M53 Lec1.6 Formal Definition of a Limit

7/27/2019 M53 Lec1.6 Formal Definition of a Limit

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Formal Definition of a Limit

Mathematics 53

Institute of Mathematics (UP Diliman)

Institute of Mathematics (UP Diliman) Limits: Formally Mathematics 53 1 / 19


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For today

1 The Formal Definition

2 Proving using the Definition



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For today





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Recall:

Intuitive Notion of a Limit

limxa

f(x) = L f(x) gets closer and closer to L as x gets closer and

closer to a (but does not reach a)



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Recall:


limxa

f(x) = L f(x) gets closer and closer to L as x gets closer and

closer to a (but does not reach a)

Alternatively,


limxa

f(x) = L we could let f(x) be as close as we would like to L by taking

values of x sufficiently close but not equal to a



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Recall: If r, s R , then |r s| = distance between the points on the number line

representing r and s.

|r s|

rs

|r s|

sr



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The Formal Definition

Definition

Let f be a function defined on

some open interval containing a

except possibly at a. The limit

of f(x) as x approaches a is L,

written

limxa

f(x) = L

a

L



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Definition





written

limxa

f(x) = L

if for every > 0 (no matter how

small)a

L

(

)

L+

L



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Definition





written

limxa

f(x) = L


small) there exists a > 0a

L

(

a a+

)

(

)

L+

L



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Definition





written

limxa

f(x) = L


small) there exists a > 0

such that if 0 < |x a| < ,

then |f(x) L| < .

a

L

(

a a+

)

(

)

L+

L



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Definition





written

limxa

f(x) = L




then |f(x) L| < .

a

L

(

a a+

)

(

)

L+

L



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Definition





written

limxa

f(x) = L




then |f(x) L| < .

a

L

(

a a+

)

(

)

L+

L



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Definition





written

limxa

f(x) = L




then |f(x) L| < .

a

L

(

a a+

)

(

)

L+

L



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Definition




of f(x) as x approaches a is L,written

limxa

f(x) = L




then |f(x) L| < .

a

L

(

a a+

)

(

)

L+

L



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Definition





limxa

f(x) = L




then |f(x) L| < .

a

L

(

a a+

)

(

)

L+

L



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Definition





limxa

f(x) = L




then |f(x) L| < .

a

L

(

a a+

)

(

)

L+

L



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Definition





limxa

f(x) = L




then |f(x) L| < .

a

L

(

a a+

)

(

)

L+

L



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Definition





limxa

f(x) = L




then |f(x) L| < .

a

L

(

a a+

)

(

)

L+

L



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a

L



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a

L



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a

L


fi


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a

L


Th F l D fi i i


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a

L


Th F l D fi iti


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a

L


Th F l D fi iti


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a

L




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a

L




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a

L




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a

L




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a

L




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a

L




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a

L?




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a

L?




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a

L?




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a

L?




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a

L?




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a

L?




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a

L?




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a

L?




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a

L?




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a

L?




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a

L?




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a

L?




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a

L?




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a

L?




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a

L?




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a

L?




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a

L?




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a

L?




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a

L?




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a

L?


For today


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Proving using the Definition


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Two steps in proving limits that limxa

f(x) = L using the definition:




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I. Choosing a suitable




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I. Choosing a suitable

II. Verifying that the chosen works.


Example


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Prove that limx1(3x + 2) = 5.


Example


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Proof. Let > 0.


Example


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Proof. Let > 0.

I. GOAL: Find > 0 such that


Example


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Prove thatlimx1(

3x +

2) =

5.

Proof. Let > 0.


if 0 < |x 1| < ,


Example


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Prove thatlimx1(

3x +

2) =

5.

Proof. Let > 0.


if 0 < |x 1| < , then


Example


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Prove thatlimx1(

3x +

2) =

5.

Proof. Let > 0.


if 0 < |x 1| < , then |(3x + 2) 5| <


Example


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Prove that limx1(

3x +

2) =

5.

Proof. Let > 0.


if 0 < |x 1| < , then |(3x + 2) 5| <

|3x 3| <


Example


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Prove that limx1(

3x +

2) =

5.

Proof. Let > 0.


if 0 < |x 1| < , then |(3x + 2) 5| <

|3x 3| <

3|x 1| <


Example


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Prove that limx1(

3x +

2) =

5.

Proof. Let > 0.


if 0 < |x 1| < , then |(3x + 2) 5| <

|3x 3| <

3|x 1| <

|x 1| 0.


if 0 < |x 1| < , then |(3x + 2) 5| <

|3x 3| <

3|x 1| <

|x 1| 0, choose =

3.


Example


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II. Given > 0, choose =

3. Then, if

0 < |x 1| < ,


Example


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3. Then, if

0 < |x 1| < ,

we have

|(3x + 2) 5|


Example


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3. Then, if

0 < |x 1| < ,

we have

|(3x + 2) 5| = 3|x 1|


Example


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3. Then, if

0 < |x 1| 0, choose =

3. Then, if

0 0, choose =

3. Then, if

0 0, choose =

3. Then, if

0 0.


Example


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Prove that limx2

(5x + 6) = 4.

Proof. Let > 0.



Example


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Prove that limx2

(5x + 6) = 4.

Proof. Let > 0.


if 0 < |x (2)| < ,


Example


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Prove that limx2

(5x + 6) = 4.

Proof. Let > 0.


if 0 < |x (2)| < , then


Example


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Prove that limx2

(5x + 6) = 4.

Proof. Let > 0.


if 0 < |x (2)| < , then |(5x + 6) (4)| <


Example


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Prove that limx2

(5x + 6) = 4.

Proof. Let > 0.


if 0 < |x (2)| < , then |(5x + 6) (4)| < |5x + 10| <


Example


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Prove that limx2

(5x + 6) = 4.

Proof. Let > 0.


if 0 < |x (2)| < , then |(5x + 6) (4)| < |5x + 10| <

5|x + 2| <


Example


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Prove that limx2

(5x + 6) = 4.

Proof. Let > 0.


if 0 < |x (2)| < , then |(5x + 6) (4)| < |5x + 10| <

5|x + 2| <

|x + 2| 0.


if 0 < |x (2)| < , then |(5x + 6) (4)| < |5x + 10| <

5|x + 2| <

|x + 2| 0, choose =

5.


Example


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5. Then if

0 < |x

(2)| < ,


Example


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5. Then if

0 < |x (2)| < ,

we have

|(5x + 6) (4)|


Example


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5. Then if

0 < |x (2)| < ,

we have

|(5x + 6) (4)| = 5|x + 2|


Example


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5. Then if

0 < |x (2)| < ,

we have

|(5x + 6) (4)| = 5|x + 2| < 5 =


Example


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5. Then if

0 < |x (2)| < ,

we have

|(5x + 6) (4)| = 5|x + 2| < 5 = 5

5


Example


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5. Then if

0 < |x (2)| < ,

we have

|(5x + 6) (4)| = 5|x + 2| < 5 = 5

5

= .


Example


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5. Then if

0 < |x (2)| < ,

we have

|(5x + 6) (4)| = 5|x + 2| < 5 = 5

5

= .

By definition, we have limx1

(5x + 6) = 4.


Example


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Prove that lim

x0

(4 3x) = 4.


Example


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Prove that lim

x0

(4 3x) = 4.

Proof. Let > 0.


Example

l ( )


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Prove that lim

x0

(4 3x) = 4.

Proof. Let > 0.



Example

P h li (4 3 ) 4


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Prove that lim

x0

(4 3x) = 4.

Proof. Let > 0.


if 0 0.


if 0 0.


if0 0.


if0 0.


if0 0.


if0 0 such that

if 0 0, choose =

3. So if

0 < |x 0| <


Example

Prove that lim(4 3x) = 4.


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ax0

( 3 )


3. So if

0 < |x 0| <

then

|(4 3x) 4| =


Example

Prove that lim(4 3x) = 4.


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x0( )


3. So if

0 < |x 0| <

then

|(4 3x) 4| = 3|x|


Example

Prove that lim0(4 3x) = 4.


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x0( )


3. So if

0 < |x 0| <

then

|(4 3x) 4| = 3|x| < 3


Example

Prove that lim0(4 3x) = 4.


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x0( )


3. So if

0 < |x 0| <

then

|(4 3x) 4| = 3|x| < 3 = 3

3


Example

Prove that lim0

(4 3x) = 4.


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x0


3. So if

0 < |x 0| <

then

|(4 3x) 4| = 3|x| < 3 = 3

3

= .


Example

Prove that limx0

(4 3x) = 4.


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x0


3. So if

0 < |x 0| <

then

|(4 3x) 4| = 3|x| < 3 = 3

3

= .

That is,

if 0 0, choose =

3. So if

0 < |x 0| <

then

|(4 3x) 4| = 3|x| < 3 = 3

3

= .

That is,

if 0 0 works such that

if 0 < |x a| < then |f(x) L| <


Remark

The value of is NOT unique!


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If a given > 0 works such that

if 0 < |x a| < then |f(x) L| <

then any smaller 0 <

will also work


Remark

The value of is NOT unique!


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If a given > 0 works such that

if 0 < |x a| < then |f(x) L| <

then any smaller 0 <

will also work since

if 0 < |x a| <

then

|f(x) L| < .


Announcements


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Reminder: First Exam is on 5 December 2012, Wednesday.

Math 53 Module for Units 2-3 is available (Php 40.00).

Copies of the Math 53 Module will be made available in the math library.


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M53 Lec1.6 Formal Definition of a Limit

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