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 $f(a)$ .

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