Introduction to Limits: Tables & Graphs

A limit is the value a function approaches as x gets closer and closer to a number. We write $\lim_{x \to a} f(x) = L$ when f(x) can be made as close to L as we like by taking x close enough to a, from both sides. The function does not even have to be defined at a.

Before this lesson: Function Notation, Graphing Rational Functions, Piecewise Functions and Difference Quotient

What a limit means

A limit asks where a function is headed, not where it lands. Imagine walking toward a door. Your limit is the door, even if the door is locked and you never step through it.

That is why limits are useful at points where a function has a hole. The function $f(x) = \frac{x^2 - 4}{x - 2}$ is undefined at x = 2, because it gives $\frac{0}{0}$ . Its values still close in on 4 as x gets near 2, so its limit there is 4.

For a limit to exist, the function must approach the same value from the left (x less than a) and from the right (x greater than a). If the two sides disagree, or the values grow without bound, the limit does not exist.

Graph of a function that follows the line y = x + 2 with an open circle at (1, 3) and a filled dot at (1, 5). Arrows along the line approach the open circle from the left and from the right, so the limit as x approaches 1 is 3, even though f(1) = 5. — The graph heads toward the open circle at height 3 from both sides, so the limit is 3. The dot at (1, 5) is the function value, which does not change the limit.

How to estimate a limit

Method	How it works	Watch for
From a table	Pick x-values closer and closer to a from both sides, such as 1.9, 1.99, 1.999 and 2.1, 2.01, 2.001, and see what f(x) approaches	Use values on both sides
From a graph	Trace the curve toward x = a from the left and from the right and read the height it heads toward	Ignore any dot exactly at x = a
By substitution	For polynomials and other smooth functions, $\lim_{x \to a} f(x) = f(a)$	If you get $\frac{0}{0}$ , use a table, a graph or algebra instead

The limit $\lim_{x \to a} f(x)$ and the value f(a) are different questions. They are equal for many functions, but a hole, a jump or a moved point can make them differ.

Introduction to limits examples

Example 1: Estimate a limit from a table

Estimate $\lim_{x \to 2} \frac{x^2 - 4}{x - 2}$ .

x	1.9	1.99	1.999	2	2.001	2.01	2.1
f(x)	3.9	3.99	3.999	undefined	4.001	4.01	4.1

f(1.9) = \frac{1.9^2 - 4}{1.9 - 2}

Find one table value to see how the table is made. Substitute 1.9.

f(1.9) = \frac{3.61 - 4}{-0.1}

Square 1.9 and subtract in the denominator.

f(1.9) = \frac{-0.39}{-0.1}

Subtract in the numerator.

f(1.9) = 3.9

Divide. A negative over a negative is positive.

Answer: from both sides the values approach 4, so $\lim_{x \to 2} \frac{x^2 - 4}{x - 2} = 4$ . Algebra agrees: the fraction simplifies to x + 2 when $x \ne 2$ , and 2 + 2 = 4.

Example 2: Read a limit from a graph

Use the graph above. Find $\lim_{x \to 1} f(x)$ and f(1).

\lim_{x \to 1^-} f(x) = 3

Trace the line from the left toward x = 1. It heads to the open circle at height 3.

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

Trace from the right. It heads to the same height.

\lim_{x \to 1} f(x) = 3

Both sides agree, so the limit is 3.

f(1) = 5

The filled dot shows the actual value at x = 1.

Answer: the limit is 3, but f(1) = 5. The limit only cares about values near 1, not the value at 1.

Example 3: A famous trig limit

Estimate $\lim_{x \to 0} \frac{\sin x}{x}$ , with x in radians.

x	−0.1	−0.01	−0.001	0	0.001	0.01	0.1
f(x)	0.99833	0.99998	0.9999998	undefined	0.9999998	0.99998	0.99833

f(0.1) = \frac{\sin 0.1}{0.1}

Substitute 0.1. Set the calculator to radian mode (see radians).

f(0.1) \approx \frac{0.099833}{0.1}

Find the sine.

f(0.1) \approx 0.99833

Divide.

Answer: the values approach 1 from both sides, so $\lim_{x \to 0} \frac{\sin x}{x} = 1$ . Substituting x = 0 gives $\frac{0}{0}$ , which tells you nothing, so the table is the way in. Calculus 1 proves this limit exactly.

Example 4: A limit that does not exist

Estimate $\lim_{x \to 3} \frac{1}{(x - 3)^2}$ .

x	2.9	2.99	2.999	3	3.001	3.01	3.1
f(x)	100	10,000	1,000,000	undefined	1,000,000	10,000	100

f(2.99) = \frac{1}{(2.99 - 3)^2}

Substitute 2.99.

f(2.99) = \frac{1}{(-0.01)^2}

Subtract inside the parentheses.

f(2.99) = \frac{1}{0.0001}

Square. The result is positive.

f(2.99) = 10000

Divide.

Answer: the values grow without bound, so the limit does not exist. The graph has a vertical asymptote at x = 3. (Some books write $\lim_{x \to 3} f(x) = \infty$ to say how it fails to exist.)

Example 5: Word problem with speed

A ball is thrown upward, and its height after t seconds is $s(t) = 64t - 16t^2$ feet. Its average velocity from t = 1 to $t = 1 + \Delta t$ is $\frac{s(1 + \Delta t) - s(1)}{\Delta t}$ . Estimate the limit as $\Delta t \to 0$ , the ball’s velocity at t = 1.

Δt	0.1	0.01	0.001
Average velocity (ft/s)	30.4	31.84	31.984

s(1) = 64(1) - 16(1)^2 = 48

Find the height at t = 1.

s(1.1) = 64(1.1) - 16(1.1)^2

Find the height at t = 1.1, which is Δt = 0.1.

s(1.1) = 70.4 - 19.36 = 51.04

Multiply, then subtract.

\frac{s(1.1) - s(1)}{0.1} = \frac{51.04 - 48}{0.1}

Substitute into the average velocity.

\frac{s(1.1) - s(1)}{0.1} = 30.4

Subtract, then divide. This is the first table entry; the others are found the same way.

Answer: the averages approach 32, so the velocity at t = 1 is 32 ft/s. This is how calculus defines instantaneous speed, using the difference quotient.

Common mistakes

Mistake	Example of the error	Fix
Thinking the limit must equal f(a)	Saying $\lim_{x \to 1} f(x) = 5$ because f(1) = 5 in Example 2	The limit is where the graph heads, 3. The value at the point does not matter.
Stopping at 0/0	Saying $\lim_{x \to 2} \frac{x^2 - 4}{x - 2}$ does not exist because it gives $\frac{0}{0}$	$\frac{0}{0}$ means “look closer.” A table or algebra gives 4.
Checking only one side	Using only x = 1.9, 1.99, 1.999	Check x-values greater than a too; both sides must agree.
Radian-mode errors	Getting 0.01745 for $\frac{\sin 0.1}{0.1}$	That is degree mode. Use radians for limits like this.
Taking too few values	Guessing from x = 1.9 alone	Use several values that get steadily closer to a.
Calling an infinite limit a number	Writing a limit of 1,000,000 in Example 4	The values never settle, so the limit does not exist.

Related methods

Direct substitution

For polynomials, $\lim_{x \to a} p(x) = p(a)$ . For example, $\lim_{x \to 3} (x^2 + 1) = 3^2 + 1 = 10$ . The limit laws behind this, and algebra methods such as factoring, are in Calculus 1, Unit 1.

Left and right sides

The notation $x \to a^-$ means x approaches a from the left, and $x \to a^+$ means from the right. The next lesson covers these one-sided limits and jumps, where the two sides disagree.

Practice problems

Use a table with x = 0.9, 0.99, 1.01 and 1.1 to estimate $\lim_{x \to 1} \frac{x^2 - 1}{x - 1}$ .
Use a table to estimate $\lim_{x \to 0} \frac{e^x - 1}{x}$ .
The graph of f has an open circle at (2, −1) and a filled dot at (2, 3), and otherwise is a smooth curve through the open circle. Find $\lim_{x \to 2} f(x)$ and f(2).
Find $\lim_{x \to 4} (3x - 5)$ by substitution.
Find $\lim_{x \to -3} (x^2 + 2x)$ by substitution.
Use a table to estimate $\lim_{x \to 5} \frac{x - 5}{x^2 - 25}$ .
Describe $\lim_{x \to 0} \frac{1}{x^2}$ .
Use a table to estimate $\lim_{x \to 0} \frac{1 - \cos x}{x}$ , with x in radians.
Let g(x) = x + 1 for $x \ne 2$ and g(2) = 7. Find $\lim_{x \to 2} g(x)$ and g(2).
Water fills a tank so the volume after t minutes is $V(t) = t^3$ liters. Estimate the rate of filling at t = 2 by finding $\frac{V(2 + h) - V(2)}{h}$ for h = 0.1, 0.01 and 0.001.

Show answers

f(0.9) = \frac{0.81 - 1}{-0.1} = 1.9

Substitute 0.9: $0.9^2 = 0.81$ .

f(0.99) = \frac{0.9801 - 1}{-0.01} = 1.99

Substitute 0.99.

f(1.01) = \frac{1.0201 - 1}{0.01} = 2.01

Substitute 1.01.

f(1.1) = \frac{1.21 - 1}{0.1} = 2.1

Substitute 1.1.

Answer: both sides approach 2, so the limit is 2.

f(-0.1) \approx 0.95163

$\frac{e^{-0.1} - 1}{-0.1}$ ; see the number e.

f(-0.01) \approx 0.99502

Substitute −0.01.

f(0.01) \approx 1.00502

Substitute 0.01.

f(0.1) \approx 1.05171

Substitute 0.1.

Answer: both sides approach 1, so the limit is 1.

\lim_{x \to 2} f(x) = -1

The curve heads to the open circle from both sides.

f(2) = 3

The filled dot gives the value.

Answer: the limit is −1 and f(2) = 3.

\lim_{x \to 4} (3x - 5) = 3(4) - 5

A polynomial, so substitute x = 4.

\lim_{x \to 4} (3x - 5) = 7

Simplify.

Answer: 7

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

Substitute x = −3.

\lim_{x \to -3} (x^2 + 2x) = 9 - 6 = 3

Simplify.

Answer: 3

f(4.9) = \frac{-0.1}{-0.99} \approx 0.10101

Substitute 4.9: $4.9^2 - 25 = -0.99$ .

f(4.99) \approx 0.10010

Substitute 4.99.

f(5.01) \approx 0.09990

Substitute 5.01.

f(5.1) = \frac{0.1}{1.01} \approx 0.09901

Substitute 5.1: $5.1^2 - 25 = 1.01$ .

Answer: both sides approach 0.1, so the limit is $\frac{1}{10}$ . (It simplifies to $\frac{1}{x + 5}$ for $x \ne 5$ .)

f(0.1) = \frac{1}{0.01} = 100

Try x = 0.1.

f(0.01) = \frac{1}{0.0001} = 10000

Try x = 0.01. Negative x-values give the same results, since $x^2$ is positive.

Answer: the values grow without bound, so the limit does not exist.

f(0.1) \approx 0.04996

$\frac{1 - \cos 0.1}{0.1}$ in radian mode.

f(0.01) \approx 0.00500

Substitute 0.01.

f(-0.01) \approx -0.00500

Substitute −0.01.

f(-0.1) \approx -0.04996

Substitute −0.1.

Answer: both sides approach 0, so the limit is 0.

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

Near 2, g(x) = x + 1, which heads to 3.

g(2) = 7

The value at 2 is defined separately.

Answer: the limit is 3 and g(2) = 7.

10.

\frac{(2.1)^3 - 8}{0.1} = \frac{9.261 - 8}{0.1} = 12.61

For h = 0.1.

\frac{(2.01)^3 - 8}{0.01} = \frac{8.120601 - 8}{0.01} = 12.0601

For h = 0.01.

\frac{(2.001)^3 - 8}{0.001} \approx 12.006

For h = 0.001.

Answer: the values approach 12, so the tank is filling at about 12 liters per minute at t = 2.

Introduction to limits FAQ

What is a limit in math?

It is the value a function gets closer to as the input gets closer to a certain number. It describes where the function is headed near that number.

Can a limit exist if the function is undefined at that point?

Yes. $\frac{x^2 - 4}{x - 2}$ is undefined at 2, but its limit there is 4. A limit only depends on values near the point.

When does a limit not exist?

It does not exist when the left and right sides approach different values, when the values grow without bound, or when they keep jumping around without settling.

How do you find a limit from a table?

Use x-values that get closer to a from both sides and see what number the outputs approach. If both sides approach the same number, that is the limit.

Summary

$\lim_{x \to a} f(x) = L$ means f(x) gets as close to L as we like when x is close enough to a.
Estimate limits with tables from both sides, with graphs, or by substitution for polynomials.
The limit can differ from f(a), and f(a) may not even exist.
$\frac{0}{0}$ after substituting means “look closer,” not “no limit.”
If the sides disagree or the values grow without bound, the limit does not exist.

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Next lesson: One-Sided Limits

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