Lesson 4: Calculating Limits (Section 21 slides)

Section 1.4
Calculating Limits

V63.0121.021, Calculus I

New York University

September 16, 2010

Announcements

First written homework due today (put it in the envelope) Remember
to put your lecture and recitation section numbers on your paper

Announcements

First written homework due
today (put it in the
envelope) Remember to put
your lecture and recitation
section numbers on your
paper

V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 2 / 45

Yoda on teaching a concepts course

“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.


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

Recall: The concept of limit

Basic Limits

Limit Laws
The direct substitution property

Limits with Algebra
Two more limit theorems

Two important trigonometric limits


Heuristic Definition of a Limit

Definition
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 counting 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 (vertical) strip must
also be inside the green (horizontal) strip.
If Emerson shrinks the error, Dana can still win.

Limit FAIL: Jump
y

1

x

Part of graph in-
−1 side blue is not
inside green


Limit FAIL: Jump
y

Part of graph in-
side blue is not 1
inside green

x

−1


Limit FAIL: Jump
y

Part of graph in-
side blue is not 1
inside green

x

−1

|x|
So lim does not exist.
x→0 x


Limit FAIL: unboundedness
y

1
lim+ does not exist
x→0 x
because the function is
unbounded near 0

L?

x
0


Limit EPIC FAIL

π
Here is a graph of the function 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


Outline


Basic Limits

Limit Laws

Limits with Algebra



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 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 corresponding
x→a
to the error 1/c.
Dana 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, limits of quotients where numerator and denominator both
tend to 0 are exactly where the magic happens.
more about this later



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



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



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



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



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



Fact (Continued)
f (x) L
5. lim = , if M = 0.
x→a g (x) M
n
x→a x→a
7. lim x n = an (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 additionally assume
x→a x→a
that lim f (x) > 0)
x→a


Applying the limit laws

Example
Find lim x 2 + 2x + 4 .
x→3



Example
Find lim x 2 + 2x + 4 .
x→3

Solution
By applying the limit laws repeatedly:

lim x 2 + 2x + 4
x→3



Example
Find lim x 2 + 2x + 4 .
x→3

Solution

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



Example
Find lim x 2 + 2x + 4 .
x→3

Solution

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



Example
Find lim x 2 + 2x + 4 .
x→3

Solution

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
Find lim x 2 + 2x + 4 .
x→3

Solution

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
x 2 + 2x + 4
Find lim
x→3 x 3 + 11


Your turn

Example
x 2 + 2x + 4
Find lim
x→3 x 3 + 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
x 2 + 2x + 1
Find lim , if it exists.
x→−1 x +1



Theorem
x→a x→a

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

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


x 2 + 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+ x 2 = 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+ x 2 = 02 = 0
x→0 x→0

Likewise:
lim f (x) = lim −x = −0 = 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+ x 2 = 02 = 0
x→0 x→0

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

So lim f (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 − x 2 = 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 − x 2 = 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 − x 2 = 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 − x 2 = 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 x 2 sin = 0.
x→0 x



Example
π
Show that lim x 2 sin = 0.
x→0 x

Solution
We have for all x,
π π
−1 ≤ sin ≤ 1 =⇒ −x 2 ≤ x 2 sin ≤ x2
x x
The left and right sides go to zero as x → 0.


Illustration of the Squeeze Theorem

y h(x) = x 2

x



y h(x) = x 2

x

f (x) = −x 2



y h(x) = x 2

x
π
2
g (x) = x sin
x

f (x) = −x 2


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

sin θ θ
θ
cos θ 1



Proof.
Notice

sin θ ≤ θ tan θ

sin θ θ tan θ
θ
cos θ 1



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

sin θ θ tan θ
θ
cos θ 1



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



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

cos θ 1 sin θ
1≥ ≥ cos θ
θ



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

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


Solutions

lim = lim = lim · lim = 1 · = 1.

2. Change the variable:
sin 2θ sin 2θ sin 2θ
lim = lim 1
= 2 · lim =2·1=2
θ→0 θ 2θ→0 2θ ·
2
2θ→0 2θ

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

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


Lesson 4: Calculating Limits (Section 21 slides)

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