DO NOT READ THIS SUPPLEMENT UNLESS YOU ABSOLUTELY HAVE TO

Calculus I (part 1s): Formal The formal use of -δ -δ Limits (by Evan Dummit, 2012, v. 1.55) limits does not yield a great deal of insight into...
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Calculus I (part 1s): Formal

The formal use of

-δ

-δ

Limits

(by Evan Dummit, 2012, v. 1.55)

limits does not yield a great deal of insight into how to use limits at a beginning level. As

such, the development of the formal theory of limits has been relegated to this supplement. Or, more dramatically:

DO NOT READ THIS SUPPLEMENT UNLESS YOU ABSOLUTELY HAVE TO

Contents 0.0.1

Formal Denition and Examples

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

1

0.0.2

Proofs of the Limit Rules for Finite Limits

. . . . . . . . . . . . . . . . . . . . . . . . . . . .

3

0.0.3

Formal Arguments for Innite Limits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

5

0.0.4

Proofs of the Limit Rules for Innite Limits . . . . . . . . . . . . . . . . . . . . . . . . . . . .

6

0.0.1 Formal Denition and Examples •

Denition: A function

f (x)

has the limit

δ>0

how small) there exists a

L

x → a,

as

(depending on

)

written as

lim f (x) = L,

x→a

with the property that for all

if, for any

>0

0 < |x − a| < δ ,

(no matter

we have that

|f (x) − L| < . ◦

The symbols

δ

and



are the lowercase Greek letters delta and epsilon (respectively). Their use in the

denition of the limit is traditional. Also recall that the notation denotes the distance from



x

means the absolute value of

One way to think of this denition is as follows: suppose you claim that the function

L,

as

x

a.

gets close to

If

containing

f (x)

a,

 > 0,

and I want you to give me some open interval

with the property that

f (x)

really does stay close to the limit value

is always within

L

as

x



for

gets close to

a,

x

◦ •

f (x)

and

f (x)

has a limit

should stay near

L

when

x

is near

(a − δ, a + δ)

on the

in that interval, except possibly

then, no matter what value of

picked, you should always be able to answer my challenge with an interval around of

x,

In order to prove to me that the function really does have that limit, I challenge

you by handing you some value

x-axis at a. ◦

|x|

to zero.

a,



I

because the values

a.

Note that it is not necessary to nd the best possible

δ

 any

δ

which does the job is perfectly ne.

Important Remark: Don't worry if this formal denition seems very opaque at rst. It takes practice and experience to become comfortable with what the denition means, and to see why it really does match the intuition of how a limit should behave.



We will generally use the formal denition primarily as a tool to justify our manipulations of limits and to ground our intuition in proof.



lim c = c,

Basic Limit 1:



x→a

c

is any constant.

To show this formally, suppose we are given an

|c − c| < 



where

whenever



Now, since



So in fact here we can take any

Basic Limit 2:

 > 0,

and we want to nd a

0 < |x − a| < δ .

|c − c| = 0,

and

 > 0, δ

the inequality

|c − c| < 

is always true.

we want  any at all  and the result holds.

lim x = a.

x→a

1

δ > 0

which will make



Proof: Suppose we are given an whenever

◦ ◦

0 < |x − a| < δ ,

Let's try taking our

δ

where

to equal

This is in fact true, since if

 > 0, and h(x) = x.

.

we want to nd a

Then we need to check that

0 < |x − a| < 

then certainly

δ > 0

which will make

0 < |x − a| < δ

|x − a| < .

makes

|h(x) − a| < 

|x − a| < .

So this choice of

δ

works, and so

the limit has the value we claimed.



Note that we could have chosen

δ = /2 •

δ

to be lots of other things, and it still would have worked: for example,

would also have worked.

δ = ,

In that example it may have seemed like we just guessed that we should try

and then plugged in to

see that it would work. In general, this is how formal limit proofs work  generally, one needs some insight or observation to gure out what of

δ

to use, but all that is needed for the proof is to plug in to see that the choice



For arbitrary functions, there is not any general rule for nding



Often what is needed is to try to solve the problem backwards: i.e., to rearrange the wanted inequality

|f (x) − L| <  •

δ

actually works.

using properties of the function

Example: We prove that



(x2 − 9) = (x − 3)(x + 3), |(x − 3)(x + 3)| < .

and want to pick

δ

such that

We can factor

|x − 3| < δ already |x + 3|?

Our hypothesis

to gure out what value of

δ

will make things work.

lim x2 = 9.

We are given



.

from

x→3



>0

f (x),

δ

|x − 3| < δ

implies

2 x − 9 < .

and so we are looking to pick

tells us that we can make

|x − 3|

δ

which makes

|x − 3| · |x + 3| =

small, but what about the other

term in that product,



We are free to choose since

|x − 3| < δ ≤ 1,

δ

however we like, so (for instance) we can always insist on taking

this says

2 < x < 4,

and so

5 < x + 3 < 7.

So we can say that

|x + 3|

δ ≤ 1.

Then

is always less

than 7.



Thus, if we choose

δ ≤ 1,



If we can make this always less than

we know that

,

this work. Remembering the condition

|x − 3| · |x + 3| < |x − 3| · 7 < 7δ . δ = /7 to make δ = min(1, /7) should always

we will be done. So, for instance, we can take

δ ≤ 1,

we conclude that taking

work.



If we wanted to write this up carefully, we would do it as follows:



We claim that

∗ ∗ ∗

We are given

 > 0 and want to nd δ > 0 such that |x − 3| < δ We claim that δ = min(1, /7) will always work. To show this works, suppose that|x − 3| < δ . · · ·

∗ ◦

lim x2 = 9.

x→3

2 x − 9 < .

− 3| < 1, so −1 < x − 3 < 1. 5 < x + 3 < 7, and so |x + 3| < 7. 2 x − 9 = |x − 3| · |x + 3| < δ · 7 < 7 · (/7) = , as we wanted.

In particular, since

δ≤1

implies

we see that|x

Adding 6 to each part of this inequality gives Finally,

Therefore,

lim x2 = 9.

x→3

When we write the proof up this way, it looks very clean, but it obscures all the work we had to do in order to gure out what

δ

actually should be. This will usually be the case with formal

−δ

proofs:

they often seem like merely an excellent guess.



We can also use this denition to prove that a limit does not exist. Showing that a given limit does not exist (at all) is the same as showing that no matter what value of which we can't nd any



Example:

L

we pick, there exists some bad value of



for

δ.

Show that the step function

undened at 0, has no limit as

x , |x|

which has value

x → 0.

2

−1

for negative

x, +1

for positive

x,

and is



Here is the graph of this function: 1.0

0.5

-1.0

-0.5

0.5

1.0

-0.5

-1.0



To show the function has no limit at pick, there is a value of



Here, we can pick

x = 0,

we need to verify that that no matter what value of

L

we

which falsies the limit hypothesis.

 = 1/3

no matter what value of

L

we're trying:



Since any open interval around 0 contains both positive and negative numbers, we would need both



But the sum

|1 − L| and

∗ ∗ ◦



|−1 − L| to be less than  = 1/3. |1 − L| + |−1 − L| is always at least otherwise the sum is 2 |L|. and

L

is between

−1

This is obviously impossible, since the sum of two things each less than Therefore, the limit cannot be

L,

for any value of

Remark: One can make a similar argument for



2: if

L.

and

1/3

+1

then the sum is 2

cannot be at least 2.

In other words, the limit does not exist.

  1 f (x) = cos x

to see that it has no limit at

x = 0.

f (x) = 1 and f (x) = −1.   1 ∗ Essentially the same argument as above (with  = 1/3) will show that cos has no limit at x = 0. x In any interval around

0,

no matter how small, there are points where

0.0.2 Proofs of the Limit Rules for Finite Limits •

Let

f (x)

and

g(x)

be functions satisfying

lim f (x) = Lf

x→a

and

lim g(x) = Lg .

x→a

Then the following properties

hold:



The addition rule:



lim [f (x) + g(x)] = Lf + Lg .

x→a

Proof: Suppose we are given



 > 0.

lim g(x) = Lg , we can nd δ1 and δ2 such that |f (x) − Lf | <  for 0 < |x − a| < δ1 and |g(x) − Lg | < for 0 < |x − a| < δ2 . 2 ∗ We claim that the value δ = min(δ1 , δ2 ) will make |f (x) + g(x) − (Lf + Lg )| <  for all x with 0 < |x − a| < δ .     ∗ To verify: we know that − < f (x) − Lf < and − < g(x) − Lg < for 0 < |x − a| < min(δ1 , δ2 ), 2 2 2 2 so adding the inequalities shows − < (f (x) − Lf ) + (g(x) − Lg ) < . ∗ Or, in other words, |f (x) + g(x) − (Lf + Lg )| <  for all x with 0 < |x − a| < δ . This is what we Since we know that

 2

lim f (x) = Lf

x→a

and

x→a

wanted to show.



The subtraction rule:

◦ •

lim [f (x) − g(x)] = Lf − Lg .

x→a

Proof: The same as the addition rule, but with a minus sign instead of a plus sign.

The multiplication rule:



lim [f (x) · g(x)] = Lf · Lg .

x→a

Proof: First we show that if

∗ ∗

Suppose we are given Since we know that for all

lim h(x) = Lh

x→a

then

lim h(x)2 = L2h .

x→a

 > 0.

lim h(x) = Lh ,

x→a

then we can nd

0 < |x − a| < δ .

3

δ

such that

 |h(x) − Lh | < min 1,

 1 + 2 |Lh |





In particular, for 0 < |x − a| < δ we have |h(x) − Lh | < 1 or −1 < h(x) − Lh < 1, −1 + 2Lh < h(x) + Lh < 1 + 2Lh . Taking absolute values shows |h(x) + Lh | < 1 + 2 |Lh |.  ∗ Then h(x)2 − L2h = |h(x) − Lh | · |h(x) + Lh | < · (1 + 2 |Lh |) = , as desired. 1 + 2 |Lh |



Now, applying this result to the two particular cases where

h(x) = f (x) + g(x)



and

h(x) = f (x) − g(x) lim [f (x) + g(x)]2 =

and

shows (after an application of the addition rule and the subtraction rule) that

(Lf + Lg )2

so that

x→a

lim [f (x) − g(x)]2 = (Lf − Lg )2 .

x→a

lim [4f (x) · g(x)] = 4Lf · Lg .

Subtracting, applying the subtraction rule again, and cancelling the common terms gives

x→a

Dividing by 4 then gives the multiplication rule.





The division rule:



 f (x) Lf lim = , x→a g(x) Lg

Proof: First we show that if

Lg

provided that

lim h(x) = Lh

x→a

and

is not zero.

Lh > 0

lim

then

x→a

1 1 = . h(x) Lh

Suppose we are given

 > 0.  ∗

Since we know that all



lim h(x) = Lh ,

x→a

then we can nd

δ

such that

|h(x) − Lh | < min

 Lh , 2L2h  2

0 < |x − a| < δ . 0 < |x − a| < δ

In particular, for

we have

|h(x) − Lh |
0

is continuous there exists

|f (x) − L| < δ1

x

for all

|g(f (x)) − g(L)| < 

with

for all

for which

0 < |x − a| < δ2 .

x

with

0 < |x − a|
0 (no matter how small) there exists 0 < a − x < δ , we have that |f (x) − L| < .

following statement is true: for any with the property that for all



We say that a function

f (x)

has the limit

L

as

x→a

from above, written as

 > 0 (no matter how small) there exists 0 < x − a < δ , we have that |f (x) − L| < .

following statement is true: for any with the property that for all



a

a

lim f (x) = L,

x→a−

δ>0

(depending on

lim f (x) = L,

x→a+

δ>0

if the

)

if the

(depending on

)

The formal denitions for limits at innity are the following:



We say that a function

f (x)

has the limit

L

as

x → +∞,

written as

 > 0 (no matter how small) there x > A, we have that |f (x) − L| < .

statement is true: for any property that for all



We say that a function

f (x)

has the limit

L

as

written as

 > 0 (no matter how small) there x < −A, we have that |f (x) − L| < .

statement is true: for any property that for all

x → −∞,

5

exists an

exists an

lim f (x) = L,

x→+∞

A>0

(depending on

lim f (x) = L,

x→−∞

A>0

if the following

)

with the

if the following

(depending on

)

with the



smaller





Note: As

shrinks,

will grow very large, in contrast to when we used



Notation: The symbols

−∞

contrast with the



A

which would get smaller with

and

+∞

+∞

is often used for

lim f (x) = +∞,

if the following

mean the same thing (positive innity); the

symbol (negative innity).

The formal denitions for innite limits are the following:



We say that a function

f (x)

diverges to

+∞

x → c,

as

B > 0 (no matter how large) 0 < |x − c| < δ , we have f (x) > B .

statement is true: for any property that for all

∗ ∗ ◦

−∞

The idea of diverging to

f (x)

the property that for all



x→c

We can also formulate one-sided limits here, with

We say that a function

written as

diverges to

+∞

as

x→c

there exists a

x → ∞,

δ>0

with the

written as

lim f (x) = +∞, >0

if the following

(depending on

B)

with

f (x) > B .

have

The statements for diverging to

B)

from above or from below.

x→∞ (no matter how large) there exists an A

B >0 x > A, we

(depending on

f (x) < −B .

is analogous, except instead

statement is true: for any



δ,

.

−∞,

or the statements as

x → −∞,

are similar.

Here are some sketch-examples of innite limits evaluated using the denitions:



Example: The function it diverges to

∗ ◦

+∞

if

n

n a positive integer) diverges to +∞ as x → +∞ and, as x → −∞, −∞ if n is odd. √ A = n B , for either direction.

(for

is even and to

For this function, we can take

Example: The function



f (x) = xn

diverges to

+∞

as

x → +∞

and tends to 0 as

x → −∞.

A = ln(B). As x → −∞ we can take A = − ln(). 1 ◦ Example: The function f (x) = 2 diverges to +∞ as x → 0. x √ ∗ For this function, we can take δ = B . 1 ◦ Example: The function f (x) = diverges to −∞ as x → 0 from below, and diverges x As

x → +∞

f (x) = ex

we can take

to

+∞

as

x→0

from above.



For this function, we can take

δ = B,

for either direction.

0.0.4 Proofs of the Limit Rules for Innite Limits •

Basic Limits:



lim c = c

x→+∞

for any constant

c, lim

1

= 0,

x→+∞ x

and

lim x = ∞.

x→∞

Proof (c): Suppose we are given

 > 0. We want to nd an A > 0 such that for all x > A, we have that |c − c| < . Any value of A will work, since |c − c| = 0 is always less than  (which is positive). 1 ◦ Proof ( ): Suppose we are given  > 0. We want to nd an A > 0 such that for all x > A, we have that x 1 < . We claim that A = 1 will work. To see this, observe that if x > 1 , then after multiplying the x   1    1 1 1 inequality by we see that · x > · , or  > . Since is positive, we thus obtain < , as x x x x  x x desired.



Proof (x): We need to show that for any

(no matter how large) there exists an

B ) with f (x) = x.

have

on



Negation: If



the property that for all

lim f (x) = ∞,

x→a

then

B>0 x > A, we

lim [−f (x)] = −∞,

x→a

δ

are the same).

6

A > 0 (depending A = B , since

Here, we can simply take

and vice versa.

Proof: We need only multiply the inequality for (i.e., the values of

f (x) > B .

f (x)

by

−1

to get the necessary inequality for

−f (x)



Multiplication: If



lim f (x)

is a nite positive number or

x→a

for some positive real number



Now let we have

◦ •

B > 0 be given. g(x) > B/M .

Then for all

x→a



Now let we have

◦ •

lim f (x)

◦ •

we see that

lim g(x) = ∞,

x→a

there exists some

δ1

then

x>C

f (x)g(x) > B ,

then

lim [f (x)g(x)] = ∞.

x→a

it is true that

f (x) > M

as desired.

lim [f (x) + g(x)] = ∞.

x→a

such that for all

x

with

0 < |x − a| < δ1

it is

M.

f,

lim g(x) = ∞,

x→a

there exists some

By the hypothesis about

0 < |x − a| < δ lim f (x) = L

x→a

with

then

δ1

such that for all

f (x)

L ≥ 0)

with

0 < |x − a| < δ1

it is

M.

we see that

f (x) < , g(x)

lim g(x) = ∞,

x→a

then

as desired.

lim f (x)g(x)

x→a

is 0 if

0 < L < 1,

and



if



Proof (f

g

for 0 < L < 1): By the hypothesis about f , there exists some δ1 such that for all x with 0 < |x − a| < δ1 it is true that |f (x)| < M for some real number M with 0 < M < 1. Now let  > 0 be given: by the hypothesis about g , there exists a δ2 > 0 such that for all 0 < |x − a| < δ2 , we have g(x) > − logM . Then for all 0 < |x − a| < δ with δ = min(δ1 , δ2 ), we see that f (x)g(x) < , as desired.



Proof (f

g

L > 1.

Furthermore,



is

provided

L > 0.

δ1 such that for all x with 0 < M > 1. Now let B > 0 be given: by the hypothesis about g , there exists a δ2 > 0 such that for all 0 < |x − a| < δ2 , we have g(x) > logM B . g(x) Then for all 0 < |x − a| < δ with δ = min(δ1 , δ2 ), we see that f (x) > B , as desired. for

|x − a| < δ1

L > 1):

lim g(x)

x→a



and

x

as desired.

g , there exists a δ2 > 0 such that for all 0 < |x − a| < δ2 ,

δ = min(δ1 , δ2 ),

(where

f (x) + g(x) > B ,   f (x) lim = 0. x→a g(x)

we see that

for some (nonnegative) real number

 > 0 be given. g(x) > M/.

Then for all

δ = min(δ1 , δ2 ),

is a nite number and

|f (x)| < M

Exponentiation: If is

with

Proof: By the hypothesis about

Now let

such that for all

for some (possibly negative) real number

0 < |x − a| < δ

x→a

we have

f,

and

C

lim g(x) = ∞,

x→a

B > 0 be given. By the hypothesis about g , there exists a δ2 > 0 such that for all 0 < |x − a| < δ2 , g(x) > B − M .

true that





and

g , there exists a δ2 > 0 such that for all 0 < |x − a| < δ2 ,

δ = min(δ1 , δ2 ),

is a nite number or

f (x) > M

Then for all

Division: If



with

Proof: By the hypothesis about true that

there exists some

By the hypothesis about

0 < |x − a| < δ

lim f (x)

Addition: If



f, M.

Proof: By the hypothesis about



By the hypothesis about

it is true that

f

|f (x)| > M

f,

there exists some

for some real number

M

with

δ1 such that for all x with 0 < |x − a| < δ1 it M . Now let B > 0 be given: by the hypothesis 1/M about g , there exists a δ2 > 0 such that for all 0 < |x − a| < δ2 , we have g(x) > B . Then for all f (x) 0 < |x − a| < δ with δ = min(δ1 , δ2 ), we see that g(x) > B , as desired. Proof (g ): By the hypothesis about

is true that

|f (x)| > M

f,

there exists some

for some positive real number

Well, you're at the end of my handout. Hope it was helpful. Copyright notice: This material is copyright Evan Dummit, 2012. You may not reproduce or distribute this material without my express permission.

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