Unit 1 · Topic 1.9 Beta

Limits Across Graphs, Tables and Formulas

A limit can be read from a graph (the height the curve approaches), estimated from a table (values on both sides), found exactly from a formula (algebra), and stated in words or notation.

Practice 2: Connecting Representations

Question set for this topic

Part 1 · Hook

Why this matters

A doctor reads a heartbeat as a trace on a screen, as numbers on a chart, and as a written note. All three describe the same heart. A limit works the same way: one function can come as a graph, a table, a formula or a sentence, and you should be able to read the limit from any of them.

Part 2 · Before you start

What this builds on

Part 3 · Prerequisite check

Quick check before you start

1. On a graph, an open circle at $(2, 3)$ means

the point $(2, 3)$ is not on the graph
the function equals 3 at $x = 2$
the graph has a vertical asymptote at $x = 2$
the limit at 2 does not exist

Show the answer

An open circle marks a missing point; a closed dot marks a point that is there.

Correct: the point $(2, 3)$ is not on the graph:
the function equals 3 at $x = 2$ :
the graph has a vertical asymptote at $x = 2$ :
the limit at 2 does not exist:

2. To estimate $\lim_{x \to 2} f(x)$ from a table, you should use inputs

on both sides of 2, closer and closer to it
equal to 2
far from 2, so the pattern is clear
greater than 2 only

Show the answer

A two-sided limit needs values from both sides, approaching 2.

Correct: on both sides of 2, closer and closer to it:
equal to 2:
far from 2, so the pattern is clear:
greater than 2 only:

3. $\frac{x^2 - 9}{x - 3}$ equals which expression for $x \ne 3$ ?

$x + 3$
$x - 3$
$x^2 - 3$
$x + 9$

Show the answer

$x^2 - 9 = (x - 3)(x + 3)$ , and the $x - 3$ cancels.

Correct: $x + 3$ :
$x - 3$ :
$x^2 - 3$ :
$x + 9$ :

Part 4 · See it

See it first

Three panels for f(x) = (x² − x − 2)/(x − 2). The graph is the line y = x + 1 with an open circle at (2, 3) because f(2) is undefined. The table gives f(1.9) = 2.9, f(1.99) = 2.99, f(1.999) = 2.999 and f(2.001) = 3.001, f(2.01) = 3.01, f(2.1) = 3.1. The formula factors as (x − 2)(x + 1)/(x − 2) = x + 1 for x not 2. All three give a limit of 3 as x approaches 2. — The same limit, read from a graph, a table and a formula. LevlPrep original diagram.

Part 5 · Step by step

How it works, step by step

A graph shows the function near $c$ follow the curve toward $x = c$ from each side and read the height it approaches; ignore any dot exactly at $c$
A table lists values at inputs closing in on $c$ from both sidesthe values suggest the limit, but a table cannot prove it
A formula is availablealgebra (substitute, factor, rationalize) gives the exact limit and confirms the estimate
A limit involves a composition $f(g(x))$ find where $g(x)$ goes and from which side, then read $f$ near that input

Part 6 · Key ideas

Key ideas

The same limit can be read from a graph, a table, a formula or words. Each answer must agree.
On a graph, the limit is the height the curve approaches, not the closed dot at the point.
A table suggests a limit; a formula confirms it. Cleverly chosen inputs can make a table mislead, as with $\sin\left(\frac{\pi}{x}\right)$ .
Translate words exactly: "from values less than 4" is $x \to 4^-$ ; the number the outputs approach goes after the equals sign.
For $f(g(x))$ , find the inner limit first and note the side it comes from.

Part 7 · Misconception

A common mistake

The wrong idea: If every value in a table is close to 3, the limit is 3.

What actually happens: A table shows finitely many inputs. The function can behave differently between them; $\sin\left(\frac{\pi}{x}\right)$ is 0 at $x = 1, 0.1, 0.01$ but has no limit at 0. Use the table to estimate and a formula or a graph to confirm.

Part 8 · Check yourself

Check yourself

Exam-style questions. Anything you miss goes into your review queue.

Graph

The graph of $f$

1. Use the graph to find $\lim_{x \to -1} f(x)$ .

No calculator. Type the exact value (a fraction, π, e or a root is fine).

Show the answer

From the left, $y = x + 3$ approaches 2; from the right, $y = 1 - x$ approaches 2. Both sides agree, so the limit is 2, even though $f(-1) = 4$ .

Answer: 2

2. Which is true about $\lim_{x \to 2} f(x)$ ?

It does not exist; the sides head to $-1$ and $1$
It equals $1$ , the height of the closed dot at $x = 2$
It equals $-1$ , the height of the open circle at $x = 2$
It equals $0$ , halfway between the two one-sided limits

Show the answer

$\lim_{x \to 2^-} f(x) = -1$ and $\lim_{x \to 2^+} f(x) = 1$ . The one-sided limits differ, so the two-sided limit does not exist.

Correct: It does not exist; the sides head to $-1$ and $1$ : Left of 2 the graph follows $1 - x$ down to $-1$ ; right of 2 it starts at 1. Different, so no limit.
It equals $1$ , the height of the closed dot at $x = 2$ : The closed dot is $f(2)$ and the right-hand limit, but the left side does not head there.
It equals $-1$ , the height of the open circle at $x = 2$ : That is the left-hand limit; the right side heads to 1.
It equals $0$ , halfway between the two one-sided limits: One-sided limits are compared, never averaged.

3. Find $\lim_{x \to 0} f(f(x))$ .

No calculator. Type the exact value (a fraction, π, e or a root is fine).

Show the answer

Inside: near $x = 0$ , $f(x) = 1 - x$ , which approaches 1. Outside: for inputs near 1 (on both sides), $f(u) = 1 - u$ , which approaches 0. So $\lim_{x \to 0} f(f(x))$ $= 0$ .

Answer: 0

Table

Values of $g(x)$ near $x = 2$

Values of a function g at inputs on both sides of x = 2, rounded to five decimal places
$x$	1.9	1.99	1.999	2.001	2.01	2.1
$g(x)$	2.92564	2.99251	2.99925	3.00075	3.00751	3.07561

4. Which formula could produce the table?

$g(x) = \frac{x^2 - 4}{x - 2}$
$g(x)$ $= \frac{x^3 - 8}{x^2 - 2x}$
$g(x) = \frac{x^3 - 8}{x^2 - 4}$
$g(x) = x + 1$

Show the answer

$\frac{x^3 - 8}{x^2 - 4}$ $= \frac{(x - 2)(x^2 + 2x + 4)}{(x - 2)(x + 2)}$ $= \frac{x^2 + 2x + 4}{x + 2}$ . At $x = 1.9$ this is $\frac{11.41}{3.9} \approx 2.92564$ , matching the table.

$g(x) = \frac{x^2 - 4}{x - 2}$ : This simplifies to $x + 2$ , which is 3.9 at $x = 1.9$ and approaches 4.
$g(x)$ $= \frac{x^3 - 8}{x^2 - 2x}$ : This simplifies to $\frac{x^2 + 2x + 4}{x}$ , which approaches 6.
Correct: $g(x) = \frac{x^3 - 8}{x^2 - 4}$ : It simplifies to $\frac{x^2 + 2x + 4}{x + 2}$ , which gives 2.92564 at 1.9 and approaches 3.
$g(x) = x + 1$ : It approaches 3, but gives exactly 2.9 at $x = 1.9$ , not 2.92564.

5. A table gives $\sin\left(\frac{\pi}{x}\right) = 0$ at $x = 1, 0.1, 0.01$ and $0.001$ . What is true about $\lim_{x \to 0^+} \sin\left(\frac{\pi}{x}\right)$ ?

It equals 0, since each value in the table is 0
It does not exist; the values swing between $-1$ and $1$
It equals 1, the largest value the sine function reaches
It equals 0, but from the right side of 0 alone

Show the answer

The chosen inputs make $\frac{\pi}{x}$ a multiple of $\pi$ , where sine is 0. Other inputs, such as $x = \frac{2}{5}$ , give $\sin\left(\frac{5\pi}{2}\right) = 1$ . Near 0 the function hits $-1$ and $1$ again and again, so it has no limit.

It equals 0, since each value in the table is 0: The table's inputs were unlucky: they all land on zeros of sine. A table can mislead.
Correct: It does not exist; the values swing between $-1$ and $1$ : Inputs like $\frac{2}{5}, \frac{2}{9}, \frac{2}{13}, \ldots$ give 1, and others give $-1$ , arbitrarily close to 0.
It equals 1, the largest value the sine function reaches: The function reaches 1 near 0 but does not settle there.
It equals 0, but from the right side of 0 alone: The table is from the right side already, and the values there oscillate.

6. Let $g(x) = x^2 + 1$ . Suppose $\lim_{u \to 1^-} f(u) = 2$ and $\lim_{u \to 1^+} f(u) = 5$ . What is $\lim_{x \to 0} f(g(x))$ ?

$2$
$3.5$
$5$
It does not exist

Show the answer

As $x \to 0$ , $g(x) = x^2 + 1 \to 1$ , and $g(x) > 1$ for every $x \ne 0$ . So the inputs of $f$ approach 1 from the right, and $f(g(x)) \to 5$ .

$2$ : That would need $g(x)$ to approach 1 from below, but $x^2 + 1 \ge 1$ .
$3.5$ : Limits are not averaged.
Correct: $5$ : $x^2 + 1$ is above 1 on both sides of $x = 0$ , so only the right-hand limit of $f$ matters.
It does not exist: Both sides of $x = 0$ send $g(x)$ to 1 from above, so the limit exists and is 5.

7. A function has $f(3) = 1$ and $\lim_{x \to 3} f(x) = 4$ . What does its graph show at $x = 3$ ?

A closed dot at $(3, 4)$ and an open circle at $(3, 1)$
A closed dot at $(3, 1)$ and a vertical asymptote
Two closed dots, one at $(3, 1)$ and one at $(3, 4)$
An open circle at $(3, 4)$ and a closed dot at $(3, 1)$

Show the answer

The curve heads to height 4 from both sides but the point there is missing (open circle). The function's value, 1, sits elsewhere as a closed dot.

A closed dot at $(3, 4)$ and an open circle at $(3, 1)$ : That would make $f(3) = 4$ , and the curve would head toward 1.
A closed dot at $(3, 1)$ and a vertical asymptote: An asymptote means unbounded behavior, but the limit is the number 4.
Two closed dots, one at $(3, 1)$ and one at $(3, 4)$ : A function has one value at $x = 3$ , so there is one closed dot.
Correct: An open circle at $(3, 4)$ and a closed dot at $(3, 1)$ : The curve approaches $(3, 4)$ , which is not on the graph, and the actual point is $(3, 1)$ .

Part 9 · Summary

Summary

A limit can be read from a graph (the height the curve approaches), estimated from a table (values on both sides), found exactly from a formula (algebra), and stated in words or notation. Move between them: estimate with a table, confirm with a formula, and check the sides when one function sits inside another.

Part 10 · Up next

What comes next

Part 11 · Connections

Why this matters

What this builds on

Quick check before you start

See it first

How it works, step by step

Key ideas

A common mistake

Check yourself

The graph of ff

Values of g(x)g(x) near x=2x = 2

Summary

What comes next

Connections

The graph of $f$

Values of $g(x)$ near $x = 2$