Chapter 5 · Continuity and Differentiability

Example 2: Continuity of $f(x)=x^2$ at $x=0$

ExamplesContinuity at a point · Section 5.2

The question

Examine whether the function $f$ given by $f(x) = x^2$ is continuous at $x = 0$ ⁠.

Exam answer

f(0) = 0^2 = 0

\begin{aligned} \text{LHL} &= \lim_{x\to 0^-} f(x) \\ &= \lim_{x\to 0^-} x^2 = 0 \end{aligned}

\begin{aligned} \text{RHL} &= \lim_{x\to 0^+} f(x) \\ &= \lim_{x\to 0^+} x^2 = 0 \end{aligned}

Since $\text{LHL} = \text{RHL} = f(0)$ ⁠, $f$ is continuous at $x = 0$ ⁠.

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

The one idea this problem is built on

To test continuity at 0 we compare three numbers: $f(0)$ ⁠, the left-hand limit and the right-hand limit. Squaring wipes out the sign of $x$ ⁠, so whether $x$ is a small negative or a small positive number, $x^2$ is a very small positive number. Both sides therefore squeeze down to 0, which is exactly $f(0)$ ⁠.

Given, to find, and the plan

Given: $f(x) = x^2$ for every real $x$ ⁠.
To find: Whether $f$ is continuous at $x = 0$ ⁠.
Plan: Check that 0 is in the domain and find $f(0)$ ⁠. Compute the left-hand and right-hand limits at 0 separately. Compare the three numbers (Definition 1, Section 5.2).

Basics you need

1What a limit is

$\lim_{x\to c} f(x) = L$ means $f(x)$ can be made as close to $L$ as we like by taking $x$ close enough to $c$ (with $x \ne c$ ⁠). Basics: what a limit is

2One-sided limits

$x \to 0^-$ ⁠: $x$ approaches 0 through negative numbers. $x \to 0^+$ ⁠: through positive numbers. The limit at 0 exists exactly when both one-sided limits exist and agree. Basics: left-hand and right-hand limits

3Limit of a product (Class XI)

If $\lim f = A$ and $\lim g = B$ ⁠, then $\lim (f g) = AB$ ⁠. With $f = g = x$ this gives $\lim_{x\to c} x^2 = c\cdot c = c^2$ ⁠. It holds for one-sided limits too. Basics: limits

4Definition 1: continuity at a point

For $c$ in the domain: $f$ is continuous at $c$ when the left-hand limit, the right-hand limit and $f(c)$ all exist and are equal, that is $\lim_{x\to c} f(x) = f(c)$ ⁠. Lesson: continuity at a point

Solution

Is 0 in the domain? Find $f(0)$

$x^2$ can be computed for every real $x$ ⁠, so the domain is $\R$ and 0 belongs to it:
$f(0) = 0^2 = 0.$
Continuity is only tested at points of the domain (Definition 1), and $f(0)$ is one of the three numbers to compare.

$0^2 = 0 \times 0 = 0$ ⁠.
Left-hand limit at 0

For $x < 0$ the rule is still $x^2$ ⁠. By the product rule for limits (Basic 3),
$\lim_{x\to 0^-} f(x) = \lim_{x\to 0^-} x^2 = 0\cdot 0 = 0.$

$x$ $f(x)$ side of 0

$-0.1$ $0.01$ left

$-0.01$ $0.0001$ left

$-0.001$ $0.000001$ left

Even though $x$ is negative here, $x^2$ is positive: a negative times a negative is positive. So the graph comes down to height 0 from the left (it does not come up from below). The values 0.01, 0.0001, 0.000001 shrink towards 0.

$(-0.1)^2 = (-0.1)(-0.1) = +0.01$ ⁠. Basics: one-sided limits
Right-hand limit at 0
$\lim_{x\to 0^+} f(x) = \lim_{x\to 0^+} x^2 = 0\cdot 0 = 0.$

$x$ $f(x)$ side of 0

$0.1$ $0.01$ right

$0.01$ $0.0001$ right

$0.001$ $0.000001$ right

The right side gives exactly the same values as the left, because $(-x)^2 = x^2$ ⁠: the graph is symmetric about the $y$ -axis. Both one-sided limits are 0, so $\lim_{x\to0} x^2 = 0$ ⁠.
How close, exactly?

The distance of $f(x)$ from 0 is $|x^2 - 0| = |x|^2$ ⁠. If $|x| < 0.01$ ⁠, then
$|f(x) - 0| = |x|^2 < (0.01)^2 = 0.0001.$
In general, to get $f(x)$ within any small positive number $\varepsilon$ of 0, take $|x| < \sqrt{\varepsilon}$ ⁠.

This turns the pattern in the tables into a guarantee: $x^2$ is as close to 0 as anyone asks, from either side, once $x$ is close enough to 0. That is precisely what "the limit is 0" means (Basic 1).

For non-negative numbers, $a < b$ implies $a^2 < b^2$ ⁠; apply it with $a = |x|$ ⁠, $b = 0.01$ ⁠. Basics: what a limit is
Compare the three numbers
$\begin{aligned} \lim_{x\to 0^-} f(x) &= 0, \\ \lim_{x\to 0^+} f(x) &= 0, \\ f(0) &= 0. \end{aligned}$
All three exist and are equal, so $\lim_{x\to0} f(x) = 0 = f(0)$ ⁠. By Definition 1, $f$ is continuous at $x = 0$ .

"Limit equals value" is the whole of Definition 1. The textbook writes this as $\lim_{x\to0} f(x) = 0 = f(0)$ ⁠.

Lesson: continuity at a point

$x$	$f(x)$	side of 0
$-0.1$	$0.01$	left
$-0.01$	$0.0001$	left
$-0.001$	$0.000001$	left

$x$	$f(x)$	side of 0
$0.1$	$0.01$	right
$0.01$	$0.0001$	right
$0.001$	$0.000001$	right

The parabola

y = x^2

(same scale on both axes). From the left and from the right the curve comes down to the origin, and the origin

(0, 0)

is itself on the graph, so there is no gap at

x = 0

⁠.

Result

$f(0) = 0$ and both one-sided limits at 0 are 0, so $\lim_{x\to0} f(x) = f(0)$ ⁠: $f(x) = x^2$ is continuous at $x = 0$ . This agrees with the textbook's solution.

Check it with numbers

$x = -0.003$ and $x = 0.003$ both give $f(x) = 0.000009$ ⁠, which is less than $0.0001$ ⁠, as Step 4 promises for every $|x| < 0.01$ ⁠.

Common mistakes

Giving a negative left-hand value. $(-0.1)^2$ is $+0.01$ ⁠, not $-0.01$ ⁠. Use brackets when squaring a negative number.
Reading " $x \to 0^-$ " as " $x$ is negative, so $f(x)$ is negative". The minus sign in $0^-$ describes where $x$ comes from, not the sign of $f(x)$ ⁠.
Only checking $f(0)$ ⁠. Being defined at 0 is just one of the three conditions; the limits must also be computed and compared.

What to take away

The continuity test at a point: left-hand limit, right-hand limit and value must all exist and agree.
Nothing here was special to 0: $\lim_{x\to c} x^2 = c^2 = f(c)$ for every $c$ ⁠, so $x^2$ is continuous everywhere (a case of Example 14, every polynomial is continuous).
When a single formula is used on both sides of $c$ ⁠, the two one-sided limits are computed the same way; they can differ only when the formula changes at $c$ ⁠.

Every fact used, and where it comes from

Fact	Used in	Source
$x^2$ is defined for all real $x$	Step 1	Arithmetic
$\lim (fg) = \lim f \cdot \lim g$ ⁠; $\lim x = c$	Steps 2, 3	Algebra of limits, Class XI (Limits and Derivatives) (Basics: limits)
$(-x)^2 = x^2$	Steps 2, 3	Product of two negatives is positive
$0 \le a < b \Rightarrow a^2 < b^2$	Step 4	Inequalities of non-negative numbers
Continuous at $c$ iff $\lim_{x\to c} f(x) = f(c)$	Step 5	Definition 1, Section 5.2 (Lesson: continuity at a point)

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Example 2: Continuity of f(x)=x2f(x)=x^2f(x)=x2 at x=0x=0x=0

Basics you need

Solution

Is 0 in the domain? Find f(0)f(0)f(0)

Left-hand limit at 0

Right-hand limit at 0

How close, exactly?

Compare the three numbers

Every fact used, and where it comes from

Example 2: Continuity of $f(x)=x^2$ at $x=0$

Is 0 in the domain? Find $f(0)$