Chapter 5 · Continuity and Differentiability

Exercise 5.1 · Q3: Continuity of $x-5$ ⁠, $\frac{1}{x-5},$ $\frac{x^2-25}{x+5}$ and $|x-5|$

Exercise 5.1Continuous functions · Section 5.2

The question

Examine the following functions for continuity.

(a) $f(x) = x - 5$

(b) $f(x) = \dfrac{1}{x - 5},\ x \ne 5$

(d) $f(x) = |x - 5|$

Exam answer

(a) $f(x) = x - 5$ is defined for every real $x$ ⁠. For any real $c$ ⁠,

\begin{aligned} \lim_{x\to c} f(x) &= \lim_{x\to c}(x - 5) \\ &= c - 5 = f(c) \end{aligned}

So $f$ is continuous at every real number.

(b) $f(x) = \frac{1}{x - 5}$ is defined for $x \ne 5$ ⁠. For any $c \ne 5$ ⁠,

\begin{aligned} \lim_{x\to c} f(x) &= \lim_{x\to c}\frac{1}{x - 5} \\ &= \frac{1}{c - 5} = f(c) \end{aligned}

So $f$ is continuous at every point of its domain: a continuous function.

(c) $f(x) = \frac{x^2 - 25}{x + 5}$ is defined for $x \ne -5$ ⁠. For any $c \ne -5$ ⁠,

\begin{aligned} \lim_{x\to c} f(x) &= \lim_{x\to c}\frac{(x - 5)(x + 5)}{x + 5} \\ &= \lim_{x\to c}(x - 5) \\ &= c - 5 = f(c) \end{aligned}

So $f$ is continuous at every point of its domain: a continuous function.

(d) $f(x) = |x - 5|$ ⁠. For $x < 5$ ⁠, $f(x) = 5 - x$ ⁠, and for $x > 5$ ⁠, $f(x) = x - 5$ ⁠. Both are polynomials, so $f$ is continuous at every $x \ne 5$ ⁠.

At $x = 5$ ⁠:

\begin{aligned} \text{LHL} &= \lim_{h\to0} f(5 - h) \\ &= \lim_{h\to0}|{-h}| = 0 \\ \text{RHL} &= \lim_{h\to0} f(5 + h) \\ &= \lim_{h\to0}|h| = 0 \end{aligned}

and $f(5) = |0| = 0$ ⁠. So $\text{LHL} = \text{RHL} = f(5)$ ⁠: $f$ is continuous at $x = 5$ ⁠.

Hence $f$ is continuous at every real number. So all four functions are continuous.

Want the reason behind every step, the basics and the checks?

The one idea this problem is built on

"Examine for continuity" asks whether $f$ is a continuous function, which by Definition 2 means continuous at every point of its domain. So for each function: find the domain, then test an arbitrary point $c$ of it. Points outside the domain ( $x = 5$ in (b), $x = -5$ in (c)) are not tested at all: $f$ has no value there, so it is neither continuous nor discontinuous there in this book's sense. That is why (b) and (c) are continuous functions even though their graphs have a break or a hole.

Given, to find, and the plan

Given: Four functions, two of them with a point excluded from the domain.
To find: For each, whether it is continuous at every point of its domain.
Plan: For each part: (1) state the domain; (2) take any $c$ in the domain and compute the left-hand limit, the right-hand limit and $f(c)$ ⁠; (3) where the formula changes (the corner of $|x - 5|$ at 5), test that point separately; (4) conclude with Definition 2.

Basics you need

1Continuous at a point; continuous function

$f$ is continuous at a point $c$ of its domain if the left-hand limit, the right-hand limit and $f(c)$ are equal (Definition 1). $f$ is a continuous function if it is continuous at every point of its domain (Definition 2, Section 5.2). Lesson: continuity at a point

2One-sided limits with a small step $h$

With $h > 0$ ⁠: left-hand limit at $c$ is $\lim_{h\to0} f(c-h),$ right-hand limit is $\lim_{h\to0} f(c+h).$ Basics: left-hand and right-hand limits

3Limit of a quotient

If $p(x) \to P$ and $q(x) \to Q$ with $Q \ne 0,$ then $\dfrac{p(x)}{q(x)} \to \dfrac{P}{Q}.$ The same fact for continuous functions is Theorem 1 (4), Section 5.2.1. Basics: what a limit is

4Absolute value as a piecewise function

$|u| = u$ if $u \ge 0$ and $|u| = -u$ if $u < 0.$ So $|x - 5|$ is $x - 5$ for $x \ge 5$ and $5 - x$ for $x < 5.$ Basics: piecewise functions

5A limit only looks near the point

$\lim_{x\to c} f(x)$ depends only on the values of $f$ at points close to $c$ (and not at $c$ itself). If $f$ agrees with a simpler formula on some interval around $c,$ the limit can be computed from that formula. Basics: what a limit is

Part (a): $f(x) = x - 5$

Domain

$x - 5$ can be computed for every real $x,$ so the domain is $\R.$
Test an arbitrary point $c$

Let $c$ be any real number and $h > 0.$

Left-hand limit.
$\lim_{h\to0} f(c-h) = \lim_{h\to0}(c - 5 - h) = c - 5.$
Right-hand limit.
$\lim_{h\to0} f(c+h) = \lim_{h\to0}(c - 5 + h) = c - 5.$
Value. $f(c) = c - 5.$

All three agree, so $f$ is continuous at $c.$ Since $c$ was arbitrary, $f$ is continuous at every real number: a continuous function.

Taking $c$ arbitrary is what makes this a proof for all points at once, rather than a check at a few. (It is also a polynomial, so Example 14 gives the same conclusion.)

Definition 2: continuous at every point of the domain. Lesson: continuity at a point

Part (b): $f(x) = \frac{1}{x-5},$ $x \ne 5$

Domain: every real number except 5

The denominator $x - 5$ is zero exactly when $x = 5,$ and division by zero is undefined. The question itself excludes it. So the domain is $\R \setminus \{5\},$ that is, all $x < 5$ and all $x > 5.$

Everything that follows is about points in this domain. The point 5 is not one of them.
Test an arbitrary point $c \ne 5$

Let $c \ne 5,$ so $c - 5 \ne 0.$ As $h \to 0,$ the denominators $c - h - 5$ and $c + h - 5$ both tend to $c - 5 \ne 0.$ By the limit of a quotient (Basic 3):

Left-hand limit.
$\lim_{h\to0} f(c-h) = \lim_{h\to0} \frac{1}{(c-5) - h} = \frac{1}{c-5}.$
Right-hand limit.
$\lim_{h\to0} f(c+h) = \lim_{h\to0} \frac{1}{(c-5) + h} = \frac{1}{c-5}.$
Value. $f(c) = \dfrac{1}{c - 5}.$

All three agree, so $f$ is continuous at every $c \ne 5.$

The quotient rule for limits needs the limit of the denominator to be non-zero, and that is exactly the condition $c \ne 5.$ The same conclusion comes from Theorem 1 (4): the constant 1 and $x - 5$ are continuous (part (a)), and the denominator is not zero at $c,$ so their quotient is continuous at $c.$

Theorem 1 (4), Section 5.2.1: $f/g$ is continuous at $c$ if $f, g$ are and $g(c) \ne 0.$ Lesson: algebra of continuous functions
What about $x = 5$ ⁠? It is not in the domain

Near 5 the values blow up: $f(5.001) = 1000$ and $f(4.999) = -1000.$ But 5 is not in the domain, so Definition 1 does not apply there: 5 is not a point of discontinuity. Since $f$ is continuous at every point of its domain, $f$ is a continuous function (Definition 2).

$y = \frac{1}{x-5}$ has two branches, one for $x < 5$ and one for $x > 5.$ The pen must be lifted at the dashed line $x = 5,$ but only because $f$ is not defined there. Each branch is unbroken.

This is the same situation as $f(x) = \frac1x$ in Example 9, which the book calls a continuous function. "Continuous function" in this chapter means continuous at the points where the function is defined; it does not mean "the graph is one piece".

Part (c): $f(x) = \frac{x^2-25}{x+5},$ $x \ne -5$

Domain, and a simpler formula on it

The denominator $x + 5$ is zero only at $x = -5,$ which is excluded, so the domain is $\R \setminus \{-5\}.$ For every $x$ in the domain, factorise the numerator as a difference of squares and cancel, which is allowed because $x + 5 \ne 0$ ⁠:
$f(x) = \frac{(x-5)(x+5)}{x+5} = x - 5.$
Cancelling the factor $x + 5$ means dividing it by itself, which is only legitimate when it is not zero. That is exactly why the simplification holds on the domain and says nothing about $x = -5.$

$a^2 - b^2 = (a - b)(a + b),$ here with $a = x,$ $b = 5.$
Test an arbitrary point $c \ne -5$

Let $c \ne -5.$ For $0 < h < |c + 5|,$ the points $c - h$ and $c + h$ are also different from $-5,$ so the formula $f(x) = x - 5$ applies to them.

Left-hand limit.
$\lim_{h\to0} f(c-h) = \lim_{h\to0}(c - h - 5) = c - 5.$
Right-hand limit.
$\lim_{h\to0} f(c+h) = \lim_{h\to0}(c + h - 5) = c - 5.$
Value. $f(c) = c - 5.$

All three agree, so $f$ is continuous at every $c \ne -5,$ and therefore $f$ is a continuous function.

The restriction $h < |c + 5|$ keeps the nearby points away from $-5,$ where the simplified formula would not be valid. Since a limit only looks at points close to $c$ (Basic 5), restricting $h$ costs nothing.
The hole at $x = -5$

As $x \to -5$ the values approach $-5 - 5 = -10,$ yet $f(-5)$ does not exist: $-5$ is not in the domain, so again it is neither a point of continuity nor of discontinuity. The graph is the line $y = x - 5$ with one point removed.

On its domain $\frac{x^2-25}{x+5}$ equals $x - 5,$ so the graph is the line $y = x - 5$ with an open circle at $(-5, -10),$ the point that is missing because $x = -5$ is excluded.

Do not confuse "the limit exists at $-5$ " with "continuous at $-5$ ": continuity also needs a value $f(-5),$ and there is none. The question only asks about the function as given, on its domain.

Part (d): $f(x) = |x - 5|$

Domain and the two formulas

$|x - 5|$ is defined for every real $x,$ so the domain is $\R.$ By the definition of absolute value (Basic 4):
$f(x) = \begin{cases} 5 - x, & \text{if } x < 5 \\ x - 5, & \text{if } x \ge 5 \end{cases}$
Writing $f$ piece by piece shows where something could go wrong: only at $x = 5,$ where the formula changes. Away from 5, $f$ is a simple polynomial near each point.

Basics: piecewise functions
Points $c < 5$ and $c > 5$

Let $c < 5.$ For $0 < h < 5 - c$ both $c - h$ and $c + h$ are less than 5, so $f(x) = 5 - x$ there:
$\begin{aligned} \text{LHL} &= \lim_{h\to0} \bigl(5 - (c - h)\bigr) = 5 - c,\\ \text{RHL} &= \lim_{h\to0} \bigl(5 - (c + h)\bigr) = 5 - c, \end{aligned}$
and $f(c) = 5 - c.$ All three are equal, so $f$ is continuous at $c.$

Let $c > 5.$ For $0 < h < c - 5,$ both $c \pm h$ exceed 5, so $f(x) = x - 5$ there:
$\begin{aligned} \text{LHL} &= \lim_{h\to0} (c - h - 5) = c - 5,\\ \text{RHL} &= \lim_{h\to0} (c + h - 5) = c - 5, \end{aligned}$
and $f(c) = c - 5.$ All three are equal, so $f$ is continuous at $c.$

Choosing $h$ smaller than the distance from $c$ to 5 keeps both nearby points in the same piece as $c,$ so a single formula applies on both sides.
The corner point $c = 5$

Left-hand limit. For $h > 0,$ $5 - h < 5,$ so
$\lim_{h\to0} f(5-h) = \lim_{h\to0} \bigl(5 - (5 - h)\bigr) = \lim_{h\to0} h = 0.$
Right-hand limit. $5 + h > 5,$ so
$\lim_{h\to0} f(5+h) = \lim_{h\to0} \bigl((5 + h) - 5\bigr) = \lim_{h\to0} h = 0.$
Value. $f(5) = |0| = 0.$

All three are 0, so $f$ is continuous at 5. Together with Part (d), Step 2, $f$ is continuous at every real number: a continuous function.

$y = |x - 5|$ is a V with its corner at $(5, 0).$ A corner is not a break: both arms arrive at height 0, which is $f(5).$

A second route: $f$ is the composite of $g(x) = x - 5$ (continuous, part (a)) followed by $|\cdot|$ (continuous, Example 7). By Theorem 2, Section 5.2.1, a composite of continuous functions is continuous. The direct calculation above shows the same thing explicitly.

Theorem 2: if $g$ is continuous at $c$ and $f$ at $g(c),$ then $f \circ g$ is continuous at $c.$ Basics: composite functions

Result

All four are continuous functions.

(a) $x - 5$ ⁠: continuous at every real number.
(b) $\frac{1}{x-5}$ ⁠: continuous at every $x \ne 5,$ the whole domain.
(c) $\frac{x^2-25}{x+5}$ ⁠: continuous at every $x \ne -5,$ the whole domain.
(d) $|x - 5|$ ⁠: continuous at every real number, including the corner $x = 5.$

(Textbook answer: "(a), (b), (c) and (d) are all continuous functions".)

Check it with numbers

(b) at $c = 7$ ⁠: $f(7) = 0.5,$ $f(6.999) = \frac{1}{1.999} \approx 0.500250,$ $f(7.001) = \frac{1}{2.001} \approx 0.499750.$ Both close to 0.5.
(c) near the hole: $f(-5.001) = \frac{25.010001 - 25}{-0.001} = -10.001$ and $f(-4.999) = \frac{24.990001 - 25}{0.001} = -9.999.$ Both close to $-10,$ but there is no value $f(-5).$
(c) at $c = 0$ ⁠: $f(0) = \frac{-25}{5} = -5 = 0 - 5$ ✓ (agrees with the simplified formula).
(d) at the corner: $f(4.999) = 0.001,$ $f(5.001) = 0.001,$ $f(5) = 0.$

Common mistakes

"(b) is discontinuous at $x = 5$ ⁠." 5 is not in the domain, so the question of continuity there does not arise. The book's convention (Definition 2 and Example 9) makes (b) a continuous function.
Cancelling $x + 5$ without saying $x \ne -5.$ The simplification $f(x) = x - 5$ is valid only on the domain.
"A corner means a break." $|x - 5|$ has a sharp corner at 5 but no jump; it is continuous there. (Corners matter later, for differentiability.)
Testing only a few points. "Continuous function" is a claim about every point of the domain, so take an arbitrary $c.$

What to take away

"Examine for continuity" means: continuous at every point of the domain (Definition 2). Start by writing the domain down.
Points left out of the domain are neither continuous nor discontinuous points; a graph may break there and the function can still be a continuous function.
For $|\text{expression}|,$ split into the two cases and test the switching point with one-sided limits.
Sums, quotients (non-zero denominator) and composites of continuous functions are continuous (Theorems 1 and 2).

Every fact used, and where it comes from

Fact	Used in	Source
Continuity at $c$ ⁠: LHL, RHL and $f(c)$ exist and are equal	All parts	Definition 1, Section 5.2 (Lesson: continuity at a point)
Continuous function: continuous at every point of its domain	All parts	Definition 2, Section 5.2
$\frac1x$ is a continuous function although undefined at 0	Part (b), Step 3	Example 9
Quotient of continuous functions with $g(c) \ne 0$ is continuous	Part (b), Step 2	Theorem 1 (4), Section 5.2.1 (Lesson: algebra of continuous functions)
Limit of a quotient with non-zero denominator limit	Part (b), Step 2	Class XI Limits (Basics: what a limit is)
$a^2 - b^2 = (a-b)(a+b)$	Part (c), Step 1	Algebraic identity
$\|u\| = u$ for $u \ge 0,$ $-u$ for $u < 0$	Part (d), Step 1	Definition of modulus (Basics: piecewise functions)
$\|x\|$ is continuous	Part (d), Step 3	Example 7
Composite of continuous functions is continuous	Part (d), Step 3	Theorem 2, Section 5.2.1 (Basics: composite functions)
Polynomials are continuous	Part (a), Step 2	Example 14
Values in the check box	Check	Recomputed in the site's verification script

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Exercise 5.1 · Q3: Continuity of x−5x-5x−5⁠, 1x−5,\frac{1}{x-5},x−51​, x2−25x+5\frac{x^2-25}{x+5}x+5x2−25​ and ∣x−5∣|x-5|∣x−5∣

Basics you need

Part (a): f(x)=x−5f(x) = x - 5f(x)=x−5

Domain

Test an arbitrary point ccc

Part (b): f(x)=1x−5,f(x) = \frac{1}{x-5},f(x)=x−51​, x≠5x \ne 5x=5

Domain: every real number except 5

Test an arbitrary point c≠5c \ne 5c=5

What about x=5x = 5x=5⁠? It is not in the domain

Part (c): f(x)=x2−25x+5,f(x) = \frac{x^2-25}{x+5},f(x)=x+5x2−25​, x≠−5x \ne -5x=−5

Domain, and a simpler formula on it

Test an arbitrary point c≠−5c \ne -5c=−5

The hole at x=−5x = -5x=−5

Part (d): f(x)=∣x−5∣f(x) = |x - 5|f(x)=∣x−5∣

Domain and the two formulas

Points c<5c < 5c<5 and c>5c > 5c>5

The corner point c=5c = 5c=5

Every fact used, and where it comes from

Exercise 5.1 · Q3: Continuity of $x-5$ ⁠, $\frac{1}{x-5},$ $\frac{x^2-25}{x+5}$ and $|x-5|$

Part (a): $f(x) = x - 5$

Test an arbitrary point $c$

Part (b): $f(x) = \frac{1}{x-5},$ $x \ne 5$

Test an arbitrary point $c \ne 5$

What about $x = 5$ ⁠? It is not in the domain

Part (c): $f(x) = \frac{x^2-25}{x+5},$ $x \ne -5$

Test an arbitrary point $c \ne -5$

The hole at $x = -5$

Part (d): $f(x) = |x - 5|$

Points $c < 5$ and $c > 5$

The corner point $c = 5$