Lesson 4: Calculating Limits (handout)

V63.0121.021, Calculus I Section 1.4 : Calculating Limits September 15, 2010

Notes
Section 1.4
Calculating Limits

V63.0121.021, Calculus I

New York University

September 15, 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
Notes

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 15, 2010 2 / 45

Yoda on teaching a concepts course
Notes
“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.


1


Objectives
Notes

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
Notes

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
Notes

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.


2


The error-tolerance game
Notes

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
Notes

Still tolerance
This too this is too big
So does big
looks good

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
Notes
y

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

x

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

|x|
So lim does not exist.
x→0 x

3


Limit FAIL: unboundedness
y Notes

1
lim does not exist
The+graph escapes the
x→0 x
because the worse! is
Even
green, so no function
good
unbounded near 0

L?

x
0


Limit EPIC FAIL
Notes
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
Notes


Basic Limits

Limit Laws

Limits with Algebra



4


Really basic limits
Notes

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
Notes
y

a

x
a

Setting error equal to tolerance works!


ET game for f (x) = c
Notes
y

c

x
a

any tolerance works!


5


Really basic limits
Notes

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.


Outline
Notes


Basic Limits

Limit Laws

Limits with Algebra



Limits and arithmetic
Notes

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 (errors add)
x→a
2. lim [f (x) − g (x)] = L − M (combination of adding and scaling)
x→a
3. lim [cf (x)] = cL (error scales)
x→a
4. lim [f (x)g (x)] = L · M (more complicated, but doable)
x→a


6


Notes

Fact
x→a x→a
x→a
x→a
x→a
x→a


Justiﬁcation of the scaling law
Notes

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.


Notes

Fact
x→a x→a
x→a
x→a
x→a
x→a


7


Notes

Fact
x→a x→a
x→a
x→a
x→a
x→a


Limits and arithmetic II
Notes

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
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


Caution!
Notes

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


8


Limits and arithmetic II
Notes

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
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 additionally assume
x→a x→a
that lim f (x) > 0)
x→a


Applying the limit laws
Notes

Example
Find lim x 2 + 2x + 4 .
x→3

Solution
By applying the limit laws repeatedly:

lim x 2 + 2x + 4 = lim x 2 + lim (2x) + lim (4)
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
Notes

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

Solution
19 1
The answer is = .
38 2


9


Direct Substitution Property
Notes

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
Notes


Basic Limits

Limit Laws

Limits with Algebra



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

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


10


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

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 Notes

Example
Let

x2 x ≥0
f (x) =
−x x <0

Does lim f (x) exist?
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.
x→0

Finding limits by algebraic manipulations
Notes

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


11


Your turn
Notes
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
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
Notes

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


12


Using the Squeeze Theorem
Notes

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

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
Notes

y h(x) = x 2
π
g (x) = x 2 sin
x

x

f (x) = −x 2


Outline
Notes


Basic Limits

Limit Laws

Limits with Algebra



13


Notes

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

1 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
Notes
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.


14


Try these
Notes

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

Answer

1. 1
2. 2


Solutions
Notes

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

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
Notes

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


15

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