Chapter 5 · Continuity and Differentiability

Example 1: Continuity of $f(x)=2x+3$ at $x=1$

ExamplesContinuity at a point · Section 5.2

The question

Check the continuity of the function $f$ given by $f(x) = 2x + 3$ at $x = 1$ ⁠.

Exam answer

f(1) = 2(1) + 3 = 5

\begin{aligned} \text{LHL} &= \lim_{x\to 1^-} f(x) \\ &= \lim_{x\to 1^-}(2x + 3) \\ &= 2(1) + 3 = 5 \end{aligned}

\begin{aligned} \text{RHL} &= \lim_{x\to 1^+} f(x) \\ &= \lim_{x\to 1^+}(2x + 3) \\ &= 2(1) + 3 = 5 \end{aligned}

Since $\text{LHL} = \text{RHL} = f(1)$ ⁠, $f$ is continuous at $x = 1$ ⁠.

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

The one idea this problem is built on

"Continuous at a point" is a test with three numbers in it: the value $f(1)$ ⁠, the height the graph approaches as $x$ comes in from the left, and the height it approaches from the right. The function is continuous at 1 exactly when all three exist and are the same number. For the straight line $2x+3$ all three turn out to be 5, so this example shows the definition working in its simplest setting.

Given, to find, and the plan

Given: The function $f(x) = 2x + 3$ ⁠, one formula for every real $x$ ⁠.
To find: Whether $f$ is continuous at the point $x = 1$ ⁠.
Plan: Check that 1 is in the domain and find $f(1)$ ⁠. Then compute the left-hand limit and the right-hand limit at 1 separately. Finally compare the three numbers, as Definition 1 of Section 5.2 asks.

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 $x \ne c$ ⁠. The value $f(c)$ itself plays no part in the limit. Basics: what a limit is

2Left-hand and right-hand limits

$x \to 1^-$ means $x$ approaches 1 through numbers less than 1 (0.9, 0.99, …); $x \to 1^+$ means through numbers greater than 1 (1.1, 1.01, …). The limit at 1 exists exactly when both one-sided limits exist and are equal. Basics: left-hand and right-hand limits

3Algebra of limits (Class XI)

For a constant $k$ ⁠: $\lim_{x\to c} k = k$ and $\lim_{x\to c} x = c$ ⁠. The limit of a sum is the sum of the limits, and the limit of $k\,g(x)$ is $k$ times the limit of $g(x)$ ⁠. These rules hold for one-sided limits too. Together they say that for a polynomial the limit at $c$ is found by putting $x = c$ ⁠. Basics: limits

4Definition 1: continuity at a point

Let $c$ be in the domain of $f$ ⁠. Then $f$ is continuous at $c$ if $\lim_{x\to c} f(x) = f(c)$ ⁠. In more detail: the left-hand limit, the right-hand limit and the value $f(c)$ all exist and are equal. Lesson: continuity at a point

Solution

Is 1 in the domain? Find $f(1)$

The formula $2x + 3$ can be worked out for every real number, so the domain of $f$ is all of $\R$ ⁠, and 1 is in it. Its value there is
$f(1) = 2(1) + 3 = 5.$
Definition 1 only speaks about points of the domain: continuity at $c$ is never even asked about if $f(c)$ does not exist. So the first job is always to check that the point is in the domain. The number $f(1) = 5$ is also the first of the three numbers we will compare.

The domain of a function given by a formula (and no stated domain) is the set of all real $x$ for which the formula makes sense. A polynomial such as $2x+3$ makes sense for every real $x$ ⁠.
Left-hand limit: let $x$ approach 1 from below

For $x < 1$ the rule is still $f(x) = 2x + 3$ ⁠. Using the algebra of limits (Basic 3):
$\lim_{x\to 1^-} f(x) = \lim_{x\to 1^-}(2x+3) = 2(1) + 3 = 5.$
The table shows the same thing with actual numbers:

$x$ $f(x)$ side of 1

$0.9$ $4.8$ left

$0.99$ $4.98$ left

$0.999$ $4.998$ left

The left-hand limit asks: what height does the graph head towards as we walk towards $x = 1$ from the left? We compute it on its own because, for other functions, the two sides can behave differently (Example 10 is such a case). Here the values 4.8, 4.98, 4.998 climb towards 5.

Limit of a sum and of a constant multiple: $\lim(2x+3) = 2\lim x + \lim 3 = 2(1) + 3$ ⁠. Basics: one-sided limits
Right-hand limit: let $x$ approach 1 from above

For $x > 1$ the rule is again $f(x) = 2x + 3$ ⁠, so in the same way
$\lim_{x\to 1^+} f(x) = \lim_{x\to 1^+}(2x+3) = 2(1) + 3 = 5.$

$x$ $f(x)$ side of 1

$1.1$ $5.2$ right

$1.01$ $5.02$ right

$1.001$ $5.002$ right

From the right the values 5.2, 5.02, 5.002 come down towards 5. Both sides head for the same height, so the (two-sided) limit at 1 exists and equals 5 (Basic 2).
Why "close to 5" is guaranteed, not just suggested

A table only shows a few values. To be sure, measure exactly how far $f(x)$ is from 5:
$f(x) - 5 = (2x + 3) - 5 = 2x - 2 = 2(x - 1).$
So $|f(x) - 5| = 2\,|x - 1|$ ⁠: the distance of $f(x)$ from 5 is exactly twice the distance of $x$ from 1. If we want $f(x)$ within 0.001 of 5, it is enough to take $x$ within 0.0005 of 1; any target, however small, can be met in the same way, from either side.

This is the meaning of "the limit is 5" (Basic 1) checked directly: we can force $f(x)$ as close to 5 as anyone demands just by taking $x$ close enough to 1. It confirms Steps 2 and 3 without relying on the pattern in a table.

$|x - 1|$ is the distance between $x$ and 1 on the number line, whichever side $x$ is on. Basics: what a limit is
Compare the three numbers
$\begin{aligned} \lim_{x\to 1^-} f(x) &= 5, \\ \lim_{x\to 1^+} f(x) &= 5, \\ f(1) &= 5. \end{aligned}$
All three exist and are equal, so $\lim_{x\to 1} f(x) = 5 = f(1)$ ⁠. By Definition 1, $f$ is continuous at $x = 1$ .

Definition 1 is exactly the statement "limit equals value". The textbook writes it in one line, $\lim_{x\to1} f(x) = 5 = f(1)$ ⁠; we have simply shown where each of its pieces comes from.

On a graph: you can draw through the point $(1, 5)$ without lifting the pen. Lesson: continuity at a point

$x$	$f(x)$	side of 1
$0.9$	$4.8$	left
$0.99$	$4.98$	left
$0.999$	$4.998$	left

$x$	$f(x)$	side of 1
$1.1$	$5.2$	right
$1.01$	$5.02$	right
$1.001$	$5.002$	right

The line

y = 2x + 3

(one grid unit is the same length on both axes). Coming from either side, the graph heads straight for the point

(1, 5)

⁠, and that point is on the graph: no gap, no jump.

Result

$f(1) = 5$ and both one-sided limits at 1 equal 5, so $\lim_{x\to 1} f(x) = f(1)$ and $f$ is continuous at $x = 1$ . This is the conclusion of the textbook's own solution.

Check it with numbers

Take $x = 0.9999$ and $x = 1.0001$ ⁠, each 0.0001 away from 1. Then $f(0.9999) = 4.9998$ and $f(1.0001) = 5.0002$ ⁠: both are $0.0002 = 2\times 0.0001$ away from 5, exactly as $|f(x) - 5| = 2|x - 1|$ predicts.

Common mistakes

Stopping at " $f(1)$ exists". Being defined at the point is only one of the three conditions. Example 4 has a function defined at 0 that is not continuous there.
Thinking the limit means "put $x = 1$ ". The limit is about values near 1. Putting $x = 1$ gives the right answer here only because $2x + 3$ is a polynomial (Basic 3); for a piecewise function it can give the wrong answer.
Assuming what is to be shown. Writing $\lim_{x\to1} f(x) = f(1)$ at the start, before computing the limit, is circular. Compute the limit first, then compare.

What to take away

Continuity at $c$ is a comparison of three numbers: left-hand limit, right-hand limit and $f(c)$ ⁠.
For any line $f(x) = mx + b$ ⁠, $|f(x) - f(c)| = |m|\,|x - c|$ ⁠, so a line is continuous at every point. The same is true of every polynomial (Example 8 and Example 14).
Always check first that the point is in the domain.

Every fact used, and where it comes from

Fact	Used in	Source
Domain of a polynomial is $\R$	Step 1	A polynomial can be evaluated at every real number
$\lim k = k$ ⁠, $\lim x = c$ ⁠, limits of sums and constant multiples	Steps 2, 3	Algebra of limits, Class XI (Limits and Derivatives) (Basics: limits)
Limit exists iff both one-sided limits exist and are equal	Steps 3, 5	Left and right hand limits, Class XI, recalled in Section 5.2 after Definition 1 (Basics: one-sided limits)
$f(x) - 5 = 2(x-1)$	Step 4	Algebra
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 1: Continuity of f(x)=2x+3f(x)=2x+3f(x)=2x+3 at x=1x=1x=1

Basics you need

Solution

Is 1 in the domain? Find f(1)f(1)f(1)

Left-hand limit: let xxx approach 1 from below

Right-hand limit: let xxx approach 1 from above

Why "close to 5" is guaranteed, not just suggested

Compare the three numbers

Every fact used, and where it comes from

Example 1: Continuity of $f(x)=2x+3$ at $x=1$

Is 1 in the domain? Find $f(1)$

Left-hand limit: let $x$ approach 1 from below

Right-hand limit: let $x$ approach 1 from above