$

Exercise 1

Show that the function

$$ f(x) = \left\{\begin{array}{rcl} \frac{x^2+x}{x+1} & \text{if} & x\neq -1 \\ 1 & \text{if} & x=1 \end{array} \right. $$

is discontinuous at $a=-1$.

Solution

For the function to be continuous at $a=-1$ we must have the two-sided limit $\displaystyle\lim_{x\rightarrow -1}f(x) = f(-1) = 1$.
$$ \begin{align*} \displaystyle\lim_{x\rightarrow -1}f(x) &= \displaystyle\lim_{x\rightarrow -1} \frac{x^2+x}{x+1} \\ &= \displaystyle\lim_{x\rightarrow -1} \frac{x(x+1)}{x+1} \\ &= \displaystyle\lim_{x\rightarrow -1} x = -1 \neq 1 = f(-1) \end{align*} $$
The function $f$ is __not__ continuous at $a=-1$.

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1.8.4 More Properties of Continuous Functions

Theorem 1.8.2

a. Any polynomial function is continuous everywhere on the real line $\mathbb{R}$.

b. Any rational function is continuous everywhere in its domain; that is continuous except where the denominator is equal to zero.

Exercise 2

Show that the function

$$ f(x) = \frac{x^2}{\sqrt{x^2+1}} $$

is continuous everywhere on the real line $\mathbb{R} = (-\infty, +\infty)$.

Solution

Since $x^2 \ge 0$ for any real number $x\in\mathbb{R}$, $$ \begin{align*} x^2 + 1 &> 0 \\ \sqrt{x^2 + 1} &> 0 \end{align*} $$
Since the denominator is never equal to zero, by Theorem 1.8.2, function $f$ is continuous everywhere on the real line.

Theorem 1.8.3

The cosine and sine functions are continuous on the entire real line.

Exercise 3

Determine where the function $f(x)=\tan(x)$ is discontinuous.

Solution

The tangent function is discontinuous wherever the denominator is equal to zero.
$$ \begin{align*} \tan(x) &= \frac{\sin(x)}{\cos(x)} \\ \\ \cos(x) &= 0 \\ \\ x &= \frac{\pi}{2} + n\pi,\ n\in\mathbb{Z} \end{align*} $$

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1.8.5 The Intermediate Value Theorem

We often rely on our mathematical models having values between the ones we measure.

Example 4

If we measure the position of an asteroid at several positions at different times throughout the year, then we rely on our mathematical model to guarantee that it provides the positions of the asteroid between the moments we take down our measurements.

Figure 1

The idea here is that our mathematical model is continuous, so it also exhibits the behavior of objects in the solar system,

"They occupy the space between our observations continuously throughout their path".

The precise mathematical property of continuous functions is expressed by the following theorem.

Theorem 1.8.4 The Intermediate Value Theorem

If function $f$ is continuous on a closed interval [a,b] where $a\neq b$ and $f(a)\neq f(b)$, then for any number $L$ between $f(a)$ and $f(b)$, there exists a number $c\in(a,b)$ such that

$$ f(c) = L $$

This means that a continuous function $f$ cannot increase from $f(a)$ to $f(b)$, or decrease from $f(a)$ to $f(b)$ without also producing every number between $f(a)$ and $f(b)$ as an output of a number between $a$ and $b$.

Example 5

Consider the function $f(x) = \cos(x)$. This function is continuous everywhere. In particular it is continuous on the interval $[0, \pi]$. Now $\cos(0)=1$ and $\cos(\pi)=-1$. Therefore if we pick any real number $y$ in the interval $(-1,1)$ in the range of $f$, there must be at least one $\theta\in(0,\pi)$ so that

$$ f(\theta) = \cos(\theta) = y $$

Figure 2

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