2.1.1 Limits: what a function approaches

What a limit means

A limit describes where a function is heading. As the input $x$ gets close to a number $a$ , the output $f(x)$ may get close to a number $L$ , the limit. The limit has the units of the output: if $f$ gives a price in dollars, so does its limit.

Take $f(x) = \dfrac{x^2 - 1}{x - 1}$ . At $x = 1$ the formula reads $\frac{0}{0}$ , which is not a number, so $f(1)$ does not exist. But look at inputs close to 1, from below and from above:

$x$	0.9	0.99	0.999	1.001	1.01	1.1
$f(x)$	1.9	1.99	1.999	2.001	2.01	2.1

From both sides the outputs close in on 2. So the limit of $f(x)$ as $x$ approaches 1 is 2, even though $f(1)$ itself does not exist.

Rule: Limit

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

Read it as "the limit of $f(x)$ as $x$ approaches $a$ is $L$ ". It means $f(x)$ gets as close to $L$ as you like once $x$ is close enough to $a$ , on either side. The value at $a$ itself plays no part.

The graph of $f$ is the line $y = x + 1$ with one point missing. The missing point is drawn as an open circle, called a hole.

The function $f$ has a hole at $(1, 2)$ : the outputs head for 2 from both sides. The function $g$ is the same except that $g(1) = 3$ , and its limit at 1 is still 2.

The second graph makes the rule's last sentence concrete. It shows $g(x) = x + 1$ for every $x \neq 1$ , with $g(1) = 3$ .

Near 1 the outputs of $g$ still head for 2, so $\lim_{x \to 1} g(x) = 2$ . The single value $g(1) = 3$ does not change where the graph is heading.

Try it The cell builds the table above. round(\dots, 6) hides tiny rounding errors in the last digits; without it you would see 1.990000000000001. Calling f(1) would stop the cell with a ZeroDivisionError, which is why the table never uses 1 itself.

⌘/Ctrl + Enter runs the cell

Try it with the explorer below. Press smaller and smaller gaps to bring the two points towards 1, and turn on zoom to follow them. Both outputs close in on 2.

Explore

Pick a gap and watch the two points close in on x = 1 from the left and the right. Turn on zoom to follow them in.

Zoom in with the points

From the left: f(0.5) = 1.5
From the right: f(1.5) = 2.5
Farthest from 2: 0.5. Both sides approach 2, though f(1) itself is undefined: the limit is 2.

Finding a limit by substitution

For most functions you meet, the limit is just the value. Substitute $x = a$ into the formula, and if you get a number, that is the limit:

\lim_{x \to 3} (x^2 + 1) = 3^2 + 1 = 10

Nearby inputs agree: $x = 2.9$ gives 9.41, $x = 2.99$ gives 9.9401, $x = 3.01$ gives 10.0601 and $x = 3.1$ gives 10.61.

Substitution works for the functions of Level 1 at any input in their domain (lesson 1.1.1). These are polynomials, $\sqrt{x}$ , $e^x$ , $\ln x$ , $\sin x$ and $\cos x$ , and sums, products and ratios of them. Lesson 2.1.2 gives such functions a name: continuous.

So $\lim_{x \to -1}(x^3 + 2x^2) = -1 + 2 = 1$ , and $\lim_{x \to 4} \sqrt{x} = 2$ .

When substitution gives $\frac{0}{0}$

A $\frac{0}{0}$ is a signal, not an answer. It usually means the top and the bottom share a factor that is 0 at $a$ . Factor both, cancel the shared factor, then substitute.

For the first example, $x^2 - 1 = (x - 1)(x + 1)$ by the factor theorem (lesson 1.3.1), because 1 and $-1$ are roots. So for every $x \neq 1$ :

\begin{gathered} \frac{x^2 - 1}{x - 1} = \frac{(x - 1)(x + 1)}{x - 1} = x + 1 \\[6pt] \text{so}\quad \lim_{x \to 1} \frac{x^2 - 1}{x - 1} = 1 + 1 = 2 \end{gathered}

Cancelling $x - 1$ is allowed because a limit never uses $x = 1$ itself, so $x - 1$ is never 0. The answer matches the table.

A second example, with a square root

Take $\lim_{x \to 4} \dfrac{x - 4}{\sqrt{x} - 2}$ . Substituting gives $\frac{0}{0}$ again. The table points to 4:

$x$	3.9	3.99	3.999	4.001	4.01	4.1
value	3.974842	3.997498	3.99975	4.00025	4.002498	4.024846

For the algebra, notice that $(\sqrt{x} - 2)(\sqrt{x} + 2) = x - 4$ : multiply it out to check. So the top contains the bottom as a factor:

\begin{aligned} \frac{x - 4}{\sqrt{x} - 2} &= \frac{(\sqrt{x} - 2)(\sqrt{x} + 2)}{\sqrt{x} - 2} \\[6pt] &= \sqrt{x} + 2 \end{aligned}

Now substitute: $\sqrt{4} + 2 = 4$ . The table and the algebra agree.

The letter does not matter

Limits are often taken in a small gap called $h$ , as $h \to 0$ . Lesson 2.2.1 does this all the time. Take $\lim_{h \to 0} \frac{(1 + h)^2 - 1}{h}$ , which reads $\frac{0}{0}$ at $h = 0$ .

Expand the top first: $(1 + h)^2 - 1 = 2h + h^2$ . Then cancel $h$ to get $2 + h$ , which tends to 2. The table agrees: $h = 0.1$ gives 2.1, $h = 0.01$ gives 2.01, $h = -0.01$ gives 1.99 and $h = -0.1$ gives 1.9.

Rule: Finding a limit

\begin{gathered} \text{substitute } x = a \\ \text{if } \tfrac{0}{0}: \text{ simplify, cancel, substitute} \end{gathered}

If substituting gives a number, that is the limit. If it gives $\frac{0}{0}$ , factor or expand, cancel the shared factor, and substitute again.

Try it The loop rebuilds the square-root table. Then sp.limit(expression, x, a) computes each limit of this section exactly: 2, 4 and 2. The variable for the third is h, so it is declared as a symbol too.

import math import sympy as sp for v in [3.9, 3.99, 3.999, 4.001, 4.01, 4.1]: print(v, round((v - 4) / (math.sqrt(v) - 2), 6)) x, h = sp.symbols("x h") print(sp.limit((x**2 - 1) / (x - 1), x, 1)) print(sp.limit((x - 4) / (sp.sqrt(x) - 2), x, 4)) print(sp.limit(((1 + h)**2 - 1) / h, h, 0))

⌘/Ctrl + Enter runs the cell

When substitution gives a number over 0

A nonzero number over 0 is a different case. Take $\dfrac{1}{(x - 1)^2}$ near 1. The top stays at 1 while the bottom shrinks to 0, so the outputs grow without end:

$x$	0.9	0.99	0.999	1.001	1.01	1.1
value	100	10,000	1,000,000	1,000,000	10,000	100

No number is approached, so there is no limit. To say how it fails, write $\lim_{x \to 1} \frac{1}{(x - 1)^2} = \infty$ , read "it grows without bound".

The symbol $\infty$ is not a number, and no amount of simplifying will produce one. Lesson 2.1.2 looks at blow-ups like this from each side.

\frac{1}{(x - 1)^2}

import sympy as sp for x in [0.9, 0.99, 0.999, 1.001, 1.01, 1.1]: print(x, round(1 / (x - 1)**2)) x = sp.symbols("x") print(sp.limit(1 / (x - 1)**2, x, 1))

⌘/Ctrl + Enter runs the cell

A table suggests; it does not prove

Some limits have no factor to cancel. With $x$ in radians (lesson 1.7.1), $\frac{\sin x}{x}$ reads $\frac{0}{0}$ at 0. From both sides the table gives 0.998334 at $\pm 0.1$ and 0.999983 at $\pm 0.01$ , heading for 1.

SymPy confirms that $\lim_{x \to 0} \frac{\sin x}{x} = 1$ . A table gives evidence, and an exact method (algebra or SymPy) settles it.

A table can also mislead if you push too close. The computer stores about 16 significant digits, so $1 + h$ loses most of $h$ when $h$ is tiny. For $\frac{(1 + h)^2 - 1}{h}$ , whose limit is 2:

$h$	$10^{-8}$	$10^{-12}$	$10^{-15}$	$10^{-16}$
computed	1.99999999	2.0001778	2.22044605	0

At $h = 10^{-16}$ the computer stores $1 + h$ as exactly 1, so the top is 0. Keep the gaps of a table from 0.1 down to about 0.0001, where rounding stays invisible. The case study, lesson 2.1.5, meets the same trap with money.

\frac{\sin x}{x}

import math import sympy as sp for x in [0.1, 0.01, -0.01, -0.1]: print(x, round(math.sin(x) / x, 6)) x = sp.symbols("x") print(sp.limit(sp.sin(x) / x, x, 0)) for h in [1e-8, 1e-12, 1e-15, 1e-16]: print(h, round(((1 + h)**2 - 1) / h, 8))

⌘/Ctrl + Enter runs the cell

Common mistakes

A limit is not the value at the point. $g(1) = 3$ , yet $\lim_{x \to 1} g(x) = 2$ . And $f(1)$ need not exist at all.
$\frac{0}{0}$ is not 0, not 1, and not "no limit". It means "simplify and look again".
A number over 0 is not $\frac{0}{0}$ . Something like $\frac{1}{0}$ means the outputs blow up. There is no limit, and simplifying will not find one.
Cancel factors, not terms. $\frac{x^2 + x}{x} = x + 1$ , because $x$ is a factor of the whole top. But $\frac{x^2 + 1}{x}$ is not $x + 1$ .
Look at both sides. A limit needs the same value from the left and from the right. Lesson 2.1.2 shows what happens when they differ.
Do not push a table too close. At a gap of $10^{-12}$ rounding shows in the fourth decimal place. By $10^{-16}$ it swamps the answer.

Limits

0 of 11 correct

1.
Find $\displaystyle\lim_{x \to 3} (x^2 + 1)$
2.
Find $\displaystyle\lim_{x \to -1} (2x^2 - 3x)$
3.
Find $\displaystyle\lim_{x \to 2} \frac{x^2 - 4}{x - 2}$
4.
Find $\displaystyle\lim_{x \to -3} \frac{x^2 - 9}{x + 3}$
5.
Find $\displaystyle\lim_{x \to 1} \frac{x^3 - 1}{x - 1}$
6.
Find $\displaystyle\lim_{x \to 9} \frac{x - 9}{\sqrt{x} - 3}$
7.
Find $\displaystyle\lim_{h \to 0} \frac{(2 + h)^2 - 4}{h}$
8.
$g(x) = x + 1$ for every $x \neq 1$ , and $g(1) = 5$ . Find $\displaystyle\lim_{x \to 1} g(x)$
9.
Near which input does $\dfrac{5}{(x - 3)^2}$ grow without bound?
10.
Find $\displaystyle\lim_{x \to 0} \frac{\sin x}{x}$ , with $x$ in radians
11.
Find $\displaystyle\lim_{x \to 0} \frac{x^2 + 3x}{x}$

How to type answers

Powers with ^: x^7, 2^(1/3), a^-2.
Multiply with * or by writing side by side: 3x^2, 2(x+1).
ln(x) is the natural log, log(x, 2) is log base 2, e is Euler's number, sqrt(x) is a square root.
sin(x), cos(x) and tan(x) take radians; write pi for π and sin(x)^2 for sin²x.
Your first answer to each question is saved: it counts towards your score and brings the question back for review.
Exact answers like ln(20)/ln(3) are best. Decimals need about 4 significant figures.

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

Limits: what a function approaches

What a limit means

Rule: Limit

Finding a limit by substitution

When substitution gives $\frac{0}{0}$

A second example, with a square root

The letter does not matter

Rule: Finding a limit

When substitution gives a number over 0

A table suggests; it does not prove

Common mistakes

Limits

Summary card

Free further reading

What a limit means

Rule: Limit

Finding a limit by substitution

When substitution gives 00\frac{0}{0}00​

A second example, with a square root

The letter does not matter

Rule: Finding a limit

When substitution gives a number over 0

A table suggests; it does not prove

Common mistakes

Summary card

Free further reading

When substitution gives $\frac{0}{0}$