Chapter 5 · Continuity and Differentiability

Exercise 5.1 · Q15: Continuity of $2x$ ( $x < 0$ ⁠), $0$ ( $0 \le x \le 1$ ⁠), $4x$ ( $x > 1$ ⁠)

Exercise 5.1Continuity of piecewise functions · Section 5.2

The question

Discuss the continuity of the function $f$ ⁠, where $f$ is defined by

f(x) = \begin{cases} 2x, & \text{if } x < 0 \\ 0, & \text{if } 0 \le x \le 1 \\ 4x, & \text{if } x > 1 \end{cases}

Exam answer

For $x < 0$ ⁠, $f(x) = 2x$ ⁠; for $0 < x < 1$ ⁠, $f(x) = 0$ ⁠; for $x > 1$ ⁠, $f(x) = 4x$ ⁠. These are polynomials, so $f$ is continuous at every $x$ other than $0$ and $1$ ⁠.

At $x = 0$ ⁠:

\begin{aligned} \text{LHL} &= \lim_{h\to0} f(0 - h) \\ &= \lim_{h\to0} 2(0 - h) = 0 \\ \text{RHL} &= \lim_{h\to0} f(0 + h) \\ &= \lim_{h\to0} 0 = 0 \end{aligned}

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

At $x = 1$ ⁠:

\begin{aligned} \text{LHL} &= \lim_{h\to0} f(1 - h) \\ &= \lim_{h\to0} 0 = 0 \\ \text{RHL} &= \lim_{h\to0} f(1 + h) \\ &= \lim_{h\to0} 4(1 + h) = 4 \end{aligned}

and $f(1) = 0$ ⁠. Since $\text{LHL} \ne \text{RHL}$ ⁠, $f$ is not continuous at $x = 1$ ⁠.

Hence $x = 1$ is the only point of discontinuity of $f$ ⁠.

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

The one idea this problem is built on

Three pieces means two junctions, $x = 0$ and $x = 1$ ⁠, and each must be tested on its own. At $0$ ⁠, the line $2x$ arrives at height $2 \cdot 0 = 0$ ⁠, which is exactly the height of the middle piece, so the pieces meet. At $1$ ⁠, the middle piece is at height $0$ but the line $4x$ starts from $4 \cdot 1 = 4$ ⁠, so the graph jumps. Everywhere else $f$ is a single polynomial. So $x = 1$ is the only point of discontinuity.

Given, to find, and the plan

Given: $f(x) = 2x$ for $x < 0$ ⁠, $f(x) = 0$ for $0 \le x \le 1$ ⁠, $f(x) = 4x$ for $x > 1$ ⁠.
To find: At which points $f$ is continuous and at which it is not.
Plan: (1) Domain and junctions. (2) Points strictly inside each piece. (3) The junction $x = 0$ ⁠. (4) The junction $x = 1$ ⁠. (5) Conclude.

Basics you need

1Reading a piecewise function

Both $0$ and $1$ satisfy $0 \le x \le 1$ ⁠, so $f(0) = 0$ and $f(1) = 0$ ⁠. Basics: piecewise functions

2A limit only looks near the point

If $f$ equals a polynomial (or a constant) on an interval around $c$ ⁠, its limit at $c$ is that polynomial's value at $c$ ⁠. Basics: what a limit is

3Left-hand and right-hand limits

At a junction, each side has its own formula, so each one-sided limit is computed separately. Basics: one-sided limits

4Continuity at a point (Definition 1)

Left-hand limit $=$ right-hand limit $= f(c)$ ⁠.

Solution

Domain and junctions

Every real $x$ satisfies exactly one of $x < 0$ ⁠, $0 \le x \le 1$ ⁠, $x > 1$ ⁠, and each formula is defined for every $x$ ⁠. So the domain is $\R$ ⁠. The formula changes at $x = 0$ and at $x = 1$ ⁠: these are the two junctions.

With three pieces there are two places where the definition switches, and each needs its own test. Missing one is the most common way to get this question wrong.

Regions: $(-\infty, 0)$ ⁠, $[0, 1]$ ⁠, $(1, \infty)$ ⁠. Basics: piecewise functions
Points strictly inside a piece

If $c < 0$ ⁠: near $c$ (within distance $-c$ ⁠), $f(x) = 2x$ ⁠, so $\lim_{x\to c} f(x) = 2c = f(c)$ ⁠.

If $0 < c < 1$ ⁠: near $c$ (within distance $\min(c, 1 - c)$ ⁠), $f(x) = 0$ ⁠, so $\lim_{x\to c} f(x) = 0 = f(c)$ ⁠.

If $c > 1$ ⁠: near $c$ (within distance $c - 1$ ⁠), $f(x) = 4x$ ⁠, so $\lim_{x\to c} f(x) = 4c = f(c)$ ⁠.

So $f$ is continuous at every $c$ other than $0$ and $1$ ⁠.

Around a point strictly inside one region, a whole small interval stays inside that region, so $f$ is a single polynomial there ( $2x$ ⁠, the constant $0$ ⁠, or $4x$ ⁠), and the limit equals the value.

Polynomials, including constants, are continuous (Examples 5 and 14). Lesson: continuity at a point
The junction $x = 0$

Value. $0$ satisfies $0 \le x \le 1$ ⁠, so $f(0) = 0$ ⁠.

Left-hand limit. For $x < 0$ ⁠, $f(x) = 2x$ ⁠:
$\lim_{x\to 0^-} f(x) = \lim_{x\to 0^-} 2x = 2 \cdot 0 = 0.$
Right-hand limit. For $0 < x < 1$ ⁠, $f(x) = 0$ ⁠:
$\lim_{x\to 0^+} f(x) = \lim_{x\to 0^+} 0 = 0.$
Compare. Left-hand limit $= 0$ ⁠, right-hand limit $= 0$ ⁠, $f(0) = 0$ ⁠. All equal, so $f$ is continuous at $x = 0$ .

Approaching $0$ from the right, we only need $x$ slightly bigger than $0$ ⁠, say $0 < x < 1$ ⁠, and there the middle formula applies. The line $2x$ runs into the origin, which is where the flat middle piece begins, so the graph has no gap at $0$ ⁠.

Compute each one-sided limit with the formula valid on that side. Basics: one-sided limits
The junction $x = 1$

Value. $1$ satisfies $0 \le x \le 1$ ⁠, so $f(1) = 0$ ⁠.

Left-hand limit. For $0 < x < 1$ ⁠, $f(x) = 0$ ⁠:
$\lim_{x\to 1^-} f(x) = \lim_{x\to 1^-} 0 = 0.$
Right-hand limit. For $x > 1$ ⁠, $f(x) = 4x$ ⁠:
$\lim_{x\to 1^+} f(x) = \lim_{x\to 1^+} 4x = 4 \cdot 1 = 4.$
Compare. Left-hand limit $= 0$ ⁠, right-hand limit $= 4$ ⁠. They differ, so $\lim_{x\to 1} f(x)$ does not exist and $f$ is discontinuous at $x = 1$ .

The value $f(1) = 0$ agrees with the left side only. Just to the right of $1$ ⁠, $f(x) = 4x$ is a little more than $4$ ⁠, so the graph jumps up by $4$ at $x = 1$ ⁠.

Different one-sided limits mean no limit, hence no continuity.
Conclude

By Step 2, $f$ is continuous at every $x \ne 0, 1$ ⁠. By Step 3 it is continuous at $0$ ⁠; by Step 4 it is discontinuous at $1$ ⁠. So $x = 1$ is the only point of discontinuity of $f$ ; $f$ is continuous at every other real number.

"Discuss the continuity" asks for the complete picture: where it holds, where it fails, and the reasons. Every real number falls into Step 2, Step 3 or Step 4.

Blue:

y = 2x

for

x < 0

⁠, running into the origin. Green:

y = 0

[0, 1]

⁠, with filled ends

(0, 0)

and

(1, 0)

⁠. Gold:

y = 4x

for

x > 1

⁠, starting at the hollow point

(1, 4)

⁠. No gap at

0

⁠; a jump of 4 at

1

⁠. Horizontal scale: 1 unit = 90 px; vertical: 1 unit = 22 px.

Result

$f$ is continuous at every $x \ne 0, 1$ (polynomial pieces). At $x = 0$ ⁠: left-hand limit $0$ ⁠, right-hand limit $0$ ⁠, $f(0) = 0$ ⁠, so $f$ is continuous at $0$ ⁠. At $x = 1$ ⁠: left-hand limit $0$ ⁠, right-hand limit $4$ ⁠, so $f$ is discontinuous there. $x = 1$ is the only point of discontinuity. (Textbook answer: " $x = 1$ is the only point of discontinuity".)

Check it with numbers

Around $0$ ⁠: $f(-0.001) = 2(-0.001) = -0.002$ ⁠, $f(0) = 0$ ⁠, $f(0.001) = 0$ ⁠. All within $0.002$ of $0$ ⁠. ✓ Around $1$ ⁠: $f(0.999) = 0$ ⁠, $f(1) = 0$ ⁠, but $f(1.001) = 4.004$ ⁠. The gap of about $4$ stays however close we go: $f(1.00001) = 4.00004$ ⁠. ✓

Common mistakes

Testing only one junction. Three pieces give two junctions, $0$ and $1$ ⁠.
Using $4x$ for $f(1)$ ⁠. $1$ belongs to $0 \le x \le 1$ ⁠, so $f(1) = 0$ ⁠.
Expecting a break wherever the formula changes. At $0$ the formula changes but there is no break: $2x$ and $0$ agree at $x = 0$ ⁠.

What to take away

List every junction and test each with left-hand limit, right-hand limit and value.
A change of formula is a candidate for discontinuity, not a guarantee.
The value at a junction comes from the piece whose condition includes that point.

Every fact used, and where it comes from

Fact	Used in	Source
Constant functions and polynomials are continuous	Steps 2–4	Examples 5 and 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
Left-hand and right-hand limits; the limit exists only if they agree	Steps 3, 4	Class XI, Limits and Derivatives; recalled after Definition 1, Section 5.2
Continuity at $c$ ⁠: left-hand limit $=$ right-hand limit $= f(c)$	Steps 3–5	Definition 1, Section 5.2

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Exercise 5.1 · Q15: Continuity of 2x2x2x (x<0x < 0x<0⁠), 000 (0≤x≤10 \le x \le 10≤x≤1⁠), 4x4x4x (x>1x > 1x>1⁠)

Basics you need

Solution

Domain and junctions

Points strictly inside a piece

The junction x=0x = 0x=0

The junction x=1x = 1x=1

Conclude

Every fact used, and where it comes from

Exercise 5.1 · Q15: Continuity of $2x$ ( $x < 0$ ⁠), $0$ ( $0 \le x \le 1$ ⁠), $4x$ ( $x > 1$ ⁠)

The junction $x = 0$

The junction $x = 1$