Chapter 5 · Continuity and Differentiability

Exercise 5.1 · Q10: Points of discontinuity: $x + 1$ for $x \ge 1$ ⁠, $x^2 + 1$ 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 + 1, & \text{if } x \ge 1 \\ x^2 + 1, & \text{if } x < 1 \end{cases}

Exam answer

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

and $f(1) = 1 + 1 = 2$ ⁠. So $\text{LHL} = \text{RHL} = f(1)$ ⁠: $f$ is continuous at $x = 1$ ⁠.

Hence $f$ is continuous at every real number: there is no point of discontinuity.

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

The one idea this problem is built on

Each piece is a polynomial, and polynomials never break. So the only place a break could happen is the junction $x = 1$ ⁠, where the formula changes. There we compare three numbers: what $x^2 + 1$ approaches from the left, what $x + 1$ approaches from the right, and the actual value $f(1)$ ⁠. Both formulas give $2$ at $x = 1$ ⁠, so the two pieces meet, and $f$ has no point of discontinuity.

Given, to find, and the plan

Given: $f(x) = x + 1$ for $x \ge 1$ and $f(x) = x^2 + 1$ for $x < 1$ ⁠.
To find: Every point of the domain at which $f$ is not continuous.
Plan: (1) Find the domain and the junction. (2) Show $f$ is continuous at every $c < 1$ ⁠. (3) Same for every $c > 1$ ⁠. (4) Test $c = 1$ ⁠: left-hand limit, right-hand limit and $f(1)$ ⁠, each separately. (5) Conclude.

Basics you need

1Reading a piecewise function

Decide which condition $x$ satisfies, then use that formula only. Here $f(1)$ uses $x + 1$ ⁠, because $1 \ge 1$ ⁠. Basics: piecewise functions

2A limit only looks near the point

$\lim_{x\to c} f(x)$ depends only on the values of $f$ at points close to $c$ (other than $c$ itself). If $f$ agrees with a simpler function $g$ on a whole interval around $c$ ⁠, then the two limits are the same. Basics: what a limit is

3Polynomials are continuous

For a polynomial $p$ ⁠, $\lim_{x\to c} p(x) = p(c)$ at every real $c$ (Example 14 of the chapter). So the limit of a polynomial is found by substituting.

4Left-hand and right-hand limits

As $x \to 1^-$ we only use $x < 1$ ⁠; as $x \to 1^+$ we only use $x > 1$ ⁠. Basics: one-sided limits

5Continuity at a point (Definition 1)

$f$ is continuous at $c$ in its domain when the left-hand limit, the right-hand limit and $f(c)$ exist and are equal.

Solution

Domain and junction

Every real number satisfies exactly one of $x \ge 1$ and $x < 1$ ⁠, and both formulas are polynomials, defined for every $x$ ⁠. So the domain of $f$ is $\R$ ⁠. The formula changes at $x = 1$ ⁠, so that is the only junction.

We need the domain because continuity is tested only at points of the domain. We need the junction because it is the only point where the two formulas have to "agree"; everywhere else just one formula is in charge.

The two conditions split the line into $(-\infty, 1)$ and $[1, \infty)$ ⁠, with no overlap and no gap. Basics: piecewise functions
Every point $c < 1$

Let $c < 1$ ⁠. Then $f(c) = c^2 + 1$ ⁠. Every $x$ closer to $c$ than $1 - c$ is also less than $1$ ⁠, so near $c$ we have $f(x) = x^2 + 1$ ⁠. Hence
$\lim_{x\to c} f(x) = \lim_{x\to c} (x^2 + 1) = c^2 + 1 = f(c).$
So $f$ is continuous at every $c < 1$ ⁠.

The point $c$ sits strictly inside the region $x < 1$ ⁠, with room to spare on both sides (a distance $1 - c > 0$ up to the junction). On that whole neighbourhood $f$ is the polynomial $x^2 + 1$ ⁠, and the limit only looks at that neighbourhood. This is exactly Case 1 of Example 10 in the chapter.

$x^2 + 1$ is a polynomial, so its limit is its value (Example 14). Lesson: continuity at a point
Every point $c > 1$

Let $c > 1$ ⁠. Then $f(c) = c + 1$ ⁠. Every $x$ closer to $c$ than $c - 1$ is also greater than $1$ ⁠, so near $c$ we have $f(x) = x + 1$ ⁠. Hence
$\lim_{x\to c} f(x) = \lim_{x\to c} (x + 1) = c + 1 = f(c).$
So $f$ is continuous at every $c > 1$ ⁠.

Same reasoning as Step 2, on the other side of the junction: near any $c > 1$ ⁠, only the formula $x + 1$ is used.

$x + 1$ is a polynomial (degree 1), so it is continuous everywhere.
Test the junction $x = 1$

Value. $1 \ge 1$ ⁠, so the first formula applies: $f(1) = 1 + 1 = 2$ ⁠.

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

At the junction, the left side and the right side are governed by different formulas, so neither formula alone can give "the limit". We compute each one-sided limit with its own formula. Each is a limit of a polynomial, so we may substitute $x = 1$ ⁠.

The limit at $1$ exists because the two one-sided limits agree; continuity also needs this common value to equal $f(1)$ ⁠. Basics: one-sided limits
Conclude

$f$ is continuous at every $c < 1$ (Step 2), every $c > 1$ (Step 3) and at $c = 1$ (Step 4). That covers the whole domain $\R$ ⁠, so $f$ has no point of discontinuity.

The graph (below) has a sharp corner at $(1, 2)$ ⁠: the parabola arrives with slope $2$ and the line leaves with slope $1$ ⁠. A corner is not a break. Continuity only asks that the pieces meet; whether they meet smoothly is the question of differentiability, studied in Section 5.3.

Blue:

y = x^2 + 1

for

x < 1

⁠. Gold:

y = x + 1

for

x \ge 1

⁠. Both pieces reach the filled point

(1, 2)

⁠, so the graph is unbroken (it has a corner there). Horizontal scale: 1 unit = 60 px; vertical: 1 unit = 40 px.

Result

Both pieces are polynomials, so $f$ is continuous at every $x \ne 1$ ⁠. At $x = 1$ ⁠: left-hand limit $2$ ⁠, right-hand limit $2$ ⁠, $f(1) = 2$ ⁠. $f$ is continuous everywhere: there is no point of discontinuity. (Textbook answer: "No point of discontinuity".)

Check it with numbers

Just left of $1$ ⁠: $f(0.999) = 0.999^2 + 1 = 1.998001$ ⁠. Just right: $f(1.001) = 1.001 + 1 = 2.001$ ⁠. At $1$ ⁠: $f(1) = 2$ ⁠. Both neighbours are within $0.002$ of $2$ ⁠, and closer points get closer still ( $f(0.99999) = 1.9999800001$ ⁠, $f(1.00001) = 2.00001$ ⁠). ✓

Common mistakes

Using the wrong formula for $f(1)$ ⁠. $x = 1$ satisfies $x \ge 1$ ⁠, so $f(1) = 1 + 1$ ⁠. (Here $1^2 + 1$ happens to give the same number, but in Question 12 the choice matters.)
Calling the corner a discontinuity. The graph turns sharply at $(1, 2)$ but does not jump. Continuity is about jumps and gaps, not about smoothness.
Only checking the junction. A complete answer also says why $f$ is continuous away from $x = 1$ ⁠: there it equals a single polynomial.

What to take away

For a piecewise function built from polynomials, the only candidates for discontinuity are the junctions.
At a junction, each one-sided limit uses the formula on its own side; then compare both with the value.
Two pieces that give the same value at the junction join up; the graph may still have a corner.

Every fact used, and where it comes from

Fact	Used in	Source
Every polynomial is continuous: $\lim_{x\to c} p(x) = p(c)$	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 the same way in Example 10, Section 5.2
Left-hand and right-hand limits	Step 4	Class XI, Limits and Derivatives; recalled in Section 5.2
Continuity at $c$ ⁠: left-hand limit $=$ right-hand limit $= f(c)$	Steps 4, 5	Definition 1, Section 5.2
Slopes $2x$ and $1$ (corner remark only); differentiability is a separate, stronger property	Step 5 remark	Class XI derivatives of $x^n$ ⁠; Section 5.3

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Exercise 5.1 · Q10: Points of discontinuity: x+1x + 1x+1 for x≥1x \ge 1x≥1⁠, x2+1x^2 + 1x2+1 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 · Q10: Points of discontinuity: $x + 1$ for $x \ge 1$ ⁠, $x^2 + 1$ for $x < 1$

Every point $c < 1$

Every point $c > 1$

Test the junction $x = 1$