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1. The idea of a limit

Warm-up

Three quick questions from earlier lessons.

Discover

$f(x) = \dfrac{x^2 - 4}{x - 2}$ cannot be computed at $x = 2$ . Here are some values near 2: $f(1.9) = 3.9$ , $f(1.99) = 3.99$ , $f(2.01) = 4.01$ , $f(2.1) = 4.1$ . What is $f(x)$ getting close to?

4
0
Nothing, since $f(2)$ does not exist

Key idea

A limit describes where $f(x)$ is heading as $x$ gets close to $a$ , not what happens at $a$ .

\displaystyle\lim_{x \to a} f(x) = L

means:

f(x)

gets as close to

L

as we like when

x

is close enough to

a

(but not equal to

a

). The value

f(a)

, if it exists, plays no role.

Worked example

Find $\displaystyle\lim_{x \to 3} \dfrac{x^2 - 9}{x - 3}$ .

Substitution gives

\dfrac{0}{0}

At x = 3 we get 0/0: no value, but maybe a limit.

\dfrac{(x - 3)(x + 3)}{x - 3} = x + 3

Factor and cancel, for x ≠ 3.

\displaystyle\lim_{x \to 3} \dfrac{x^2 - 9}{x - 3} = 6

x + 3 approaches 6.

Your turn, guided

Your turn: $\displaystyle\lim_{x \to 5} \dfrac{x^2 - 25}{x - 5}$ .

\dfrac{(x - 5)(x + 5)}{x - 5} = x + 5

For x ≠ 5.

The limit is

10

5 + 5.

Practice ★

$f(x) = \dfrac{x^2 - 36}{x - 6}$ is not defined at $x = 6$ . What value does $f(x)$ approach as $x \to 6$ ?
$f(x) = \dfrac{x^2 - 1}{x - 1}$ is not defined at $x = 1$ . What value does $f(x)$ approach as $x \to 1$ ?
$f(x) = \dfrac{x^2 - 9}{x - 3}$ is not defined at $x = 3$ . What value does $f(x)$ approach as $x \to 3$ ?

Watch out!

Practice ★★

$f(x) = \dfrac{(x - p)(x + 4)}{x - p}$ for $p = 0$ : its graph is the line $y = x + 4$ with a hole. Tap the point the graph approaches as $x \to 0$ .
$f(x) = \dfrac{(x - p)(x + 1)}{x - p}$ for $p = -2$ : its graph is the line $y = x + 1$ with a hole. Tap the point the graph approaches as $x \to -2$ .

$f(x) = x^2$ for $x \ne 4$ , and $f(4) = 33$ . What is $\displaystyle\lim_{x \to 4} f(x)$ ?
1. $16$
2. $33$
3. It does not exist
$f(x) = x^2$ for $x \ne 3$ , and $f(3) = 21$ . What is $\displaystyle\lim_{x \to 3} f(x)$ ?
1. $9$
2. $21$
3. It does not exist

Exit ticket

Three questions to check you have got it.

$f(x) = \dfrac{x^2 - 36}{x - 6}$ is not defined at $x = 6$ . What value does $f(x)$ approach as $x \to 6$ ?
$f(x) = \dfrac{x^2 - 1}{x - 1}$ is not defined at $x = 1$ . What value does $f(x)$ approach as $x \to 1$ ?
$f(x) = \dfrac{x^2 - 9}{x - 3}$ is not defined at $x = 3$ . What value does $f(x)$ approach as $x \to 3$ ?

Summary card

The idea of a limit

$\displaystyle\lim_{x \to a} f(x)$ : where $f(x)$ heads as $x$ approaches $a$ .
$f(a)$ itself does not matter (it may not even exist).
$\dfrac{0}{0}$ means: simplify, then look again.

Watch out: the limit is not always $f(a)$ .

2. Compute limits with algebra

Warm-up

Three quick questions from earlier lessons.

Discover

To find $\displaystyle\lim_{x \to 1} (x^2 + 3x)$ , what can you do?

Substitute $x = 1$ : the limit is 4
You can never substitute in a limit

Key idea

First try substituting. If you get $\dfrac{0}{0}$ , simplify and try again.

If substituting gives a number, that is the limit (for polynomials, and rational functions where the denominator is not 0). If it gives

\dfrac{0}{0}

: factor and cancel, or expand, or use

(\sqrt{x} - s)(\sqrt{x} + s) = x - s^2

; then substitute.

Worked example

Find $\displaystyle\lim_{h \to 0} \dfrac{(3 + h)^2 - 9}{h}$ .

(3 + h)^2 - 9 = 6h + h^2

Substitution gives 0/0: expand instead.

\dfrac{6h + h^2}{h} = 6 + h

Divide by h (h ≠ 0).

The limit is

6

h → 0.

Your turn, guided

Your turn: $\displaystyle\lim_{x \to 4} \dfrac{x^2 - 5x + 4}{x - 4}$ .

= \displaystyle\lim_{x \to 4} (x - 1)

Factor: (x − 4)(x − 1).

= 3

4 − 1.

Practice ★

Find $\displaystyle\lim_{x \to 7} \dfrac{x^2 - 15x + 56}{x - 7}$ .
Find $\displaystyle\lim_{x \to 1} \dfrac{x^2 - 4x + 3}{x - 1}$ .
Find $\displaystyle\lim_{x \to 2} \dfrac{x^2 - 9x + 14}{x - 2}$ .
Find $\displaystyle\lim_{x \to 4} \dfrac{x^2 - 11x + 28}{x - 4}$ .

Watch out!

Practice ★★

Find $\displaystyle\lim_{h \to 0} \dfrac{(8 + h)^2 - 64}{h}$ .
Find $\displaystyle\lim_{h \to 0} \dfrac{(7 + h)^2 - 49}{h}$ .
Find $\displaystyle\lim_{h \to 0} \dfrac{(4 + h)^2 - 16}{h}$ .

Find $\displaystyle\lim_{x \to 9} \dfrac{\sqrt{x} - 3}{x - 9}$ .
Find $\displaystyle\lim_{x \to 4} \dfrac{\sqrt{x} - 2}{x - 4}$ .

Exit ticket

Three questions to check you have got it.

Find $\displaystyle\lim_{x \to 7} \dfrac{x^2 - 15x + 56}{x - 7}$ .
Find $\displaystyle\lim_{x \to 1} \dfrac{x^2 - 4x + 3}{x - 1}$ .
Find $\displaystyle\lim_{x \to 2} \dfrac{x^2 - 9x + 14}{x - 2}$ .

Summary card

Limits with algebra

Substitute. A number? That is the limit.
$\dfrac{0}{0}$ ? Factor and cancel, expand, or use conjugates.
Substitute again.

Watch out: $\dfrac{0}{0}$ is never the answer.

3. One-sided limits

Warm-up

Three quick questions from earlier lessons.

Discover

A parking garage charges 5 dollars for up to 1 hour and 9 dollars for more. As the time approaches 1 hour from below, the price is 5; from above, it is 9. Does the price approach one single number near 1 hour?

No: 5 from the left, 9 from the right
Yes: 5

Key idea

We can approach $a$ from one side only.

\displaystyle\lim_{x \to a^-} f(x)

uses

x < a

;

\displaystyle\lim_{x \to a^+} f(x)

uses

x > a

. The two-sided limit exists exactly when both one-sided limits exist and are equal.

Worked example

$f(x) = \begin{cases} x^2 & \text{if } x < 1 \\ 3 - x & \text{if } x \ge 1 \end{cases}$ . Find $\displaystyle\lim_{x \to 1} f(x)$ .

\displaystyle\lim_{x \to 1^-} f(x) = 1

x < 1: use x².

\displaystyle\lim_{x \to 1^+} f(x) = 2

x > 1: use 3 − x.

The limit does not exist

1 ≠ 2.

Your turn, guided

Your turn: same idea with $g(x) = \begin{cases} x + 4 & \text{if } x < 2 \\ 3x & \text{if } x \ge 2 \end{cases}$ .

\displaystyle\lim_{x \to 2^-} g(x) = 6

Left: 2 + 4.

\displaystyle\lim_{x \to 2^+} g(x) = 6

, so the limit exists and is 6

Right: 3 · 2.

Practice ★

$f(x) = \begin{cases} x + 3 & \text{if } x < 2 \\ 2x & \text{if } x \ge 2 \end{cases}$
$f(x) = \begin{cases} x + 7 & \text{if } x < 4 \\ 2x & \text{if } x \ge 4 \end{cases}$
$f(x) = \begin{cases} x + 1 & \text{if } x < 5 \\ 3x & \text{if } x \ge 5 \end{cases}$

Watch out!

Practice ★★

$f(x) = \begin{cases} x + 6 & \text{if } x < 5 \\ 2x & \text{if } x \ge 5 \end{cases}$ . What is $\displaystyle\lim_{x \to 5} f(x)$ ?
1. It does not exist: the left and right limits differ
2. $11$
3. $10$
$f(x) = \begin{cases} x + 7 & \text{if } x < 3 \\ 4x & \text{if } x \ge 3 \end{cases}$ . What is $\displaystyle\lim_{x \to 3} f(x)$ ?
1. It does not exist: the left and right limits differ
2. $10$
3. $12$

Find $\displaystyle\lim_{x \to 0^-} \dfrac{|3x|}{x}$ .
Find $\displaystyle\lim_{x \to 0^-} \dfrac{|5x|}{x}$ .
Find $\displaystyle\lim_{x \to 0^-} \dfrac{|8x|}{x}$ .

Exit ticket

Three questions to check you have got it.

Find $\displaystyle\lim_{x \to 0^-} \dfrac{|3x|}{x}$ .
Find $\displaystyle\lim_{x \to 0^-} \dfrac{|5x|}{x}$ .
Find $\displaystyle\lim_{x \to 0^-} \dfrac{|8x|}{x}$ .

Summary card

One-sided limits

$x \to a^-$ : from the left; $x \to a^+$ : from the right.
The limit exists only if both sides agree.

Watch out: one side alone is not the two-sided limit.

4. Limits at infinity

Warm-up

Three quick questions from earlier lessons.

Discover

What happens to $\dfrac{1}{x}$ as $x$ grows: $\dfrac{1}{10}, \dfrac{1}{100}, \dfrac{1}{1000}, \dots$ ?

It approaches 0
It has no limit, since it never reaches 0

Key idea

$\displaystyle\lim_{x \to \infty} f(x) = L$ means $f(x)$ gets close to $L$ as $x$ grows; then $y = L$ is a horizontal asymptote.

For a rational function, divide every term by the highest power of

x

in the denominator. Top degree smaller: limit 0. Equal degrees: ratio of the leading coefficients. Top degree larger: no finite limit (

\pm\infty

Worked example

Find $\displaystyle\lim_{x \to \infty} \dfrac{6x^2 - x}{2x^2 + 5}$ .

\dfrac{6 - \frac{1}{x}}{2 + \frac{5}{x^2}}

Divide every term by x².

\to \dfrac{6}{2}

1/x and 5/x² go to 0.

The limit is

3

Simplify.

Your turn, guided

Your turn: $\displaystyle\lim_{x \to \infty} \dfrac{4x + 1}{2x - 7}$ .

Ratio of leading coefficients:

\dfrac{4}{2}

Same degree.

The limit is

2

4 ÷ 2.

Practice ★

Find $\displaystyle\lim_{x \to \infty} \dfrac{ 4x^2 + 1}{ 5x^2 + 4}$ .
Find $\displaystyle\lim_{x \to \infty} \dfrac{ 3x^2 + 4}{ 2x^2 + 1}$ .
Find $\displaystyle\lim_{x \to \infty} \dfrac{ 3x^2 + 4}{ 5x^2 + 4}$ .

Watch out!

Practice ★★

Find $\displaystyle\lim_{x \to \infty} \dfrac{ 6x + 3}{x^2 + 2}$ .
Find $\displaystyle\lim_{x \to \infty} \dfrac{ 9x + 6}{x^2 + 6}$ .
Find $\displaystyle\lim_{x \to \infty} \dfrac{ 7x + 6}{x^2 + 1}$ .

What is $\displaystyle\lim_{x \to \infty} \dfrac{ 5x^3}{x^2 + 5}$ ?
1. It is $\infty$ : the fraction grows without bound
2. $5$
3. $0$
What is $\displaystyle\lim_{x \to \infty} \dfrac{ 2x^3}{x^2 + 2}$ ?
1. It is $\infty$ : the fraction grows without bound
2. $2$
3. $0$

Exit ticket

Three questions to check you have got it.

Find $\displaystyle\lim_{x \to \infty} \dfrac{ 4x^2 + 1}{ 5x^2 + 4}$ .
Find $\displaystyle\lim_{x \to \infty} \dfrac{ 3x^2 + 4}{ 2x^2 + 1}$ .
Find $\displaystyle\lim_{x \to \infty} \dfrac{ 3x^2 + 4}{ 5x^2 + 4}$ .

Summary card

Limits at infinity

Divide by the highest power in the denominator.
Top degree smaller: 0. Equal: ratio of leading coefficients. Larger: $\pm\infty$ .
A finite limit $L$ gives the asymptote $y = L$ .

Watch out: constants do not decide the limit at infinity.

5. Infinite limits and vertical asymptotes

Warm-up

Three quick questions from earlier lessons.

Discover

What happens to $\dfrac{1}{x}$ as $x$ approaches 0 from the right: $\dfrac{1}{0.1}, \dfrac{1}{0.01}, \dots$ ?

It grows without bound
It approaches 0

Key idea

Near a zero of the denominator, a fraction can blow up.

f(x)

grows without bound as

x \to a^+

(or

a^-

), we write

\lim f(x) = +\infty

(or

-\infty

): the line

x = a

is a vertical asymptote. Find the sign by testing a value just to the left or right of

a

Worked example

Find $\displaystyle\lim_{x \to 3^-} \dfrac{2}{x - 3}$ .

x - 3

is a small negative number

Just left of 3, for example x = 2.99.

The quotient is large and negative

2 divided by it.

\displaystyle\lim_{x \to 3^-} \dfrac{2}{x - 3} = -\infty

Write the infinite limit.

Your turn, guided

Your turn: $f(x) = \dfrac{x + 1}{x^2 - 16}$ . Where are the vertical asymptotes?

x = 4

x = -4

x² − 16 = 0.

Asymptotes

x = 4

and

x = -4

The numerator is not 0 there. Type the negative one.

Practice ★

What is $\displaystyle\lim_{x \to 6^+} \dfrac{ 7 }{x - 6}$ ?
1. $+\infty$
2. $-\infty$
3. $0$
What is $\displaystyle\lim_{x \to 3^+} \dfrac{ 4 }{x - 3}$ ?
1. $+\infty$
2. $-\infty$
3. $0$
What is $\displaystyle\lim_{x \to 7^+} \dfrac{ 6 }{x - 7}$ ?
1. $+\infty$
2. $-\infty$
3. $0$

Watch out!

Practice ★★

What is $\displaystyle\lim_{x \to 8^-} \dfrac{ 6 }{x - 8}$ ?
1. $+\infty$
2. $-\infty$
3. $0$
What is $\displaystyle\lim_{x \to 2^-} \dfrac{ 7 }{x - 2}$ ?
1. $+\infty$
2. $-\infty$
3. $0$
What is $\displaystyle\lim_{x \to 5^-} \dfrac{ 9 }{x - 5}$ ?
1. $+\infty$
2. $-\infty$
3. $0$

$f(x) = \dfrac{x + 8}{x^2 - 9}$ has vertical asymptotes. What is the positive one, $x = c$ ?
$f(x) = \dfrac{x + 4}{x^2 - 4}$ has vertical asymptotes. What is the positive one, $x = c$ ?

Exit ticket

Three questions to check you have got it.

What is $\displaystyle\lim_{x \to 8^-} \dfrac{ 6 }{x - 8}$ ?
1. $+\infty$
2. $-\infty$
3. $0$
What is $\displaystyle\lim_{x \to 2^-} \dfrac{ 7 }{x - 2}$ ?
1. $+\infty$
2. $-\infty$
3. $0$
What is $\displaystyle\lim_{x \to 5^-} \dfrac{ 9 }{x - 5}$ ?
1. $+\infty$
2. $-\infty$
3. $0$

Summary card

Infinite limits

Denominator → 0, numerator not: $\pm\infty$ , a vertical asymptote.
Test a value just beside $a$ for the sign.

Watch out: left and right of the asymptote, the signs can differ.

6. Continuity

Warm-up

Three quick questions from earlier lessons.

Discover

You can draw the graph of a continuous function without lifting your pencil. Which one can you draw that way near $x = 0$ ?

$y = |x|$
$y = \dfrac{1}{x}$

Key idea

Continuity at a point means: no hole, no jump, no asymptote there.

f

is continuous at $a$ when

f(a)

exists,

\displaystyle\lim_{x \to a} f(x)

exists, and they are equal. Polynomials,

\sin

\cos

e^x

and

|x|

are continuous everywhere; rational functions are continuous wherever they are defined.

Worked example

For which $k$ is $f(x) = \begin{cases} x + k & \text{if } x < 2 \\ x^2 & \text{if } x \ge 2 \end{cases}$ continuous?

f(2) = 4

f(2) uses the second piece.

\displaystyle\lim_{x \to 2^-} f(x) = 2 + k

The left limit uses the first piece.

2 + k = 4

, so

k = 2

They must be equal.

Your turn, guided

Your turn: $g(x) = \begin{cases} 3x & \text{if } x < 1 \\ x + m & \text{if } x \ge 1 \end{cases}$ .

3 = 1 + m

Left limit 3, value 1 + m.

m = 2

Solve for m.

Practice ★

For which $k$ is $f(x) = \begin{cases} x^2 + k & \text{if } x < 3 \\ 4x & \text{if } x \ge 3 \end{cases}$ continuous everywhere?
For which $k$ is $f(x) = \begin{cases} x^2 + k & \text{if } x < 3 \\ 8x & \text{if } x \ge 3 \end{cases}$ continuous everywhere?
For which $k$ is $f(x) = \begin{cases} x^2 + k & \text{if } x < 3 \\ 7x & \text{if } x \ge 3 \end{cases}$ continuous everywhere?

Watch out!

Practice ★★

Is $f(x) = \dfrac{x^2 - 25}{x - 5}$ continuous at $x = 5$ ?
1. No: $f(5)$ is not defined
2. Yes: the limit exists, it is $10$
3. No: the limit does not exist
Is $f(x) = \dfrac{x^2 - 81}{x - 9}$ continuous at $x = 9$ ?
1. No: $f(9)$ is not defined
2. Yes: the limit exists, it is $18$
3. No: the limit does not exist

Which functions are continuous for every real $x$ ? Tick all of them.
1. $x^3 - 2x + 1$
2. $\sin x$
3. $|x|$
4. $\dfrac{1}{x}$
5. $\tan x$

Exit ticket

Three questions to check you have got it.

For which $k$ is $f(x) = \begin{cases} x^2 + k & \text{if } x < 3 \\ 4x & \text{if } x \ge 3 \end{cases}$ continuous everywhere?
For which $k$ is $f(x) = \begin{cases} x^2 + k & \text{if } x < 3 \\ 8x & \text{if } x \ge 3 \end{cases}$ continuous everywhere?
For which $k$ is $f(x) = \begin{cases} x^2 + k & \text{if } x < 3 \\ 7x & \text{if } x \ge 3 \end{cases}$ continuous everywhere?

Summary card

Continuity

$f(a)$ exists.
$\displaystyle\lim_{x \to a} f(x)$ exists.
They are equal.

Watch out: a limit that exists is not enough; $f(a)$ must match it.

Answers: Limits: an introduction

1. The idea of a limit

$12$
$2$
$6$

$(0, 4)$
$(-2, -1)$

$16$
$9$

$12$
$2$
$6$

2. Compute limits with algebra

$-1$
$-2$
$-5$
$-3$

$16$
$14$
$8$

$\frac{1}{6}$
$\frac{1}{4}$

$-1$
$-2$
$-5$

3. One-sided limits

$4$
$8$
$15$

It does not exist: the left and right limits differ
It does not exist: the left and right limits differ

$-3$
$-5$
$-8$

$-3$
$-5$
$-8$

4. Limits at infinity

$\frac{4}{5}$
$\frac{3}{2}$
$\frac{3}{5}$

It is $\infty$ : the fraction grows without bound
It is $\infty$ : the fraction grows without bound

$\frac{4}{5}$
$\frac{3}{2}$
$\frac{3}{5}$

5. Infinite limits and vertical asymptotes

$+\infty$
$+\infty$
$+\infty$

$-\infty$
$-\infty$
$-\infty$

$-\infty$
$-\infty$
$-\infty$

6. Continuity

$3$
$15$
$12$

No: $f(5)$ is not defined
No: $f(9)$ is not defined

$x^3 - 2x + 1$ · $\sin x$ · $|x|$

$3$
$15$
$12$