Limit Laws — AP® Calculus AB Lesson

Part 1 · Hook

Why this matters

Graphs and tables estimate limits. Rules find them exactly. The limit laws say that limits respect arithmetic: the limit of a sum is the sum of the limits, and the same for differences, products, quotients, powers and roots. For polynomials and most everyday functions, they collapse into the easiest rule in calculus: just substitute.

Part 2 · Before you start

What this builds on

Part 3 · Prerequisite check

Quick check before you start

1. If $\lim_{x \to 2^-} f(x) = 5$ and $\lim_{x \to 2^+} f(x) = 5$ , then $\lim_{x \to 2} f(x)$ is

$5$
it does not exist
$10$
$f(2)$

Show the answer

Equal one-sided limits give the limit.

Correct: $5$ :
it does not exist:
$10$ :
$f(2)$ :

2. Let $f(x) = x^2$ and $g(x) = x + 1$ . What is $f(g(2))$ ?

$9$
$5$
$4$
$6$

Show the answer

$g(2) = 3$ and $f(3) = 9$ .

Correct: $9$ :
$5$ :
$4$ :
$6$ :

Part 4 · See it

See it first

Three curves near x = 2: f approaching 3, g approaching -1, and f + g approaching 2. A panel lists: as x approaches 2, f(x) approaches 3, g(x) approaches -1, f(x) + g(x) approaches 2, f(x) times g(x) approaches -3, and 5f(x) approaches 15. — As $x \to 2$ : $f(x) \to 3$ and $g(x) \to -1$ , so $f(x) + g(x) \to 3 + (-1)$ $= 2$ . Each law turns a new limit into limits you know. LevlPrep original diagram.

Part 5 · Step by step

How it works, step by step

$\lim_{x \to a} f(x) = L$ and $\lim_{x \to a} g(x) = M$ both existthe limit of $f \pm g$ , $fg$ and $kf$ is $L \pm M$ , $LM$ and $kL$
The denominator's limit $M$ is not 0the limit of $\frac{f}{g}$ is $\frac{L}{M}$
Polynomials and rational functions are built from x and constants with these operationstheir limits come from substitution, as long as the denominator is not 0 at $a$
A denominator heads to 0the quotient law is silent: a nonzero top means no finite limit; a top also heading to 0 needs the algebra of Topic 1.6

Part 6 · Key ideas

Key ideas

Sum, difference, constant multiple, product: if both limits exist, combine them the same way. $\lim (f + g)$ $= \lim f + \lim g$ .
Quotient: $\lim \frac{f}{g}$ $= \frac{\lim f}{\lim g}$ only when $\lim g \ne 0$ .
Powers and roots: $\lim [f(x)]^n = L^n$ , and $\lim \sqrt{f(x)} = \sqrt{L}$ when $L > 0$ .
Direct substitution: for a polynomial, or a rational function with a nonzero denominator at $a$ , the limit is the value at $a$ .
For a composite, find the inside limit first: if $g(x) \to b$ and $f$ is continuous at $b$ , then $f(g(x)) \to f(b)$ .
For a piecewise function, use each side's formula for its one-sided limit.

Part 7 · Misconception

A common mistake

The wrong idea: If neither $\lim f$ nor $\lim g$ exists, then $\lim (f + g)$ does not exist.

What actually happens: The laws need both limits to exist; when they do not, the laws are silent, not negative. Check the sum one side at a time: two jumps can cancel, as with $f$ and $g$ at $x = 1$ , where the sum's limit is 3.

Part 8 · Check yourself

Check yourself

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

Graph

The graphs of $f$ and $g$

Both functions are defined for

-1 \le x \le 4

and are made of line segments.

f

(crimson):

f(x) = x + 1

for

x < 1

, ending at an open circle

(1, 2)

, and

f(x) = x - 1

for

x \ge 1

, starting at the filled point

(1, 0)

.

g

(blue):

g(x) = 2 - x

for

x < 1

, ending at an open circle

(1, 1)

, and

g(x) = x + 2

for

x \ge 1

, starting at the filled point

(1, 3)

.

1. What is $\lim_{x \to 1} \left[f(x) + g(x)\right]$ ?

$3$
It does not exist, because neither $\lim_{x \to 1} f(x)$ nor $\lim_{x \to 1} g(x)$ exists.
$5$
$2$

Show the answer

Work one side at a time. From the left, $f(x) \to 2$ and $g(x) \to 1$ , so $f(x) + g(x) \to 3$ . From the right, $f(x) \to 0$ and $g(x) \to 3$ , so the sum $\to 3$ . Both sides give 3, so the limit is 3, even though $f$ and $g$ each jump.

Correct: $3$ : Left: $2 + 1$ . Right: $0 + 3$ . Both 3.
It does not exist, because neither $\lim_{x \to 1} f(x)$ nor $\lim_{x \to 1} g(x)$ exists.: The sum law needs both limits to exist, so it cannot be used here; but the sum can still have a limit. Check each side of the sum directly.
$5$ : 5 mixes sides: $f$ 's left-hand limit 2 plus $g$ 's right-hand limit 3. Each one-sided limit of the sum uses the same side for both functions.
$2$ : 2 is the left-hand limit of $f$ alone; the question asks about the sum $f + g$ .

2. What is $\lim_{x \to 1} f(x)g(x)$ ?

It does not exist.
$2$ , from $2 \cdot 1$
$0$
$3$

Show the answer

From the left, $f(x)g(x) \to 2 \cdot 1 = 2$ . From the right, $f(x)g(x) \to 0 \cdot 3 = 0$ . The one-sided limits differ, so the limit does not exist.

Correct: It does not exist.: Left 2, right 0.
$2$ , from $2 \cdot 1$ : 2 is only the left-hand limit of the product.
$0$ : 0 is only the right-hand limit (and $f(1)g(1)$ ).
$3$ : 3 is the limit of the sum, not the product.

3. Read the needed values from the graphs and use the quotient law to find $\lim_{x \to 2} \frac{f(x)}{g(x)}$ .

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

Show the answer

Near $x = 2$ both graphs are unbroken lines: $f(x) \to 1$ and $g(x) \to 4$ . Since $4 \ne 0$ , the quotient law gives $\frac{1}{4}$ .

Answer: $\frac{1}{4}$ ≈ 0.250

4. Use the graphs to evaluate the limit of the composite function, $\lim_{x \to 0} f(g(x))$ .

$1$
$2$
$3$
$0$

Show the answer

As $x \to 0$ , $g(x) = 2 - x \to 2$ . Near an input of 2, $f$ is the unbroken line $x - 1$ , so $f(g(x)) \to f(2) = 1$ .

Correct: $1$ : Inside first: $g(x) \to 2$ ; then $f$ near 2 gives 1.
$2$ : 2 is the inside limit, $\lim_{x \to 0} g(x)$ . Then apply $f$ .
$3$ : That is $g(1)$ , not the composite near $x = 0$ .
$0$ : That is $f(1)$ from the right; the input to $f$ is near 2, not 1.

5. Find $\lim_{x \to -1} \frac{x^3 + 2x^2 - 5}{x^2 + 1}$ .

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

Show the answer

A rational function whose denominator is not 0 at $x = -1$ : substitute. Numerator $(-1)^3 + 2(-1)^2 - 5 = -4$ ; denominator $(-1)^2 + 1 = 2$ . Limit: $\frac{-4}{2} = -2$ .

Answer: -2

6. Which limit can be found by direct substitution?

$\lim_{x \to 0} \frac{x^2 + 1}{x + 4}$
$\lim_{x \to 3} \frac{x^2 - 9}{x - 3}$
$\lim_{x \to 2} \frac{x + 1}{x - 2}$
$\lim_{x \to 1} \sqrt{x - 5}$

Show the answer

A rational function can be evaluated by substitution where its denominator is not 0. At $x = 0$ the denominator is 4, so the limit is $\frac{1}{4}$ .

Correct: $\lim_{x \to 0} \frac{x^2 + 1}{x + 4}$ : Denominator 4 at $x = 0$ : substitute.
$\lim_{x \to 3} \frac{x^2 - 9}{x - 3}$ : At $x = 3$ the denominator is 0, so substitution gives no value; Topic 1.6 handles this.
$\lim_{x \to 2} \frac{x + 1}{x - 2}$ : At $x = 2$ the denominator is 0.
$\lim_{x \to 1} \sqrt{x - 5}$ : $\sqrt{x - 5}$ is not defined for $x$ near 1.

7. What can you conclude about $\lim_{x \to 2} \frac{x + 1}{x - 2}$ ?

It is not a finite number: the top approaches 3 and the bottom approaches 0.
It equals 0, because the bottom approaches 0 and the quotient shrinks with it.
It equals 3, the limit of the top.
It equals $\frac{3}{0}$ .

Show the answer

The quotient law fails because the denominator's limit is 0. With the top near 3 and the bottom near 0, the quotient grows without bound in size, so it has no finite limit.

Correct: It is not a finite number: the top approaches 3 and the bottom approaches 0.: Nonzero over a number near 0: unbounded.
It equals 0, because the bottom approaches 0 and the quotient shrinks with it.: A small denominator makes a quotient large.
It equals 3, the limit of the top.: The denominator still matters.
It equals $\frac{3}{0}$ .: $\frac{3}{0}$ is not a number; that is not a limit value.

Part 9 · Summary

Summary

The limit laws let you split a limit over sums, differences, constant multiples, products, quotients (with a nonzero denominator limit), powers, roots and compositions. Together they justify direct substitution for polynomials and for rational functions where the denominator is not 0. When a denominator heads to 0, look closer: a nonzero numerator means no finite limit.

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 graphs of ff and gg

Summary

What comes next

Connections

The graphs of $f$ and $g$