Lesson 4: Calcuating Limits (slides)

Sec on 1.4
Calcula ng Limits
V63.0121.001: Calculus I
Professor Ma hew Leingang
New York University

February 2, 2011

Announcements
First wri en HW due today
.

Announcements

First wri en HW due
today
Get-to-know-you survey
and photo deadline is
February 11

Objectives
Know basic limits like lim x = a
x→a
and lim c = c.
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.

Yoda on teaching course concepts
You must unlearn
what you have
learned.
In other words, we are
building up concepts and
allowing ourselves only to
speak in terms of what we
personally have produced.

Outline
Recall: The concept of limit

Basic Limits

Limit Laws
The direct subs tu on property

Limits with Algebra
Two more limit theorems

Two important trigonometric limits

Heuristic Definition of a Limit
Defini on
We write
lim f(x) = L
x→a
and say

“the limit of f(x), as x approaches a, equals L”

if we can make the values of f(x) arbitrarily close to L (as close to L
as we like) by taking x to be sufficiently close to a (on either side of
a) but not equal to a.

The error-tolerance game
A game between two players (Dana and Emerson) to decide if a limit
lim f(x) exists.
x→a
Step 1 Dana proposes L to be the limit.
Step 2 Emerson challenges with an “error” level around L.
Step 3 Dana chooses a “tolerance” level around a so that points x
within that tolerance of a (not coun ng a itself) are taken to
values y within the error level of L. If Dana cannot, Emerson
wins and the limit cannot be L.
Step 4 If Dana’s move is a good one, Emerson can challenge again
or give up. If Emerson gives up, Dana wins and the limit is L.

The error-tolerance game

L

.
a
To be legit, the part of the graph inside the blue (ver cal) strip
must also be inside the green (horizontal) strip.
Even if Emerson shrinks the error, Dana can s ll move.

Limit FAIL: Jump
y

1

. x

Part of graph
−1 inside blue
is not inside
green

Limit FAIL: Jump
y
Part of graph
inside blue
is not inside
1
green

. x

−1

Limit FAIL: Jump
y
Part of graph |x|
So lim does not
inside blue x→0 x
is not inside exist.
1
green

. x

−1

Limit FAIL: unboundedness
y

1
lim+ does not exist be-
x→0 x
cause the func on is un-
bounded near 0
L?

. x
0

Limit EPIC FAIL (π )
Here is a graph of the func on f(x) = sin :
x
y
1

. x

−1

For every y in [−1, 1], there are inﬁnitely many points x arbitrarily
close to zero where f(x) = y. So lim f(x) cannot exist.
x→0

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
Let c be a constant and a a real number.
(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
Se ng error equal to
tolerance works!

. x
a

ET game for f(x) = c

.

y

. x

y

c

. x
a

y

c
any tolerance works!

. x
a

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 Emerson gives us an error of 1 (for instance), Dana 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.
Dana wins the round.

Fact
x→a x→a
x→a
2. lim [f(x) − g(x)] = L − M (combina on 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 (Con nued)
f(x) L
5. lim = , if M ̸= 0.
x→a g(x) M

Caution!
The quo ent 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, limits of quo ents where numerator and denominator
both tend to 0 are exactly where the magic happens.

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

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

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

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

Fact (Con nued)
f(x) L
5. lim = , if M ̸= 0.
x→a g(x) 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 (Con nued)
f(x) L
5. lim = , if M ̸= 0.
x→a g(x) 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
√ √
n
9. lim f(x) = n lim f(x) (If n is even, we must addi onally
x→a x→a
assume that lim f(x) > 0)
x→a

Applying the limit laws
Example
( )
Find lim x2 + 2x + 4 .
x→3

Example
( )
x→3

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

Example
( )
x→3

Solu on
( ) ( )
lim x2 + 2x + 4 = lim x2 + lim (2x) + lim (4)
x→3 x→3 x→3 x→3

Example
( )
x→3

Solu on
( ) ( )
x→3 x→3 x→3 x→3
( )2
= lim x + 2 · lim (x) + 4
x→3 x→3

Example
( )
x→3

Solu on
( ) ( )
x→3 x→3 x→3 x→3
( )2
= lim x + 2 · lim (x) + 4
x→3 x→3
2
= (3) + 2 · 3 + 4

Example
( )
x→3

Solu on
( ) ( )
x→3 x→3 x→3 x→3
( )2
= lim x + 2 · lim (x) + 4
x→3 x→3
2
= (3) + 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

Solu on
19 1
The answer is = .
38 2

Direct Substitution Property
As a direct consequence of the limit laws and the really basic limits
we have:
Theorem (The Direct Subs tu on Property)
If f is a polynomial or a ra onal func on and a is in the domain of f,
then
lim f(x) = f(a)
x→a

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

Example of the MTP principle
Example
x2 + 2x + 1
Find lim , if it exists.
x→−1 x+1

Example of the MTP principle
Example
x2 + 2x + 1
Find lim , if it exists.
x→−1 x+1

Solu on
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 piecewise function
Example
{
x2 x ≥ 0
Let f(x) = . Does lim f(x) exist?
−x x < 0 x→0

Solu on

.

Example
{
x2 x ≥ 0
−x x < 0 x→0

Solu on
MTP DSP
We have lim+ f(x) = lim+ x2 = 02 = 0
x→0 x→0
.

Example
{
x2 x ≥ 0
−x x < 0 x→0

Solu on
MTP DSP
x→0 x→0
Likewise: lim− f(x) = lim− −x = −0 = 0 .
x→0 x→0

Example
{
x2 x ≥ 0
−x x < 0 x→0

Solu on
MTP DSP
x→0 x→0
Likewise: lim− f(x) = lim− −x = −0 = 0 .
x→0 x→0
So lim f(x) = 0.
x→0

Finding limits by algebra
Example
√
x−2
Find lim .
x→4 x−4

Example
√
x−2
Find lim .
x→4 x−4
Solu on
√ 2 √ √
Write the denominator as x − 4 = x − 4 = ( x − 2)( x + 2).

Example
√
x−2
Find lim .
x→4 x−4
Solu on
√ 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
{
1 − x2 x≥1
Let f(x) = . Find lim f(x) if it exists.
2x x<1 x→1

Solu on

Your turn
Example
{
1 − x2 x≥1
2x x<1 x→1

Solu on
( ) DSP
lim+ f(x) = lim+ 1 − x2 = 0
x→1 x→1

Your turn
Example
{
1 − x2 x≥1
2x x<1 x→1

Solu on
( ) DSP
lim+ f(x) = lim+ 1 − x2 = 0
x→1 x→1

.
1

Your turn
Example
{
1 − x2 x≥1
2x x<1 x→1

Solu on
( ) DSP
lim+ f(x) = lim+ 1 − x2 = 0
x→1 x→1
DSP
lim− f(x) = lim− (2x) = 2
x→1 x→1
.
1

Your turn
Example
{
1 − x2 x≥1
2x x<1 x→1

Solu on
( ) DSP
lim+ f(x) = lim+ 1 − x2 = 0
x→1 x→1
DSP
lim− f(x) = lim− (2x) = 2
x→1 x→1
The le - and right-hand limits disagree, so the .
limit does not exist. 1

A message from
the Mathematical Grammar Police

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

A message from

x→a
Too many verbs

A message from

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

(as usual, provided these limits lim f(x) = lim h(x) = L,
x→a x→a
exist).
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 )
2
Show that lim x sin = 0.
x→0 x

Example
(π )
2
Show that lim x sin = 0.
x→0 x
Solu on
We have for all x,
(π ) (π )
−1 ≤ sin ≤ 1 =⇒ −x ≤ x sin
2 2
≤ x2
x x
The le and right sides go to zero as x → 0.

Illustrating the Squeeze Theorem
y h(x) = x2

. x

y h(x) = x2

. x

f(x) = −x2

y h(x) = x2
(π )
2
g(x) = x sin
x
. x

f(x) = −x2

Two trigonometric limits

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.

No ce θ

θ
. θ
1

Proof.

No ce sin θ ≤ θ

sin θ θ
. θ
cos θ 1

Proof.

No ce sin θ ≤ θ tan θ

sin θ θ tan θ
. θ
cos θ 1

Proof.
θ
No ce sin θ ≤ θ ≤ 2 tan ≤ tan θ
2

sin θ θ tan θ
. θ
cos θ 1

Proof.
θ
2
θ 1
Divide by sin θ: 1 ≤ ≤
sin θ cos θ
sin θ θ tan θ
. θ
cos θ 1

Proof.
θ
2
θ 1
sin θ cos θ
sin θ θ tan θ sin θ
. θ Take reciprocals: 1 ≥ ≥ cos θ
θ
cos θ 1

Proof.
θ
2
θ 1
sin θ cos θ
sin θ θ tan θ sin θ
. θ Take reciprocals: 1 ≥ ≥ cos θ
θ
cos θ 1
As θ → 0, the le 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
= ·
θ θ 1 + cos θ

Proof.

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

Proof.

= · =
2
sin θ
=
θ(1 + cos θ)

Proof.

= · =
2
sin θ sin θ sin θ
= = ·
θ(1 + cos θ) θ 1 + cos θ

Proof.

= · =
2
= = ·
So
( ) ( )
1 − cos θ sin θ sin θ
lim = lim · lim
θ→0 θ θ→0 θ θ→0 1 + cos θ

Proof.

= · =
2
= = ·
So
( ) ( )
1 − cos θ sin θ sin θ 0
lim = lim · lim =1· = 0.
θ→0 θ θ→0 θ θ→0 1 + cos θ 2

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 deﬁni on 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 iden ty:
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

y
The limit laws allow us to
compute limits
reasonably.
BUT we cannot make up . x
extra laws otherwise we
get into trouble.

Lesson 4: Calcuating Limits (slides)

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