Lesson 3: Limit Laws

Section 1.4
Calculating Limits

V63.0121.002.2010Su, Calculus I

New York University

May 18, 2010

Announcements

WebAssign Class Key: nyu 0127 7953
Office Hours: MR 5:00–5:45, TW 7:50–8:30, CIWW 102 (here)
Quiz 1 Thursday on 1.1–1.4

. . . . . .

Announcements

WebAssign Class Key: nyu
0127 7953
Office Hours: MR
5:00–5:45, TW 7:50–8:30,
CIWW 102 (here)
Quiz 1 Thursday on
1.1–1.4

. . . . . .

V63.0121.002.2010Su, Calculus I (NYU) Section 1.4 Calculating Limits May 18, 2010 2 / 37

Objectives

Know basic limits like
lim x = a and lim c = c.
x→a x→a
Use the limit laws to
compute elementary limits.
Use algebra to simplify
limits.
Understand and state the
Squeeze Theorem.
Use the Squeeze Theorem
to demonstrate a limit.

. . . . . .


Outline

Basic Limits

Limit Laws
The direct substitution property

Limits with Algebra
Two more limit theorems

Two important trigonometric limits

. . . . . .


Really basic limits

Fact
Let c be a constant and a a real number.
(i) lim x = a
x→a
(ii) lim c = c
x→a

. . . . . .


Really basic limits

Fact
(i) lim x = a
x→a
(ii) lim c = c
x→a

Proof.
The first is tautological, the second is trivial.

. . . . . .


ET game for f(x) = x

y
.

. x
.

. . . . . .


y
.

. .
a

. . x
.
a
.

. . . . . .


y
.

. .
a

. . x
.
a
.

Setting error equal to tolerance works!
. . . . . .


ET game for f(x) = c

.

. . . . . .


y
.

. x
.

. . . . . .


y
.

.
c
.

. . x
.
a
.

. . . . . .


y
.

.
c
.

. . x
.
a
.

any tolerance works!
. . . . . .


Really basic limits

Fact
(i) lim x = a
x→a
(ii) lim c = c
x→a

Proof.
The first is tautological, the second is trivial.

. . . . . .


Outline

Basic Limits

Limit Laws

Limits with Algebra


. . . . . .


Limits and arithmetic

Fact
Suppose lim f(x) = L and lim g(x) = M and c is a constant. Then
x→a x→a
1. lim [f(x) + g(x)] = L + M
x→a

. . . . . .



Fact
x→a x→a
1. lim [f(x) + g(x)] = L + M (errors add)
x→a

. . . . . .



Fact
x→a x→a
x→a
2. lim [f(x) − g(x)] = L − M
x→a

. . . . . .



Fact
x→a x→a
x→a
2. lim [f(x) − g(x)] = L − M
x→a
3. lim [cf(x)] = cL
x→a

. . . . . .



Fact
x→a x→a
x→a
2. lim [f(x) − g(x)] = L − M
x→a
3. lim [cf(x)] = cL (error scales)
x→a

. . . . . .


Justification of the scaling law

errors scale: If f(x) is e away from L, then

(c · f(x) − c · L) = c · (f(x) − L) = c · e

That is, (c · f)(x) is c · e away from cL,
So if Player 2 gives us an error of 1 (for instance), Player 1 can
use the fact that lim f(x) = L to find a tolerance for f and g
x→a
corresponding to the error 1/c.
Player 1 wins the round.

. . . . . .



Fact
x→a x→a
x→a
2. lim [f(x) − g(x)] = L − M (combination of adding and scaling)
x→a
x→a

. . . . . .



Fact
x→a x→a
x→a
x→a
x→a
4. lim [f(x)g(x)] = L · M
x→a

. . . . . .



Fact
x→a x→a
x→a
x→a
x→a
4. lim [f(x)g(x)] = L · M (more complicated, but doable)
x→a

. . . . . .


Limits and arithmetic II

Fact (Continued)
f(x) L
5. lim = , if M ̸= 0.
x→a g(x) M

. . . . . .


Caution!

The quotient rule for limits says that if lim g(x) ̸= 0, then
x→a

f(x) limx→a f(x)
lim =
x→a g(x) limx→a g(x)

It does NOT say that if lim g(x) = 0, then
x→a

f(x)
lim does not exist
x→a g(x)

In fact, it can.
more about this later

. . . . . .



Fact (Continued)
f(x) L
5. lim = , if M ̸= 0.
g(x)
x→a M
[ ]n
n
6. lim [f(x)] = lim f(x)
x→a x→a

. . . . . .



Fact (Continued)
f(x) L
5. lim = , if M ̸= 0.
g(x)
x→a M
[ ]n
n
6. lim [f(x)] = lim f(x) (follows from 4 repeatedly)
x→a x→a

. . . . . .



Fact (Continued)
f(x) L
5. lim = , if M ̸= 0.
g(x)
x→a M
[ ]n
n
x→a x→a
n n
7. lim x = a
x→a

. . . . . .



Fact (Continued)
f(x) L
5. lim = , if M ̸= 0.
g(x)
x→a M
[ ]n
n
x→a x→a
n n
7. lim x = a
x→a
√ √
8. lim n x = n a
x→a

. . . . . .



Fact (Continued)
f(x) L
5. lim = , if M ̸= 0.
g(x)
x→a M
[ ]n
n
x→a x→a
n n
7. lim x = a (follows from 6)
x→a
√ √
8. lim n x = n a
x→a

. . . . . .



Fact (Continued)
f(x) L
5. lim = , if M ̸= 0.
g(x)
x→a M
[ ]n
n
x→a x→a
n n
7. lim x = a (follows from 6)
x→a
√ √
8. lim n x = n a
x→a
√ √
9. lim n f(x) = n lim f(x) (If n is even, we must additionally assume
x→a x→a
that lim f(x) > 0)
x→a

. . . . . .


Applying the limit laws

Example
( )
Find lim x2 + 2x + 4 .
x→3

. . . . . .



Example
( )
x→3

Solution
By applying the limit laws repeatedly:
( )
lim x2 + 2x + 4
x→3

. . . . . .



Example
( )
x→3

Solution
( ) ( )
lim x2 + 2x + 4 = lim x2 + lim (2x) + lim (4)
x→3 x→3 x→3 x→3

. . . . . .



Example
( )
x→3

Solution
( ) ( )
x→3 x→3 x→3 x→3
( )2
= lim x + 2 · lim (x) + 4
x→3 x→3

. . . . . .



Example
( )
x→3

Solution
( ) ( )
x→3 x→3 x→3 x→3
( )2
= lim x + 2 · lim (x) + 4
x→3 x→3

= (3)2 + 2 · 3 + 4

. . . . . .



Example
( )
x→3

Solution
( ) ( )
x→3 x→3 x→3 x→3
( )2
= lim x + 2 · lim (x) + 4
x→3 x→3

= (3)2 + 2 · 3 + 4
= 9 + 6 + 4 = 19.

. . . . . .


Your turn

Example
x2 + 2x + 4
Find lim
x→3 x3 + 11

. . . . . .


Your turn

Example
x2 + 2x + 4
Find lim
x→3 x3 + 11

Solution
19 1
The answer is = .
38 2

. . . . . .


Direct Substitution Property

Theorem (The Direct Substitution Property)
If f is a polynomial or a rational function and a is in the domain of f, then

lim f(x) = f(a)
x→a

. . . . . .


Outline

Basic Limits

Limit Laws

Limits with Algebra


. . . . . .


Limits do not see the point! (in a good way)

Theorem
If f(x) = g(x) when x ̸= a, and lim g(x) = L, then lim f(x) = L.
x→a x→a

. . . . . .



Theorem
x→a x→a

Example
x2 + 2x + 1
Find lim , if it exists.
x→−1 x+1

. . . . . .



Theorem
x→a x→a

Example
x2 + 2x + 1
Find lim , if it exists.
x→−1 x+1

Solution
x2 + 2x + 1
Since = x + 1 whenever x ̸= −1, and since
x+1
x2 + 2x + 1
lim x + 1 = 0, we have lim = 0.
x→−1 x→−1 x+1

. . . . . .


x2 + 2x + 1
ET game for f(x) =
x+1

y
.

. . x
.
−
. 1

Even if f(−1) were something else, it would not effect the limit.

. . . . . .

Limit of a function defined piecewise at a boundary
point

Example
Let
{
x2 x ≥ 0 .
f(x) =
−x x < 0

Does lim f(x) exist?
x→0

. . . . . .


point

Example
Let
{
x2 x ≥ 0 .
f(x) =
−x x < 0

x→0

Solution
We have
MTP DSP
lim+ f(x) = lim+ x2 = 02 = 0
x→0 x→0

. . . . . .


point

Example
Let
{
x2 x ≥ 0 .
f(x) =
−x x < 0

x→0

Solution
We have
MTP DSP
lim+ f(x) = lim+ x2 = 02 = 0
x→0 x→0

Likewise:
lim f(x) = lim −x = −0 = 0
x→0− x→0− . . . . . .


Finding limits by algebraic manipulations

Example
√
x−2
Find lim .
x→4 x − 4

. . . . . .



Example
√
x−2
Find lim .
x→4 x − 4

Solution
√ 2 √ √
Write the denominator as x − 4 = x − 4 = ( x − 2)( x + 2).

. . . . . .



Example
√
x−2
Find lim .
x→4 x − 4

Solution
√ 2 √ √
Write the denominator as x − 4 = x − 4 = ( x − 2)( x + 2). So
√ √
x−2 x−2
lim = lim √ √
x→4 x − 4 x→4 ( x − 2)( x + 2)
1 1
= lim √ =
x→4 x+2 4

. . . . . .


Your turn

Example
Let
{
1 − x2 x≥1
f(x) =
2x x<1

Find lim f(x) if it exists.
x→1

. . . . . .


Your turn

Example
Let
{
1 − x2 x≥1
f(x) =
2x x<1

Find lim f(x) if it exists.
x→1

Solution
We have
( )
DSP
lim+ f(x) = lim+ 1 − x2 = 0
x→1 x→1

. . . . . .


Your turn

Example
Let
{
1 − x2 x≥1
f(x) =
2x x<1
. .
Find lim f(x) if it exists. 1
.
x→1

Solution
We have
( )
DSP
lim+ f(x) = lim+ 1 − x2 = 0
x→1 x→1

. . . . . .


Your turn

Example
Let
{
1 − x2 x≥1
f(x) =
2x x<1
. .
.
x→1

Solution
We have
( )
DSP
lim+ f(x) = lim+ 1 − x2 = 0
x→1 x→1
DSP
lim f(x) = lim (2x) = 2
− −
x→1 x→1

. . . . . .


Your turn

Example
.
Let
{
1 − x2 x≥1
f(x) =
2x x<1
. .
.
x→1

Solution
We have
( )
DSP
lim+ f(x) = lim+ 1 − x2 = 0
x→1 x→1
DSP
lim f(x) = lim (2x) = 2
− −
x→1 x→1

. . . . . .


Your turn

Example
.
Let
{
1 − x2 x≥1
f(x) =
2x x<1
. .
.
x→1

Solution
We have
( )
DSP
lim+ f(x) = lim+ 1 − x2 = 0
x→1 x→1
DSP
lim f(x) = lim (2x) = 2
− −
x→1 x→1

The left- and right-hand limits disagree, so the limit does not exist. . . . . . .


A message from the Mathematical Grammar Police

Please do not say “ lim f(x) = DNE.” Does not compute.
x→a

. . . . . .



x→a
Too many verbs

. . . . . .



x→a
Too many verbs
Leads to FALSE limit laws like “If lim f(x) DNE and lim g(x) DNE,
x→a x→a
then lim (f(x) + g(x)) DNE.”
x→a

. . . . . .


Two More Important Limit Theorems

Theorem
If f(x) ≤ g(x) when x is near a (except possibly at a), then

lim f(x) ≤ lim g(x)
x→a x→a

(as usual, provided these limits exist).

Theorem (The Squeeze/Sandwich/Pinching Theorem)
If f(x) ≤ g(x) ≤ h(x) when x is near a (as usual, except possibly at a),
and
lim f(x) = lim h(x) = L,
x→a x→a

then
lim g(x) = L.
x→a

. . . . . .


Using the Squeeze Theorem

We can use the Squeeze Theorem to replace complicated expressions
with simple ones when taking the limit.

. . . . . .



Example
(π )
Show that lim x2 sin = 0.
x→0 x

. . . . . .



Example
(π )
Show that lim x2 sin = 0.
x→0 x

Solution
We have for all x,
(π ) (π )
−1 ≤ sin ≤ 1 =⇒ −x2 ≤ x2 sin ≤ x2
x x
The left and right sides go to zero as x → 0.

. . . . . .


Illustration of the Squeeze Theorem

y
. . (x) = x2
h

. x
.

. . . . . .



y
. . (x) = x2
h

. x
.

.(x) = −x2
f

. . . . . .



y
. . (x) = x2
h
(π )
. (x) = x2 sin
g
x

. x
.

.(x) = −x2
f

. . . . . .


Outline

Basic Limits

Limit Laws

Limits with Algebra


. . . . . .



Theorem
The following two limits hold:
sin θ
lim =1
θ→0 θ
cos θ − 1
lim =0
θ→0 θ

. . . . . .


Proof of the Sine Limit

Proof.
Notice

θ

.
θ
.
θ
.
1
.

. . . . . .



Proof.
Notice

sin θ ≤ θ

. in θ .
s θ
.
θ
.
c
. os θ 1
.

. . . . . .



Proof.
Notice

sin θ ≤ θ tan θ

. in θ . .an θ
s θ t
.
θ
.
c
. os θ 1
.

. . . . . .



Proof.
Notice
θ
sin θ ≤ θ ≤ 2 tan ≤ tan θ
2

. in θ . .an θ
s θ t
.
θ
.
c
. os θ 1
.

. . . . . .



Proof.
Notice
θ
2
Divide by sin θ:

θ 1
1≤ ≤
sin θ cos θ
. in θ . .an θ
s θ t
.
θ
.
c
. os θ 1
.

. . . . . .



Proof.
Notice
θ
2
Divide by sin θ:

θ 1
1≤ ≤
sin θ cos θ
. in θ . .an θ
s θ t
.
θ Take reciprocals:
.
c
. os θ 1
. sin θ
1≥ ≥ cos θ
θ

. . . . . .



Proof.
Notice
θ
2
Divide by sin θ:

θ 1
1≤ ≤
sin θ cos θ
. in θ . .an θ
s θ t
.
θ Take reciprocals:
.
c
. os θ 1
. sin θ
1≥ ≥ cos θ
θ

As θ → 0, the left and right sides tend to 1. So, then, must the middle
expression.
. . . . . .


Proof of the Cosine Limit

Proof.

1 − cos θ 1 − cos θ 1 + cos θ 1 − cos2 θ
= · =
θ θ 1 + cos θ θ(1 + cos θ)
sin2 θ sin θ sin θ
= = ·
θ(1 + cos θ) θ 1 + cos θ

So
( ) ( )
1 − cos θ sin θ sin θ
lim = lim · lim
θ→0 θ θ→0 θ θ→0 1 + cos θ

= 1 · 0 = 0.

. . . . . .


Try these

Example
tan θ
1. lim
θ→0 θ
sin 2θ
2. lim
θ→0 θ

. . . . . .


Try these

Example
tan θ
1. lim
θ→0 θ
sin 2θ
2. lim
θ→0 θ

Answer

1. 1
2. 2

. . . . . .


Solutions

1. Use the basic trigonometric limit and the definition of tangent.

tan θ sin θ sin θ 1 1
lim = lim = lim · lim = 1 · = 1.
θ→0 θ θ→0 θ cos θ θ→0 θ θ→0 cos θ 1

. . . . . .


Solutions


lim = lim = lim · lim = 1 · = 1.

2. Change the variable:

sin 2θ sin 2θ sin 2θ
lim = lim = 2 · lim =2·1=2
θ→0 θ 2θ→0 2θ · 1 2θ→0 2θ
2

. . . . . .


Solutions


lim = lim = lim · lim = 1 · = 1.

2. Change the variable:

sin 2θ sin 2θ sin 2θ
lim = lim = 2 · lim =2·1=2
θ→0 θ 2θ→0 2θ · 1 2θ→0 2θ
2

OR use a trigonometric identity:

sin 2θ 2 sin θ cos θ sin θ
lim = lim = 2· lim · lim cos θ = 2·1·1 = 2
θ→0 θ θ→0 θ θ→0 θ θ→0

. . . . . .


Summary

Limits laws say limits play
well with the rules of
arithmetic
When limit laws do not
work we can be
algebraically creative
When algebra does not
work we can try
Squeezing.

. . . . . .


Lesson 3: Limit Laws

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