lesson 3: the limit of a function (slides)

..

Sec on 1.3The Limit of a Func on

V63.0121.011: Calculus IProfessor Ma hew Leingang

New York University

January 31, 2011

Announcements

I First wri en HW due Wednesday February 2I Get-to-know-you survey and photo deadline is February 11

Announcements

I First wri en HW dueWednesday February 2

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

Guidelines for written homework

I Papers should be neat and legible. (Use scratch paper.)I Label with name, lecture number (011), recita on number,date, assignment number, book sec ons.

I Explain your work and your reasoning in your own words. Usecomplete English sentences.

RubricPoints Descrip on of Work3 Work is completely accurate and essen ally perfect.

Work is thoroughly developed, neat, and easy to read.Complete sentences are used.

2 Work is good, but incompletely developed, hard toread, unexplained, or jumbled. Answers which arenot explained, even if correct, will generally receive 2points. Work contains “right idea” but is flawed.

1 Work is sketchy. There is some correct work, but mostof work is incorrect.

0 Work minimal or non-existent. Solu on is completelyincorrect.

Written homework: Don’t

Written homework: Do

Written homework: DoWritten Explanations

Written homework: DoGraphs

ObjectivesI Understand and state theinformal defini on of alimit.

I Observe limits on agraph.

I Guess limits by algebraicmanipula on.

I Guess limits by numericalinforma on.

..

Limit

Yoda on teaching course concepts

You must unlearnwhat you havelearned.

In other words, we arebuilding up concepts andallowing ourselves only tospeak in terms of what wepersonally have produced.

Zeno’s Paradox

That which is in locomo on mustarrive at the half-way stage beforeit arrives at the goal.

(Aristotle Physics VI:9, 239b10)

Outline

Heuris cs

Errors and tolerances

Examples

Precise Defini on of a Limit

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.

Outline

Heuris cs


Examples


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.


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 xwithin 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.



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.


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




.

.


.

S ll too big

.

This looks good

.

So does this

.a

.

L



Outline

Heuris cs


Examples


Playing the E-T GameExample

Describe how the the Error-Tolerance game would be played todetermine lim

x→0x2.

Solu on



x→0x2.

Solu on

I Dana claims the limit is zero.

I If Emerson challenges with an error level of 0.01, Dana needsto guarantee that−0.01 < x2 < 0.01 for all x sufficiently closeto zero.

I If−0.1 < x < 0.1, then 0 ≤ x2 < 0.01, so Dana wins the round.



x→0x2.

Solu on

I Dana claims the limit is zero.I If Emerson challenges with an error level of 0.01, Dana needsto guarantee that−0.01 < x2 < 0.01 for all x sufficiently closeto zero.

I If−0.1 < x < 0.1, then 0 ≤ x2 < 0.01, so Dana wins the round.



x→0x2.

Solu on

I If Emerson re-challenges with an error level of 0.0001 = 10−4,what should Dana’s tolerance be?

I A tolerance of 0.01 works because|x| < 10−2 =⇒

∣∣x2∣∣ < 10−4.



x→0x2.

Solu on

I Dana has a shortcut: By se ng tolerance equal to the squareroot of the error, Dana can win every round. Once Emersonrealizes this, Emerson must give up.

Graphical version of E-T gamewith x2

....x

...

y

No ma er how small anerror Emerson picks,Dana can find a fi ngtolerance band.

A piecewise-defined functionExample

Find limx→0

|x|x

if it exists.

Solu on

The error-tolerance game fails, but

limx→0+

f(x) = 1 limx→0−

f(x) = −1


Find limx→0

|x|x

if it exists.

Solu onThe func on can also be wri en as

|x|x

=

{1 if x > 0;−1 if x < 0

What would be the limit?

The error-tolerance game fails, but

limx→0+

f(x) = 1 limx→0−

f(x) = −1

The E-T game with a piecewisefunction

Find limx→0

|x|x

if it exists.

.... x...

y

..

−1

..

1

....

.I think the limit is 1

. Can you fit an error of 0.5?.How aboutthis for a tol-erance?

.No. Part ofgraph insideblue is notinside green

.

Oh, I guessthe limit isn’t1 .

I think the limitis −1 . Can you fit an

error of 0.5?.

No. Part ofgraph insideblue is notinside green

.

Oh, I guessthe limit isn’t−1 .

I think the limitis 0 . Can you fit an

error of 0.5?. No. None of

graph insideblue is insidegreen

.Oh, I guessthe limit isn’t0

.I give up! Iguess there’sno limit!


Find limx→0

|x|x

if it exists.

.... x...

y

..

−1

..

1

.....I think the limit is 1



.



error of 0.5?.


.








Find limx→0

|x|x

if it exists.

.... x...

y

..

−1

..

1

.....I think the limit is 1

. Can you fit an error of 0.5?

.How aboutthis for a tol-erance?


.



error of 0.5?.


.








Find limx→0

|x|x

if it exists.

.... x...

y

..

−1

..

1

....





.



error of 0.5?.


.








Find limx→0

|x|x

if it exists.

.... x...

y

..

−1

..

1

....




.

Oh, I guessthe limit isn’t1

.I think the limitis −1 . Can you fit an

error of 0.5?.


.








Find limx→0

|x|x

if it exists.

.... x...

y

..

−1

..

1

....




.


.I think the limitis −1

. Can you fit anerror of 0.5?

.


.








Find limx→0

|x|x

if it exists.

.... x...

y

..

−1

..

1

....




.


.I think the limitis −1 . Can you fit an

error of 0.5?

.


.








Find limx→0

|x|x

if it exists.

.... x...

y

..

−1

..

1

....





.


I think the limitis −1


.


.








Find limx→0

|x|x

if it exists.

.... x...

y

..

−1

..

1

....





.



error of 0.5?

.


.








Find limx→0

|x|x

if it exists.

.... x...

y

..

−1

..

1

....




.



error of 0.5?

.


.

Oh, I guessthe limit isn’t−1

.I think the limitis 0 . Can you fit an






Find limx→0

|x|x

if it exists.

.... x...

y

..

−1

..

1

....




.



error of 0.5?.


.


.I think the limitis 0


. No. None ofgraph insideblue is insidegreen




Find limx→0

|x|x

if it exists.

.... x...

y

..

−1

..

1

....




.



error of 0.5?.


.


.I think the limitis 0 . Can you fit an

error of 0.5?





Find limx→0

|x|x

if it exists.

.... x...

y

..

−1

..

1

....





.



error of 0.5?.


.


I think the limitis 0






Find limx→0

|x|x

if it exists.

.... x...

y

..

−1

..

1

....





.



error of 0.5?.


.



error of 0.5?





Find limx→0

|x|x

if it exists.

.... x...

y

..

−1

..

1

....




.



error of 0.5?.


.



error of 0.5?





Find limx→0

|x|x

if it exists.

.... x...

y

..

−1

..

1

....




.



error of 0.5?.


.







One-sided limitsDefini onWe write

limx→a+

f(x) = L

and say

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

if we can make the values of f(x) arbitrarily close to L (as close to L aswe like) by taking x to be sufficiently close to a and greater than a.

One-sided limitsDefini onWe write

limx→a−

f(x) = L

and say

“the limit of f(x), as x approaches a from the le , 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 and less than a.

The error-tolerance gameFind lim

x→0+

|x|x

and limx→0−

|x|x

if they exist.

.. x.

y

..

−1

..

1

..

.

Part of graphinside blue isinside green

.


I So limx→0+

f(x) = 1 and limx→0−

f(x) = −1


x→0+

|x|x

and limx→0−

|x|x

if they exist.

.. x.

y

..

−1

..

1

...


.


I So limx→0+


f(x) = −1


x→0+

|x|x

and limx→0−

|x|x

if they exist.

.. x.

y

..

−1

..

1

..

.


.


I So limx→0+


f(x) = −1


Find limx→0

|x|x

if it exists.

Solu onThe error-tolerance game fails, but

limx→0+

f(x) = 1 limx→0−

f(x) = −1

Another ExampleExample

Find limx→0+

1xif it exists.

Solu onThe limit does not exist because the func on is unbounded near 0.Next week we will understand the statement that

limx→0+

1x= +∞

The error-tolerance game with 1/x

Find limx→0+

1xif it exists.

.. x.

y

.0..

L?

.

The graph escapesthe green, so nogood

.

Even worse!

.

The limit does not existbecause the func on isunbounded near 0

Another ExampleExample

Find limx→0+

1xif it exists.

Solu onThe limit does not exist because the func on is unbounded near 0.Next week we will understand the statement that

limx→0+

1x= +∞

Weird, wild stuffExample

Find limx→0

sin(πx

)if it exists.

Solu on

I f(x) = 0 when x =

1kfor any integer k

I f(x) = 1 when x =

24k+ 1

for any integer k

I f(x) = −1 when x =

24k− 1

for any integer k

Function valuesx π/x sin(π/x)

1 π 01/2 2π 01/k kπ 02 π/2 1

2/5 5π/2 12/9 9π/2 1

2/(4k+ 1) ((4k+ 1)π)/2 12/3 3π/2 − 12/7 7π/2 − 1

2/(4k− 1) ((4k− 1)π)/2 − 1

...

π/2

..π ..

3π/2

.. 0


1

π 0

1/2 2π 01/k kπ 02 π/2 1

2/5 5π/2 12/9 9π/2 1

2/(4k+ 1) ((4k+ 1)π)/2 12/3 3π/2 − 12/7 7π/2 − 1

2/(4k− 1) ((4k− 1)π)/2 − 1

...

π/2

..π ..

3π/2

.. 0


1

π 0

1/2

2π 0

1/k kπ 02 π/2 1

2/5 5π/2 12/9 9π/2 1

2/(4k+ 1) ((4k+ 1)π)/2 12/3 3π/2 − 12/7 7π/2 − 1

2/(4k− 1) ((4k− 1)π)/2 − 1

...

π/2

..π ..

3π/2

.. 0


1

π 0

1/2

2π 0

1/k

kπ 0

2 π/2 12/5 5π/2 12/9 9π/2 1

2/(4k+ 1) ((4k+ 1)π)/2 12/3 3π/2 − 12/7 7π/2 − 1

2/(4k− 1) ((4k− 1)π)/2 − 1

...

π/2

..π ..

3π/2

.. 0

Function valuesx π/x sin(π/x)1 π 0

1/2

2π 0

1/k

kπ 0

2 π/2 12/5 5π/2 12/9 9π/2 1

2/(4k+ 1) ((4k+ 1)π)/2 12/3 3π/2 − 12/7 7π/2 − 1

2/(4k− 1) ((4k− 1)π)/2 − 1

...

π/2

..π ..

3π/2

.. 0


1/2 2π 0

1/k

kπ 0

2 π/2 12/5 5π/2 12/9 9π/2 1

2/(4k+ 1) ((4k+ 1)π)/2 12/3 3π/2 − 12/7 7π/2 − 1

2/(4k− 1) ((4k− 1)π)/2 − 1

...

π/2

..π ..

3π/2

.. 0


1/2 2π 01/k kπ 0

2 π/2 12/5 5π/2 12/9 9π/2 1

2/(4k+ 1) ((4k+ 1)π)/2 12/3 3π/2 − 12/7 7π/2 − 1

2/(4k− 1) ((4k− 1)π)/2 − 1

...

π/2

..π ..

3π/2

.. 0


1/2 2π 01/k kπ 0

2

π/2 1

2/5 5π/2 12/9 9π/2 1

2/(4k+ 1) ((4k+ 1)π)/2 12/3 3π/2 − 12/7 7π/2 − 1

2/(4k− 1) ((4k− 1)π)/2 − 1

...

π/2

..π ..

3π/2

.. 0


1/2 2π 01/k kπ 0

2

π/2 1

2/5

5π/2 1

2/9 9π/2 12/(4k+ 1) ((4k+ 1)π)/2 1

2/3 3π/2 − 12/7 7π/2 − 1

2/(4k− 1) ((4k− 1)π)/2 − 1

...

π/2

..π ..

3π/2

.. 0


1/2 2π 01/k kπ 0

2

π/2 1

2/5

5π/2 1

2/9

9π/2 1

2/(4k+ 1) ((4k+ 1)π)/2 12/3 3π/2 − 12/7 7π/2 − 1

2/(4k− 1) ((4k− 1)π)/2 − 1

...

π/2

..π ..

3π/2

.. 0


1/2 2π 01/k kπ 0

2

π/2 1

2/5

5π/2 1

2/9

9π/2 1

2/(4k+ 1)

((4k+ 1)π)/2 1

2/3 3π/2 − 12/7 7π/2 − 1

2/(4k− 1) ((4k− 1)π)/2 − 1

...

π/2

..π ..

3π/2

.. 0


1/2 2π 01/k kπ 02 π/2 1

2/5

5π/2 1

2/9

9π/2 1

2/(4k+ 1)

((4k+ 1)π)/2 1

2/3 3π/2 − 12/7 7π/2 − 1

2/(4k− 1) ((4k− 1)π)/2 − 1

...

π/2

..π ..

3π/2

.. 0


1/2 2π 01/k kπ 02 π/2 1

2/5 5π/2 1

2/9

9π/2 1

2/(4k+ 1)

((4k+ 1)π)/2 1

2/3 3π/2 − 12/7 7π/2 − 1

2/(4k− 1) ((4k− 1)π)/2 − 1

...

π/2

..π ..

3π/2

.. 0


1/2 2π 01/k kπ 02 π/2 1

2/5 5π/2 12/9 9π/2 1

2/(4k+ 1)

((4k+ 1)π)/2 1

2/3 3π/2 − 12/7 7π/2 − 1

2/(4k− 1) ((4k− 1)π)/2 − 1

...

π/2

..π ..

3π/2

.. 0


1/2 2π 01/k kπ 02 π/2 1

2/5 5π/2 12/9 9π/2 1

2/(4k+ 1) ((4k+ 1)π)/2 1

2/3 3π/2 − 12/7 7π/2 − 1

2/(4k− 1) ((4k− 1)π)/2 − 1

...

π/2

..π ..

3π/2

.. 0


1/2 2π 01/k kπ 02 π/2 1

2/5 5π/2 12/9 9π/2 1

2/(4k+ 1) ((4k+ 1)π)/2 1

2/3

3π/2 − 1

2/7 7π/2 − 12/(4k− 1) ((4k− 1)π)/2 − 1

...

π/2

..π ..

3π/2

.. 0


1/2 2π 01/k kπ 02 π/2 1

2/5 5π/2 12/9 9π/2 1

2/(4k+ 1) ((4k+ 1)π)/2 1

2/3

3π/2 − 1

2/7

7π/2 − 1

2/(4k− 1) ((4k− 1)π)/2 − 1

...

π/2

..π ..

3π/2

.. 0


1/2 2π 01/k kπ 02 π/2 1

2/5 5π/2 12/9 9π/2 1

2/(4k+ 1) ((4k+ 1)π)/2 1

2/3

3π/2 − 1

2/7

7π/2 − 1

2/(4k− 1)

((4k− 1)π)/2 − 1

...

π/2

..π ..

3π/2

.. 0


1/2 2π 01/k kπ 02 π/2 1

2/5 5π/2 12/9 9π/2 1

2/(4k+ 1) ((4k+ 1)π)/2 12/3 3π/2 − 1

2/7

7π/2 − 1

2/(4k− 1)

((4k− 1)π)/2 − 1

...

π/2

..π ..

3π/2

.. 0


1/2 2π 01/k kπ 02 π/2 1

2/5 5π/2 12/9 9π/2 1

2/(4k+ 1) ((4k+ 1)π)/2 12/3 3π/2 − 12/7 7π/2 − 1

2/(4k− 1)

((4k− 1)π)/2 − 1

...

π/2

..π ..

3π/2

.. 0


1/2 2π 01/k kπ 02 π/2 1

2/5 5π/2 12/9 9π/2 1

2/(4k+ 1) ((4k+ 1)π)/2 12/3 3π/2 − 12/7 7π/2 − 1

2/(4k− 1) ((4k− 1)π)/2 − 1

...

π/2

..π ..

3π/2

.. 0


Find limx→0

sin(πx

)if it exists.

Solu on

I f(x) = 0 when x =

1kfor any integer k

I f(x) = 1 when x =

24k+ 1

for any integer k


24k− 1

for any integer k


Find limx→0

sin(πx

)if it exists.

Solu on

I f(x) = 0 when x =1kfor any integer k

I f(x) = 1 when x =

24k+ 1

for any integer k


24k− 1

for any integer k


Find limx→0

sin(πx

)if it exists.

Solu on


I f(x) = 1 when x =2

4k+ 1for any integer k


24k− 1

for any integer k


Find limx→0

sin(πx

)if it exists.

Solu on


I f(x) = 1 when x =2

4k+ 1for any integer k

I f(x) = −1 when x =2

4k− 1for any integer k

GraphHere is a graph of the func on:

.. x.

y

..

−1

..

1

There are infinitely many points arbitrarily close to zero where f(x) is0, or 1, or−1. So the limit cannot exist.

What could go wrong?Summary of Limit Pathologies

How could a func on fail to have a limit? Some possibili es:I le - and right- hand limits exist but are not equalI The func on is unbounded near aI Oscilla on with increasingly high frequency near a

Meet the MathematicianAugustin Louis Cauchy

I French, 1789–1857I Royalist and CatholicI made contribu ons in geometry,calculus, complex analysis,number theory

I created the defini on of limitwe use today but didn’tunderstand it

Outline

Heuris cs


Examples


Precise Definition of a LimitNo, this is not going to be on the test

Let f be a func on defined on an some open interval that containsthe number a, except possibly at a itself. Then we say that the limitof f(x) as x approaches a is L, and we write

limx→a

f(x) = L,

if for every ε > 0 there is a corresponding δ > 0 such that

if 0 < |x− a| < δ, then |f(x)− L| < ε.

The error-tolerance game = ε, δ

.

.

L+ ε

.

L− ε

.a− δ

.a+ δ

.

This δ is too big

.a− δ

.a+ δ

.

This δ looks good

.a− δ

.a+ δ

.

So does this δ

.a

.

L

The error-tolerance game = ε, δ

..

L+ ε

.

L− ε

.a− δ

.a+ δ

.

This δ is too big

.a− δ

.a+ δ

.

This δ looks good

.a− δ

.a+ δ

.

So does this δ

.a

.

L

SummaryMany perspectives on limits

I Graphical: L is the value the func on“wants to go to” near a

I Heuris cal: f(x) can be made arbitrarilyclose to L by taking x sufficiently closeto a.

I Informal: the error/tolerance gameI Precise: if for every ε > 0 there is acorresponding δ > 0 such that if0 < |x− a| < δ, then |f(x)− L| < ε.

I Algebraic: next me

.. x.

y

..

−1

..

1

.

FAIL

lesson 3: the limit of a function (slides)

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