lesson 4: calculating limits (handout)

.

.

..

Sec on 1.4Calcula ng Limits

V63.0121.001: Calculus IProfessor Ma hew Leingang

New York University

February 2, 2011

Announcements

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

.

Announcements

I First wri en HW duetoday

I Get-to-know-you surveyand photo deadline isFebruary 11

.

ObjectivesI Know basic limits likelimx→a

x = a and limx→a

c = c.I Use the limit laws tocompute elementarylimits.

I Use algebra to simplifylimits.

I Understand and state theSqueeze Theorem.

I Use the SqueezeTheorem to demonstratea limit.

.

Notes

.

Notes

.

Notes

. 1.

. Sec on 1.4: Limits. V63.0121.001: Calculus I . February 2, 2011

.

.

..

Limit

.

OutlineRecall: The concept of limit

Basic Limits

Limit LawsThe direct subs tu on property

Limits with AlgebraTwo more limit theorems

Two important trigonometric limits

.

Heuristic Definition of a LimitDefini onWe write

limx→a

f(x) = L

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 Las we like) by taking x to be sufficiently close to a (on either side ofa) but not equal to a.

.

Notes

.

Notes

.

Notes

. 2.


.

.

The error-tolerance gameA game between two players (Dana and Emerson) to decide if a limitlimx→a

f(x) exists.

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 tovalues y within the error level of L. If Dana cannot, Emersonwins and the limit cannot be L.

Step 4 If Dana’s move is a good one, Emerson can challenge againor give up. If Emerson gives up, Dana wins and the limit is L.

.

The error-tolerance game

..

This tolerance is too big

.

S ll too big

.

This looks good

.

So does this

.a

.

L

I To be legit, the part of the graph inside the blue (ver cal) stripmust also be inside the green (horizontal) strip.

I Even if Emerson shrinks the error, Dana can s ll move.

.


.

.


.

S ll too big

.

This looks good

.

So does this

.a

.

L



.

Notes

.

Notes

.

Notes

. 3.


.

.


.

.


.

S ll too big

.

This looks good

.

So does this

.a

.

L



.


.

.


.

S ll too big

.

This looks good

.

So does this

.a

.

L



.

Limit FAIL: Jump

.. x.

y

..

−1

..

1

...

Part of graphinside blueis not insidegreen

.


I So limx→0

|x|x

does notexist.

I But limx→0+

f(x) = 1

andlimx→0−

f(x) = −1.

.

Notes

.

Notes

.

Notes

. 4.


.

.

Limit FAIL: Jump

.. x.

y

..

−1

..

1

..

.


.


I So limx→0

|x|x

does notexist.

I But limx→0+

f(x) = 1

andlimx→0−

f(x) = −1.

.

Limit FAIL: Jump

.. x.

y

..

−1

..

1

..

.


.


I So limx→0

|x|x

does notexist.

I But limx→0+

f(x) = 1

andlimx→0−

f(x) = −1.

.

Limit FAIL: unboundedness

.. x.

y

.0..

L?

.

The graph escapesthe green, so nogood

.

Even worse!

.

limx→0+

1xdoes not exist be-

cause the func on is un-bounded near 0

.

Notes

.

Notes

.

Notes

. 5.


.

.

Limit EPIC FAILHere is a graph of the func on f(x) = sin

(πx

):

.. x.

y

..

−1

..

1

For every y in [−1, 1], there are infinitely many points x arbitrarilyclose to zero where f(x) = y. So lim

x→0f(x) cannot exist.

.


Basic Limits




.

Really basic limits

FactLet c be a constant and a a real number.(i) lim

x→ax = a

(ii) limx→a

c = c

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

.

Notes

.

Notes

.

Notes

. 6.


.

.

ET game for f(x) = x

.. x.

y

..a

..

a

I Se ng error equal to tolerance works!

.

ET game for f(x) = c

.. x.

y

..a

..

c

I any tolerance works!

.


Basic Limits




.

Notes

.

Notes

.

Notes

. 7.


.

.

Limits and arithmetic

FactSuppose lim

x→af(x) = L and lim

x→ag(x) = M and c is a constant. Then

1. limx→a

[f(x) + g(x)] = L+M (errors add)

2. limx→a

[f(x)− g(x)] = L−M (combina on of adding and scaling)

3. limx→a

[cf(x)] = cL (error scales)

4. limx→a

[f(x)g(x)] = L ·M (more complicated, but doable)

.


FactSuppose lim



1. limx→a


2. limx→a


3. limx→a


4. limx→a


.

Justification of the scaling lawI 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,I So if Emerson gives us an error of 1 (for instance), Dana can usethe fact that lim

x→af(x) = L to find a tolerance for f and g

corresponding to the error 1/c.I Dana wins the round.

.

Notes

.

Notes

.

Notes

. 8.


.

.


FactSuppose lim



1. limx→a


2. limx→a


3. limx→a


4. limx→a


.


FactSuppose lim



1. limx→a


2. limx→a


3. limx→a


4. limx→a


.

Limits and arithmetic IIFact (Con nued)

5. limx→a

f(x)g(x)

=LM

, if M ̸= 0.

6. limx→a

[f(x)]n =[limx→a

f(x)]n

(follows from 4 repeatedly)

7. limx→a

xn = an (follows from 6)

8. limx→a

n√x = n

√a

9. limx→a

n√

f(x) = n

√limx→a

f(x) (If n is even, we must addi onallyassume that lim

x→af(x) > 0)

.

Notes

.

Notes

.

Notes

. 9.


.

.

Caution!I The quo ent rule for limits says that if lim

x→ag(x) ̸= 0, then

limx→a

f(x)g(x)

=limx→a f(x)limx→a g(x)

I It does NOT say that if limx→a

g(x) = 0, then

limx→a

f(x)g(x)

does not exist

In fact, limits of quo ents where numerator and denominatorboth tend to 0 are exactly where the magic happens.

I more about this later

.

Limits and arithmetic IIFact (Con nued)

5. limx→a

f(x)g(x)

=LM

, if M ̸= 0.

6. limx→a

[f(x)]n =[limx→a

f(x)]n

(follows from 4 repeatedly)

7. limx→a

xn = an (follows from 6)

8. limx→a

n√x = n

√a

9. limx→a

n√

f(x) = n

√limx→a

f(x) (If n is even, we must addi onallyassume that lim

x→af(x) > 0)

.

Applying the limit lawsExample

Find limx→3

(x2 + 2x+ 4

).

Solu onBy applying the limit laws repeatedly:

limx→3

(x2 + 2x+ 4

)= lim

x→3

(x2)+ lim

x→3(2x) + lim

x→3(4)

=(limx→3

x)2

+ 2 · limx→3

(x) + 4

= (3)2 + 2 · 3+ 4= 9+ 6+ 4 = 19.

.

Notes

.

Notes

.

Notes

. 10.


.

.

Your turnExample

Find limx→3

x2 + 2x+ 4x3 + 11

Solu on

The answer is1938

=12.

.

Direct Substitution Property

As a direct consequence of the limit laws and the really basic limitswe 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

limx→a

f(x) = f(a)

.


Basic Limits




.

Notes

.

Notes

.

Notes

. 11.


.

.

Limits do not see the point!

TheoremIf f(x) = g(x) when x ̸= a, and lim

x→ag(x) = L, then lim

x→af(x) = L.

.

Example of the MTP principleExample

Find limx→−1

x2 + 2x+ 1x+ 1

, if it exists.

Solu on

Sincex2 + 2x+ 1

x+ 1= x+ 1 whenever x ̸= −1, and since

limx→−1

x+ 1 = 0, we have limx→−1

x2 + 2x+ 1x+ 1

= 0.

.

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

x + 1

.. x.

y

...−1

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

.

Notes

.

Notes

.

Notes

. 12.


.

.

Limit of a function definedpiecewise at a boundary pointExample

Let

f(x) =

{x2 x ≥ 0−x x < 0

Does limx→0

f(x) exist?

..

Solu onWe have

limx→0+

f(x) MTP= lim

x→0+x2 DSP

= 02 = 0

Likewise:limx→0−

f(x) = limx→0−

−x = −0 = 0

So limx→0

f(x) = 0.

.

Finding limits by algebraicmanipulations

Example

Find limx→4

√x− 2x− 4

.

Solu on

Write the denominator as x− 4 =√x2 − 4 = (

√x− 2)(

√x+ 2). So

limx→4

√x− 2x− 4

= limx→4

√x− 2

(√x− 2)(

√x+ 2)

= limx→4

1√x+ 2

=14

.

Your turnExample

Let

f(x) =

{1− x2 x ≥ 12x x < 1

Find limx→1

f(x) if it exists.

...1

..

Solu on

We have

limx→1+

f(x) = limx→1+

(1− x2

) DSP= 0

limx→1−

f(x) = limx→1−

(2x) DSP= 2

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

.

Notes

.

Notes

.

Notes

. 13.


.

.

Two Important Limit TheoremsTheoremIf f(x) ≤ g(x) when x is near a (except possibly at a), then

limx→a

f(x) ≤ limx→a

g(x)

(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 ata), and

limx→a

f(x) = limx→a

h(x) = L,

thenlimx→a

g(x) = L.

.

Using the Squeeze TheoremWe can use the Squeeze Theorem to replace complicatedexpressions with simple ones when taking the limit.Example

Show that limx→0

x2 sin(πx

)= 0.

Solu onWe have for all x,

−1 ≤ sin(πx

)≤ 1 =⇒ −x2 ≤ x2 sin

(πx

)≤ x2

The le and right sides go to zero as x → 0.

.

Illustrating the Squeeze Theorem

.. x.

y

.

h(x) = x2

.

f(x) = −x2

.

g(x) = x2 sin(πx

)

.

Notes

.

Notes

.

Notes

. 14.


.

.


Basic Limits




.

Two important trigonometriclimits

TheoremThe following two limits hold:

I limθ→0

sin θθ

= 1

I limθ→0

cos θ − 1θ

= 0

.

Proof of the Sine LimitProof.

.. θ.sin θ

.cos θ

.θ

.tan θ

.−1

.1

No ce

sin θ ≤ θ ≤ 2 tanθ

2≤ tan θ

Divide by sin θ:

1 ≤ θ

sin θ≤ 1

cos θ

Take reciprocals:

1 ≥ sin θθ

≥ cos θ

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

.

Notes

.

Notes

.

Notes

. 15.


.

.

Proof of the Cosine LimitProof.

1− cos θθ

=1− cos θ

θ· 1+ cos θ1+ cos θ

=1− cos2 θθ(1+ cos θ)

=sin2 θ

θ(1+ cos θ)=

sin θθ

· sin θ1+ cos θ

So

limθ→0

1− cos θθ

=

(limθ→0

sin θθ

)·(limθ→0

sin θ1+ cos θ

)= 1 · 0 = 0.

.

Try theseExample

1. limθ→0

tan θθ

2. limθ→0

sin 2θθ

Answer

1. 12. 2

.

Summary

I The limit laws allow us tocompute limitsreasonably.

I BUT we cannot make upextra laws otherwise weget into trouble.

.. x.

y

.

Notes

.

Notes

.

Notes

. 16.


lesson 4: calculating limits (handout)

Technology

limit lim fx

limit of fx

x lim fx

lim fx gx

lim gx

c fx c

lim cfx

lim fxgx