Chapter 5 · Continuity and Differentiability

Exercise 5.1 · Q13: Is $x + 5$ for $x \le 1$ ⁠, $x - 5$ for $x > 1$ continuous?

Exercise 5.1Continuity of piecewise functions · Section 5.2

The question

Is the function defined by

f(x) = \begin{cases} x + 5, & \text{if } x \le 1 \\ x - 5, & \text{if } x > 1 \end{cases}

a continuous function?

Exam answer

For $x < 1$ ⁠, $f(x) = x + 5$ ⁠, and for $x > 1$ ⁠, $f(x) = x - 5$ ⁠. Both are polynomials, so $f$ is continuous at every $x \ne 1$ ⁠.

At $x = 1$ ⁠:

\begin{aligned} \text{LHL} &= \lim_{h\to0} f(1 - h) \\ &= \lim_{h\to0}\bigl((1 - h) + 5\bigr) = 6 \\ \text{RHL} &= \lim_{h\to0} f(1 + h) \\ &= \lim_{h\to0}\bigl((1 + h) - 5\bigr) = -4 \end{aligned}

and $f(1) = 1 + 5 = 6$ ⁠. Since $\text{LHL} \ne \text{RHL}$ ⁠, $f$ is not continuous at $x = 1$ ⁠.

No. $f$ is discontinuous at $x = 1$ ⁠, so $f$ is not a continuous function.

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

The one idea this problem is built on

"A continuous function" means continuous at every point of its domain (Definition 2). So one bad point is enough to answer "no". Each piece is a straight line, continuous on its own, but at $x = 1$ the left line arrives at height $1 + 5 = 6$ and the right line starts from height $1 - 5 = -4$ ⁠. The graph drops by $10$ ⁠: the one-sided limits differ, so $f$ is not continuous at $1$ ⁠, and therefore $f$ is not a continuous function.

Given, to find, and the plan

Given: $f(x) = x + 5$ for $x \le 1$ and $f(x) = x - 5$ for $x > 1$ ⁠.
To find: Whether $f$ is continuous at every point of its domain.
Plan: (1) Domain and junction. (2) Continuity at every $c < 1$ and every $c > 1$ ⁠. (3) At $c = 1$ ⁠: left-hand limit, right-hand limit and $f(1)$ ⁠, each separately. (4) Apply Definition 2 to answer the question.

Basics you need

1Reading a piecewise function

$x = 1$ satisfies $x \le 1$ ⁠, so $f(1) = 1 + 5$ ⁠. Basics: piecewise functions

2A limit only looks near the point

If $f$ equals a polynomial on an interval around $c$ ⁠, its limit at $c$ is the polynomial's value at $c$ ⁠. Basics: what a limit is

3Left-hand and right-hand limits

If they are different, the limit does not exist and $f$ is not continuous there. Basics: one-sided limits

4Continuous at a point vs. a continuous function

Definition 1: continuous at $c$ means left-hand limit $=$ right-hand limit $= f(c)$ ⁠. Definition 2: a continuous function is one that is continuous at every point of its domain.

Solution

Domain and junction

Each real $x$ satisfies exactly one of $x \le 1$ and $x > 1$ ⁠, and $x + 5$ ⁠, $x - 5$ are defined for every $x$ ⁠. So the domain is $\R$ ⁠, and the formula changes only at $x = 1$ ⁠.

Definition 2 asks about every point of the domain, so we first need to know what the domain is: here, all real numbers.

Regions: $(-\infty, 1]$ and $(1, \infty)$ ⁠. Basics: piecewise functions
Every point other than $1$

If $c < 1$ ⁠: all $x$ within distance $1 - c$ of $c$ are less than $1$ ⁠, so there $f(x) = x + 5$ ⁠, and
$\lim_{x\to c} f(x) = \lim_{x\to c} (x + 5) = c + 5 = f(c).$
If $c > 1$ ⁠: all $x$ within distance $c - 1$ of $c$ are greater than $1$ ⁠, so there $f(x) = x - 5$ ⁠, and
$\lim_{x\to c} f(x) = \lim_{x\to c} (x - 5) = c - 5 = f(c).$
So $f$ is continuous at every $c \ne 1$ ⁠.

Around any point other than the junction, only one formula is in use, and each formula is a polynomial of degree 1. Exactly the same reasoning is Cases 1 and 2 of Example 10 in the chapter, where $f(x) = x + 2$ for $x \le 1$ and $x - 2$ for $x > 1$ ⁠.

Polynomials are continuous (Example 14). Lesson: continuity at a point
Test the junction $x = 1$

Value. $1 \le 1$ ⁠, so $f(1) = 1 + 5 = 6$ ⁠.

Left-hand limit. For $x < 1$ ⁠, $f(x) = x + 5$ ⁠:
$\lim_{x\to 1^-} f(x) = \lim_{x\to 1^-} (x + 5) = 1 + 5 = 6.$
Right-hand limit. For $x > 1$ ⁠, $f(x) = x - 5$ ⁠:
$\lim_{x\to 1^+} f(x) = \lim_{x\to 1^+} (x - 5) = 1 - 5 = -4.$
Compare. Left-hand limit $= 6$ ⁠, right-hand limit $= -4$ ⁠. They differ, so $\lim_{x\to 1} f(x)$ does not exist and $f$ is not continuous at $x = 1$ .

The value $f(1) = 6$ agrees with the left side, but continuity needs both sides. The right-hand values sit near $-4$ ⁠, ten units lower, so there is a genuine jump.

Unequal one-sided limits mean no limit, hence no continuity. Basics: one-sided limits
Answer the question

The point $x = 1$ is in the domain of $f$ ⁠, and $f$ is not continuous there (Step 3). A continuous function must be continuous at every point of its domain (Definition 2). So $f$ is not a continuous function. Its only point of discontinuity is $x = 1$ (Step 2 shows it is continuous everywhere else).

The question asks a yes-or-no question about the whole function, so we answer "no" and say exactly where and why it fails. Being continuous at every other point does not rescue it: one point of discontinuity in the domain is enough.

Definition 2 of the chapter: a real function is continuous if it is continuous at every point in its domain.

Blue:

y = x + 5

for

x \le 1

⁠, ending at the filled point

(1, 6)

⁠. Gold:

y = x - 5

for

x > 1

⁠, starting at the hollow point

(1, -4)

⁠, which is not on the graph. The graph drops by 10 at

x = 1

⁠. Horizontal scale: 1 unit = 45 px; vertical: 1 unit = 24 px.

Result

No, $f$ is not a continuous function. It is continuous at every $x \ne 1$ ⁠, but at $x = 1$ the left-hand limit is $6$ and the right-hand limit is $-4$ ⁠, so $f$ is discontinuous at $x = 1$ ⁠. (Textbook answer: " $f$ is not continuous at $x = 1$ ".)

Check it with numbers

$f(0.999) = 0.999 + 5 = 5.999$ and $f(1) = 6$ ⁠, but $f(1.001) = 1.001 - 5 = -3.999$ ⁠. Moving from $0.999$ to $1.001$ ⁠, a step of only $0.002$ ⁠, changes $f$ by $-3.999 - 5.999 = -9.998$ ⁠. Closer in, $f(0.99999) = 5.99999$ and $f(1.00001) = -3.99999$ ⁠: the difference stays near $-10$ ⁠. ✓

Common mistakes

"Both pieces are lines, so $f$ is continuous." Each piece is continuous on its own region; the junction still has to be tested.
Answering only "No". Say where the discontinuity is and show the one-sided limits.
Confusing "continuous at most points" with "a continuous function". Definition 2 needs every point of the domain.

What to take away

"Is $f$ a continuous function?" means "is $f$ continuous at every point of its domain?"
One point of discontinuity in the domain makes the answer "no".
For piecewise polynomials, test the junctions with one-sided limits and the value.

Every fact used, and where it comes from

Fact	Used in	Source
Every polynomial is continuous	Steps 2, 3	Example 14, Section 5.2 (Lesson: continuity at a point)
A limit at $c$ depends only on values near $c$	Step 2	Class XI, Limits and Derivatives; Example 10, Section 5.2
The limit exists only if the one-sided limits are equal	Step 3	Class XI, Limits and Derivatives; recalled after Definition 1, Section 5.2
Continuity at a point	Step 3	Definition 1, Section 5.2
A continuous function is continuous at every point of its domain	Step 4	Definition 2, Section 5.2

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Exercise 5.1 · Q13: Is x+5x + 5x+5 for x≤1x \le 1x≤1⁠, x−5x - 5x−5 for x>1x > 1x>1 continuous?

Basics you need

Solution

Domain and junction

Every point other than 111

Test the junction x=1x = 1x=1

Answer the question

Every fact used, and where it comes from

Exercise 5.1 · Q13: Is $x + 5$ for $x \le 1$ ⁠, $x - 5$ for $x > 1$ continuous?

Every point other than $1$

Test the junction $x = 1$