Chapter 5 · Continuity and Differentiability

Exercise 5.1 · Q12: Points of discontinuity: $x^{10} - 1$ for $x \le 1$ ⁠, $x^2$ for $x > 1$

Exercise 5.1Continuity of piecewise functions · Section 5.2

The question

Find all points of discontinuity of $f$ ⁠, where $f$ is defined by

f(x) = \begin{cases} x^{10} - 1, & \text{if } x \le 1 \\ x^2, & \text{if } x > 1 \end{cases}

Exam answer

For $x < 1$ ⁠, $f(x) = x^{10} - 1$ ⁠, and for $x > 1$ ⁠, $f(x) = x^2$ ⁠. 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)^{10} - 1\bigr) = 0 \\ \text{RHL} &= \lim_{h\to0} f(1 + h) \\ &= \lim_{h\to0}(1 + h)^2 = 1 \end{aligned}

and $f(1) = 1^{10} - 1 = 0$ ⁠. 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

Inside each region $f$ is a polynomial, so it is continuous there. At the junction $x = 1$ the left formula arrives at $1^{10} - 1 = 0$ ⁠, but the right formula starts from $1^2 = 1$ ⁠. The graph jumps up by $1$ ⁠. The two one-sided limits differ, so the limit at $1$ does not exist, and $x = 1$ is a point of discontinuity, the only one.

Given, to find, and the plan

Given: $f(x) = x^{10} - 1$ for $x \le 1$ and $f(x) = x^2$ for $x > 1$ ⁠.
To find: Every point of the domain at which $f$ is not continuous.
Plan: (1) Domain and junction. (2) Continuity at every $c < 1$ ⁠. (3) Continuity at every $c > 1$ ⁠. (4) At $c = 1$ ⁠: left-hand limit, right-hand limit and $f(1)$ ⁠, each separately. (5) Conclude.

Basics you need

1Reading a piecewise function

$x = 1$ satisfies $x \le 1$ ⁠, so $f(1)$ uses $x^{10} - 1$ ⁠, not $x^2$ ⁠. 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 that polynomial's limit. Basics: what a limit is

3Polynomials are continuous

$\lim_{x\to c} p(x) = p(c)$ (Example 14). Note $1^n = 1$ for every power $n$ ⁠.

4When the limit fails to exist

If the left-hand and right-hand limits at $c$ are different numbers, $\lim_{x\to c} f(x)$ does not exist, so $f$ cannot be continuous at $c$ ⁠, whatever $f(c)$ is. Basics: one-sided limits

5Continuity at a point (Definition 1)

At $c$ in the domain: continuous exactly when left-hand limit $=$ right-hand limit $= f(c)$ ⁠.

Solution

Domain and junction

Each real $x$ satisfies exactly one of $x \le 1$ and $x > 1$ ⁠; both formulas are polynomials, defined everywhere. So the domain of $f$ is $\R$ and the only junction is $x = 1$ ⁠.

We test continuity only at points of the domain, and the junction is the only place where a break can come from switching formulas.

The regions are $(-\infty, 1]$ and $(1, \infty)$ ⁠. Basics: piecewise functions
Every point $c < 1$

Let $c < 1$ ⁠, so $f(c) = c^{10} - 1$ ⁠. All $x$ within distance $1 - c$ of $c$ are less than $1$ ⁠, so there $f(x) = x^{10} - 1$ ⁠. Hence
$\lim_{x\to c} f(x) = \lim_{x\to c} (x^{10} - 1) = c^{10} - 1 = f(c).$
So $f$ is continuous at every $c < 1$ ⁠.

Near any point strictly left of $1$ ⁠, only the formula $x^{10} - 1$ is used, and it is a polynomial. (Its graph is almost flat at height $-1$ between about $-0.8$ and $0.8$ ⁠, because $x^{10}$ is tiny there, but "flat" has nothing to do with "broken".)

$x^{10} - 1$ is a polynomial, so it is continuous (Example 14). Lesson: continuity at a point
Every point $c > 1$

Let $c > 1$ ⁠, so $f(c) = c^2$ ⁠. All $x$ within distance $c - 1$ of $c$ are greater than $1$ ⁠, so there $f(x) = x^2$ ⁠. Hence
$\lim_{x\to c} f(x) = \lim_{x\to c} x^2 = c^2 = f(c).$
So $f$ is continuous at every $c > 1$ ⁠.

The same argument as Step 2, on the right of the junction.

$x^2$ is a polynomial, continuous everywhere (Example 2 shows it at $0$ ⁠; Example 14 in general).
Test the junction $x = 1$

Value. $1 \le 1$ ⁠, so $f(1) = 1^{10} - 1 = 1 - 1 = 0$ ⁠.

Left-hand limit. For $x < 1$ ⁠, $f(x) = x^{10} - 1$ ⁠:
$\lim_{x\to 1^-} f(x) = \lim_{x\to 1^-} (x^{10} - 1) = 1 - 1 = 0.$
Right-hand limit. For $x > 1$ ⁠, $f(x) = x^2$ ⁠:
$\lim_{x\to 1^+} f(x) = \lim_{x\to 1^+} x^2 = 1^2 = 1.$
Compare. Left-hand limit $= 0$ but right-hand limit $= 1$ ⁠. They are different, so $\lim_{x\to 1} f(x)$ does not exist, and $f$ is not continuous at $x = 1$ .

Notice that the value $f(1) = 0$ matches the left-hand limit: $f$ is continuous "from the left" at $1$ ⁠. That is not enough. Definition 1 needs the limit from both sides to exist and equal $f(1)$ ⁠, and the right-hand side jumps to $1$ ⁠. This is the same situation as the function in Fig 5.1 at the start of the chapter: the value agrees with one side only.

Different one-sided limits mean the limit does not exist, so continuity fails at once. Basics: one-sided limits
Conclude

$f$ is continuous at every $c < 1$ (Step 2) and every $c > 1$ (Step 3), and discontinuous at $c = 1$ (Step 4). So $x = 1$ is the only point of discontinuity of $f$ .

"Find all points" asks for a complete list, so we must say both where $f$ fails (only at $1$ ⁠) and why it does not fail anywhere else (Steps 2 and 3).

Blue:

y = x^{10} - 1

for

x \le 1

⁠, ending at the filled point

(1, 0)

⁠, which is

f(1)

⁠. Gold:

y = x^2

for

x > 1

⁠, starting at the hollow point

(1, 1)

⁠, which is not on the graph. The jump of height 1 at

x = 1

is the discontinuity. Horizontal scale: 1 unit = 110 px; vertical: 1 unit = 46 px.

Result

$f$ is continuous at every $x \ne 1$ ⁠, because near such a point it is a single polynomial. At $x = 1$ ⁠: left-hand limit $0$ ⁠, right-hand limit $1$ ⁠, $f(1) = 0$ ⁠; the one-sided limits differ. $x = 1$ is the only point of discontinuity. (Textbook answer: " $f$ is discontinuous at $x = 1$ ".)

Check it with numbers

Left of $1$ ⁠: $f(0.999) = 0.999^{10} - 1$ $\approx 0.990045 - 1$ $= -0.009955$ ⁠, close to $0$ ⁠. Right of $1$ ⁠: $f(1.001) = 1.001^2 = 1.002001$ ⁠, close to $1$ ⁠. However close to $1$ we go, the values on the two sides stay about $1$ apart: $f(0.99999) \approx -0.0001$ and $f(1.00001) \approx 1.00002$ ⁠. The gap does not close. ✓

Common mistakes

Using $x^2$ for $f(1)$ ⁠. $x = 1$ belongs to $x \le 1$ ⁠, so $f(1) = 0$ ⁠. (Either way the one-sided limits differ, so the conclusion stands, but the value must be right.)
" $f(1)$ equals the left-hand limit, so $f$ is continuous at 1." Both one-sided limits must equal $f(1)$ ⁠.
Thinking $1^{10}$ is $10$ ⁠. Any power of $1$ is $1$ ⁠.
Listing only the junction without justifying the rest. State why every other point is a point of continuity.

What to take away

For piecewise polynomials, only the junctions can be points of discontinuity.
If the one-sided limits at a junction differ, the function is discontinuous there, no matter which side the value belongs to.
A complete answer lists every point of discontinuity and explains why there are no others.

Every fact used, and where it comes from

Fact	Used in	Source
Every polynomial is continuous	Steps 2–4	Example 14, Section 5.2 (Lesson: continuity at a point)
A limit at $c$ depends only on values near $c$	Steps 2, 3	Class XI, Limits and Derivatives; used in Example 10, Section 5.2
The limit exists only if the one-sided limits are equal	Step 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 4, 5	Definition 1, Section 5.2
Value agreeing with one side only (Fig 5.1)	Step 4 remark	Opening discussion of Section 5.2

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Exercise 5.1 · Q12: Points of discontinuity: x10−1x^{10} - 1x10−1 for x≤1x \le 1x≤1⁠, x2x^2x2 for x>1x > 1x>1

Basics you need

Solution

Domain and junction

Every point c<1c < 1c<1

Every point c>1c > 1c>1

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

Conclude

Every fact used, and where it comes from

Exercise 5.1 · Q12: Points of discontinuity: $x^{10} - 1$ for $x \le 1$ ⁠, $x^2$ for $x > 1$

Every point $c < 1$

Every point $c > 1$

Test the junction $x = 1$