calci proofslimits

Upload: chessgeneral

Post on 04-Jun-2018

219 views

Category:

Documents


0 download

TRANSCRIPT

  • 8/13/2019 CalcI ProofsLimits

    1/17

    PrefaceHere are my online notes for my Calculus I course that I teach here at Lamar University. Despite

    the fact that these are my class notes, they should be accessible to anyone wanting to learn

    Calculus I or needing a refresher in some of the early topics in calculus.

    Ive tried to make these notes as self contained as possible and so all the information needed to

    read through them is either from an Algebra or Trig class or contained in other sections of the

    notes.

    Here are a couple of warnings to my students who may be here to get a copy of what happened on

    a day that you missed.

    1. Because I wanted to make this a fairly complete set of notes for anyone wanting to learncalculus I have included some material that I do not usually have time to cover in classand because this changes from semester to semester it is not noted here. You will need tofind one of your fellow class mates to see if there is something in these notes that wasntcovered in class.

    2. Because I want these notes to provide some more examples for you to read through, Idont always work the same problems in class as those given in the notes. Likewise, evenif I do work some of the problems in here I may work fewer problems in class than are

    presented here.

    3. Sometimes questions in class will lead down paths that are not covered here. I try toanticipate as many of the questions as possible when writing these up, but the reality isthat I cant anticipate all the questions. Sometimes a very good question gets asked inclass that leads to insights that Ive not included here. You should always talk tosomeone who was in class on the day you missed and compare these notes to their notes

    and see what the differences are.

    4. This is somewhat related to the previous three items, but is important enough to merit itsown item. THESE NOTES ARE NOT A SUBSTITUTE FOR ATTENDING CLASS!!Using these notes as a substitute for class is liable to get you in trouble. As already noted

    not everything in these notes is covered in class and often material or insights not in thesenotes is covered in class.

  • 8/13/2019 CalcI ProofsLimits

    2/17

    Calculus I

    2007 Paul Dawkins 2 http://tutorial.math.lamar.edu/terms.aspx

    ProofofVariousLimitPropertiesIn this section we are going to prove some of the basic properties and facts about limits that we

    saw in the Limitschapter. Before proceeding with any of the proofs we should note that many of

    the proofs use theprecise definitionof the limit and it is assumed that not only have you read that

    section but that you have a fairly good feel for doing that kind of proof. If youre not verycomfortable using the definition of the limit to prove limits youll find many of the proofs in this

    section difficult to follow.

    The proofs that well be doing here will not be quite as detailed as those in the precise definition

    of the limit section. The proofs that we did in that section first did some work to get a guess for

    the d and then we verified the guess. The reality is that often the work to get the guess is not

    shown and the guess for d is just written down and then verified. For the proofs in this section

    where a d is actually chosen well do it that way. To make matters worse, in some of the proofs

    in this section work very differently from those that were in the limit definition section.

    So, with that out of the way, lets get to the proofs.

    Limit Properties

    In the Limit Propertiessection we gave several properties of limits. Well prove most of them

    here. First, lets recall the properties here so we have them in front of us. Well also be making a

    small change to the notation to make the proofs go a little easier. Here are the properties for

    reference purposes.

    Assume that ( )limx a

    f x K

    = and ( )limx a

    g x L

    = exist and that cis any constant. Then,

    1. ( ) ( )lim limx a x acf x c f x cK = =

    2. ( ) ( ) ( ) ( )lim lim limx a x a x a

    f x g x f x g x K L

    = =

    3. ( ) ( ) ( ) ( )lim lim limx a x a x a

    f x g x f x g x KL

    = =

    4. ( )( )

    ( )

    ( ) ( )

    limlim , provided lim 0

    lim

    x a

    x a x a

    x a

    f xf x KL g x

    g x g x L

    = = =

    5. ( ) ( )lim lim , where is any real number nn nx a x a

    f x f x K n

    = =

    6. ( ) ( )lim limn nx a x a

    f x f x

    =

    http://tutorial.math.lamar.edu/terms.aspxhttp://tutorial.math.lamar.edu/terms.aspx
  • 8/13/2019 CalcI ProofsLimits

    3/17

    Calculus I

    2007 Paul Dawkins 3 http://tutorial.math.lamar.edu/terms.aspx

    7. limx a

    c c

    =

    8. limx a

    a

    =

    9. lim n nx a a = Note that we added values (K,L, etc.) to each of the limits to make the proofs much easier. In

    these proofs well be using the fact that we know ( )limx a

    f x K

    = and ( )limx a

    x L

    = well use

    the definition of the limit to make a statement about ( )f x K- and ( )g x L- which will then

    be used to prove what we actually want to prove. When you see these statements do not worry

    too much about why we chose them as we did. The reason will become apparent once the proof

    is done.

    Also, were not going to be doing the proofs in the order they are written above. Some of the

    proofs will be easier if weve got some of the others proved first.

    Proof of 7

    This is a very simple proof. To make the notation a little clearer lets define the function

    ( )f x c= then what were being asked to prove is that ( )limx a

    f x c

    = . So lets do that.

    Let 0e> and we need to show that we can find a 0d > so that

    ( ) whenever 0f x c x ae d- < < - <

    The left inequality is trivially satisfied for anyxhowever because we defined ( )f x c= . So

    simply choose 0d> to be any number you want (you generally cant do this with these proofs).Then,

    ( ) 0f x c c c e- = - = <

    Proof of 1

    There are several ways to prove this part. If you accept 3And 7then all you need to do is let

    ( )g x c= and then this is a direct result of 3and 7. However, wed like to do a more rigorous

    mathematical proof. So here is that proof.

    First, note that if 0c= then ( ) 0cf x = and so,

    ( ) ( )lim 0 lim0 0 0x a x a

    f x f x

    = = =

    http://tutorial.math.lamar.edu/terms.aspxhttp://tutorial.math.lamar.edu/terms.aspx
  • 8/13/2019 CalcI ProofsLimits

    4/17

    Calculus I

    2007 Paul Dawkins 4 http://tutorial.math.lamar.edu/terms.aspx

    The limit evaluation is a special case of 7(with 0c= ) which we just proved Therefore we know

    1is true for 0c= and so we can assume that 0c for the remainder of this proof.

    Let 0e> then because ( )limx a

    f x K

    = by the definition of the limit there is a 1 0d > such that,

    ( ) 1whenever 0f x K x ace d- < < - <

    Now choose 1d d= and we need to show that

    ( ) whenever 0cf x cK x ae d- < < - <

    and well be done. So, assume that 0 x a d< - < and then,

    ( ) ( )cf x cK c f x K cc

    ee- = - < =

    Proof of 2

    Note that well need something called the triangle inequalityin this proof. The triangle

    inequality states that,

    a b a b+ +

    Heres the actual proof.

    Well be doing this proof in two parts. First lets prove ( ) ( )limx a

    f x g x K L

    + = + .

    Let 0e> then because ( )limx a

    f x K

    = and ( )limx a

    x L

    = there is a 1 0d > and a 2 0d > such

    that,

    ( )

    ( )

    1

    2

    whenever 02

    whenever 02

    f x K x a

    g x L x a

    ed

    ed

    - < < - and a 2 0d > such that,

    ( )( )( )( )

    1

    2

    0 whenever 0

    0 whenever 0

    f x K x a

    g x L x a

    e d

    e d

    - - < < - and let d e= . Then, if 0 x a d e< - < = we have,

    ( )f x a x a d e- = - < =

    So, weve proved that limx a

    x a

    = .

    http://tutorial.math.lamar.edu/terms.aspxhttp://tutorial.math.lamar.edu/terms.aspx
  • 8/13/2019 CalcI ProofsLimits

    9/17

    Calculus I

    2007 Paul Dawkins 9 http://tutorial.math.lamar.edu/terms.aspx

    Proof of 9

    This is just a special case of property 5with ( )f x x= and so we wont prove it here.

    Facts, Infinite Limits

    Given the functions ( )f x and ( )g x suppose we have,

    ( ) ( )lim limx c x c

    f x g x L

    = =

    for some real numbers candL. Then,

    1. ( ) ( )limx c

    f x g x

    =

    2.If 0L> then ( ) ( )limx c f x g x =

    3.If 0L< then ( ) ( )limx c

    f x g x

    = -

    4.( )( )

    lim 0x c

    g x

    f x=

    Partial Proof of 1

    We will prove ( ) ( )limx c f x g x + = here. The proof of ( ) ( )limx c f x g x - = is nearlyidentical and is left to the reader.

    Let 0M> then because we know ( )limx c

    f x

    = there exists a 1 0d > such that if

    10 x c d< - < we have,

    ( ) 1f x M L> - +

    Also, because we know ( )limx c

    x L

    = there exists a 2 0d > such that if 20 x c d< - < we

    have,

    ( ) ( ) ( )0 1 1 1 1 1x L g x L L g x L< - < - < - < - < < +

    Now, let { }1 2min ,d d d= and so if 0 x c d< - < we know from the above statements that wewill have both,

    ( ) ( )1 1f x M L g x L> - + > -

    http://tutorial.math.lamar.edu/terms.aspxhttp://tutorial.math.lamar.edu/terms.aspx
  • 8/13/2019 CalcI ProofsLimits

    10/17

    Calculus I

    2007 Paul Dawkins 10 http://tutorial.math.lamar.edu/terms.aspx

    This gives us,

    ( ) ( )

    ( ) ( )

    1 1f x g x M L L

    M f x g x M

    + > - + + -

    = + >

    So, weve proved that ( ) ( )limx c

    f x g x

    + = .

    Proof of 2

    Let 0M> then because we know ( )limx c

    f x

    = there exists a 1 0d > such that if

    10 x c d< - < we have,

    ( ) 2MLf x >

    Also, because we know ( )limx c

    x L

    = there exists a 2 0d > such that if 20 x c d< - < we

    have,

    ( ) ( ) ( ) 32 2 2 2 20 LL L L Lg x L g x L g x< - < - < - < < <

    Note that because we know that 0L> choosing2

    L in the first inequality above is a valid choice

    because it will also be positive as required by the definition of the limit.

    Now, let { }1 2min ,d d d= and so if 0 x c d< - < we know from the above statements that wewill have both,

    ( ) ( )2 2M LL

    f x g x> >

    This gives us,

    ( ) ( ) ( ) ( )

    ( ) ( )

    22

    M LL

    f x g x

    f x g x M

    >

    = >

    So, weve proved that ( ) ( )limx c

    f x g x

    = .

    Proof of 3

    Let 0M> then because we know ( )limx c

    f x

    = there exists a 1 0d > such that if

    http://tutorial.math.lamar.edu/terms.aspxhttp://tutorial.math.lamar.edu/terms.aspx
  • 8/13/2019 CalcI ProofsLimits

    11/17

    Calculus I

    2007 Paul Dawkins 11 http://tutorial.math.lamar.edu/terms.aspx

    10 x c d< - < we have,

    ( ) 2MLf x ->

    Note that because 0L< in the case we will have 2 0ML

    - > here.

    Next, because we know ( )limx c

    x L

    = there exists a 2 0d > such that if 20 x c d< - < we

    have,

    ( ) ( ) ( )32 2 2 2 20 LL L L Lg x L g x L g x< - < - < - < - < <

    Again, because we know that 0L< we will have2

    0L- > . Also, for reasons that will shortly be

    apparent, multiply the final inequality by a minus sign to get,

    ( ) 32 2LL g x- < - < -

    Now, let { }1 2min ,d d d= and so if 0 x c d< - < we know from the above statements that wewill have both,

    ( ) ( )2 2M L

    Lf x g x-> - > -

    This gives us,

    ( ) ( ) ( ) ( )

    ( )( )

    ( ) ( )

    22

    M LL

    f x g x f x g x

    f x g x M

    - = -

    > - -

    = - >

    This may seem to not be what we needed however multiplying this by a minus sign gives,

    ( ) ( )f x g x M< -

    and because we originally chose 0M> we have now proven that ( ) ( )limx c

    f x g x

    = - .

    Proof of 4

    Well need to do this in three cases. Lets start with the easiest case.

    Case 1 : 0L=

    Let 0e> then because we know ( )limx c

    f x

    = there exists a 1 0d > such that if

    10 x c d< - < we have,

    ( ) 1 0f xe

    > >

    http://tutorial.math.lamar.edu/terms.aspxhttp://tutorial.math.lamar.edu/terms.aspx
  • 8/13/2019 CalcI ProofsLimits

    12/17

    Calculus I

    2007 Paul Dawkins 12 http://tutorial.math.lamar.edu/terms.aspx

    Next, because we know ( )lim 0x c

    g x

    = there exists a 2 0d > such that if 20 x c d< - < we

    have,

    ( )0 g x e< <

    Now, let { }1 2min ,d d d= and so if 0 x c d< - < we know from the above statements that wewill have both,

    ( ) ( )1f x g xe

    e> <

    This gives us,

    ( )

    ( )

    ( )

    ( ) ( ) 1g xg x

    f x f x f x

    e ee

    e

    = < < =

    In the second step we could remove the absolute value bars from ( )f x because we know it is

    positive.

    So, we proved that( )( )

    lim 0x c

    g x

    f x= if 0L= .

    Case 2 : 0L>

    Let 0e> then because we know ( )limx c

    f x

    = there exists a 1 0d > such that if

    10 x c d< - < we have,

    ( ) 32 0Lf xe

    > >

    Next, because we know ( )limx c

    x L

    = there exists a 2 0d > such that if 20 x c d< - < we

    have,

    ( ) ( ) ( ) 32 2 2 2 20 LL L L Lg x L g x L g x< - < - < - < < <

    Also, because we are assuming that 0L> it is safe to assume that for2

    0 x c d< - < we have

    ( ) 0g x > .

    Now, let { }1 2min ,d d d= and so if 0 x c d< - < we know from the above statements that wewill have both,

    ( ) ( )3 32 2L Lf x g xe

    > <

    http://tutorial.math.lamar.edu/terms.aspxhttp://tutorial.math.lamar.edu/terms.aspx
  • 8/13/2019 CalcI ProofsLimits

    13/17

    Calculus I

    2007 Paul Dawkins 13 http://tutorial.math.lamar.edu/terms.aspx

    This gives us,

    ( )

    ( )

    ( )

    ( ) ( )

    3 32 2

    32

    L Lg x g x

    Lf x f x f xe

    e

    = < < =

    In the second step we could remove the absolute value bars because we know or can safely

    assume (as noted above) that both functions were positive.

    So, we proved that( )( )

    lim 0x c

    g x

    f x= if 0L> .

    Case 3 : 0L< .

    Let 0e> then because we know ( )limx c

    f x

    = there exists a 1 0d > such that if

    10 x c d< - < we have,

    ( ) 32 0Lf xe

    -> >

    Next, because we know ( )limx c

    x L

    = there exists a 2 0d > such that if 20 x c d< - < we

    have,

    ( ) ( ) ( )32 2 2 2 20 LL L L Lg x L g x L g x< - < - < - < - < <

    Next, multiply this be a negative sign to get,

    ( ) 32 2LL g x- < - < -

    Also, because we are assuming that 0L< it is safe to assume that for 20 x c d< - < we have

    ( ) 0g x < .

    Now, let { }1 2min ,d d d= and so if 0 x c d< - < we know from the above statements that wewill have both,

    ( ) ( )3 32 2L Lf x g xe

    - -> - <

    This gives us,

    ( )

    ( )

    ( )

    ( ) ( )

    3 32 2

    32

    L Lg x g x

    Lf x f x f xe

    e

    - --= < < =

    -

    In the second step we could remove the absolute value bars by adding in the negative because we

    http://tutorial.math.lamar.edu/terms.aspxhttp://tutorial.math.lamar.edu/terms.aspx
  • 8/13/2019 CalcI ProofsLimits

    14/17

    Calculus I

    2007 Paul Dawkins 14 http://tutorial.math.lamar.edu/terms.aspx

    know that ( ) 0f x > and can safely assume that ( ) 0g x < (as noted above).

    So, we proved that( )( )

    lim 0x c

    g x

    f x= if 0L< .

    Fact 1, Limits At Infinity, Part 1

    1.If ris a positive rational number and cis any real number then,

    lim 0rx

    c

    =

    2.If ris a positive rational number, cis any real number andxr

    is defined for 0x< then,

    lim 0rx

    c

    x-=

    Proof of 1

    This is actually a fairly simple proof but well need to do three separate cases.

    Case 1 :Assume that 0c> . Next, let 0e> and define rc

    Me

    = . Note that because cand e

    are both positive we know that this root will exist. Now, assume that we have rc

    x Me> = .

    Give this assumption we have,

    get rid of the root

    rearrange things a little

    0 everything is positive so we can add absolute value bars

    r

    r

    r

    r

    cx

    cx

    c

    x

    c

    x

    e

    e

    e

    e

    >

    >

    . Then

    using Case 1 and the fact that we can factor constants out of a limit we get,

    lim lim lim 0 0r r rx x x

    c k k

    x x x

    -= = - = - =

    Proof of 2

    This is very similar to the proof of 1so well just do the first case (as its the hardest) and leave

    the other two cases up to you to prove.

    Case 1 :Assume that 0c> . Next, let 0e> and define rc

    Ne

    = - . Note that because cand e

    are both positive we know that this root will exist. Now, assume that we have rc

    x Ne

    < = - .

    Note that this assumption also tells us thatxwill be negative. Give this assumption we have,

    take absolute value of both sides

    get rid of the root

    rearrange things a little and use the fact that 0

    0 rewrite things a little

    r

    r

    r

    r

    r

    cx

    cx

    cx

    c

    x

    c

    x

    e

    e

    e

    e e e

    e

    < -

    >

    >

    < = >

    - weve proven that lim 0rx

    c

    x= . Note that the main difference here is that we

    need to take the absolute value first to deal with the minus sign. Because both sides are negative

    we know that when we take the absolute value of both sides the direction of the inequality will

    have to switch as well.

    Case 2, Case 3 :As noted above these are identical to the proof of the corresponding cases in the

    http://tutorial.math.lamar.edu/terms.aspxhttp://tutorial.math.lamar.edu/terms.aspx
  • 8/13/2019 CalcI ProofsLimits

    16/17

    Calculus I

    2007 Paul Dawkins 16 http://tutorial.math.lamar.edu/terms.aspx

    first proof and so are omitted here.

    Fact 2, Limits At Infinity, Part I

    If ( ) 11 1 0n n

    n np x a x a x a x a-

    -= + + + +L is a polynomial of degree n(i.e. 0na ) then,

    ( ) ( )lim lim lim limn nn nx x x x

    p x a x p x a x - -

    = =

    Proof of ( )lim lim nnx x

    p x a x

    =

    Were going to prove this in an identical fashion to the problems that we worked in this section

    involving polynomials. Well first factor outn

    na x from the polynomial and then make a giant

    use of Fact 1(which we just proved above) and the basic properties of limits.

    ( ) ( )11 1 0

    1 01

    1

    lim lim

    lim 1

    n n

    n nx x

    n nn n nx

    n n n

    p x a x a x a x a

    a aaa x

    a x a x a x

    --

    --

    = + + + +

    = + + + +

    L

    L

    Now, clearly the limit of the second term is one and the limit of the first term will be either or

    - depending upon the sign of na . Therefore by the Factsfrom the Infinite Limits section we

    can see that the limit of the whole polynomial will be the same as the limit of the first term or,

    ( ) ( )

    1

    1 1 0lim lim limn n n

    n n nx x x

    x a x a x a x a a x--

    = + + + + =L

    Proof of ( )lim lim nnx x

    p x a x- -

    =

    The proof of this part is literally identical to the proof of the first part, with the exception that all

    s are changed to - , and so is omitted here.

    Fact 2, Continuity

    If ( )f x is continuous at b= and ( )limx a

    g x b

    = then,

    ( )( ) ( )( ) ( )lim limx a x a

    f g x f g x f b

    = =

    http://tutorial.math.lamar.edu/terms.aspxhttp://tutorial.math.lamar.edu/terms.aspx
  • 8/13/2019 CalcI ProofsLimits

    17/17

    Calculus I

    2007 Paul Dawkins 17 http://tutorial.math.lamar.edu/terms.aspx

    Proof

    Let 0e> then we need to show that there is a 0d> such that,

    ( )( ) ( ) whenever 0f g x f b x ae d- < < - <

    Lets start with the fact that ( )f x is continuous at b= . Recall that this means that

    ( ) ( )limx b

    f x f b

    = and so there must be a 1 0d > so that,

    ( ) ( ) 1whenever 0f x f b x be d- < < - <

    Now, lets recall that ( )limx a

    x b

    = . This means that there must be a 0d> so that,

    ( ) 1 whenever 0g x b x ad d- < < - <

    But all this means that were done.

    Lets summarize up. First assume that 0 x a d< - < . This then tells us that,

    ( ) 1g x b d- <

    But, we also know that if 10 x b d< - < then we must also have ( ) ( )f x f b e- < . What this

    is telling us is that if a number is within a distance of 1d of bthen we can plug that number into

    ( )f x and well be within a distance of e of ( )f b .

    So, ( ) 1g x b d- < is telling us that ( )x is within a distance of 1d of band so if we plug it

    into ( )f x well get,

    ( )( ) ( )f g x f b e- < and this is exactly what we wanted to show.

    http://tutorial.math.lamar.edu/terms.aspxhttp://tutorial.math.lamar.edu/terms.aspx