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Thread: Limits

  1. #1
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    Limits

    Can someone please check this? I have my doubts, and I need to be solid at this stuff.

    Prove lim_(x -> 3-) = 2(x+3)/|x-3| = -2 = L

    Where that limit reads as x approaches 3 from the left.
    I'll use f(x) = 2(x + 3)/|x-3|

    Note,
    |f(x) - L | = | 2(x + 3)/|x - 3| + 2 | = |2(x + 3) + 2(x - 3) | / |x - 3| = 2 |(x - 3) + 2|x - 3||/ |x - 3|

    But if |x - 3| < 1, then -1 < x - 3 < 1 and x - 3 < 1

    So, we are left with,
    |f(x) - L| < 2 | 1 + 2|x - 3|| / 1 = 2 + 4|x - 3|

    Thus, let e > 0 be arbitrary. Let d = min{1, (e - 2) / 4}. Then if 3 - d < x < 3, then, |x - 3| < (e - 2)/4

    |f(x) - L| = 2 |(x - 3) + 2|x - 3||/ |x - 3| < 2 + 4|x - 3| < 2 + (e - 2)/4 = e

    QED

  2. #2
    Honeybadger Moderator cw823's Avatar
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    Quote Originally Posted by Frodo Baggins
    Can someone please check this? I have my doubts, and I need to be solid at this stuff.

    Prove lim_(x -> 3-) = 2(x+3)/|x-3| = -2 = L

    Where that limit reads as x approaches 3 from the left.
    I'll use f(x) = 2(x + 3)/|x-3|

    Note,
    |f(x) - L | = | 2(x + 3)/|x - 3| + 2 | = |2(x + 3) + 2(x - 3) | / |x - 3| = 2 |(x - 3) + 2|x - 3||/ |x - 3|

    But if |x - 3| < 1, then -1 < x - 3 < 1 and x - 3 < 1

    So, we are left with,
    |f(x) - L| < 2 | 1 + 2|x - 3|| / 1 = 2 + 4|x - 3|

    Thus, let e > 0 be arbitrary. Let d = min{1, (e - 2) / 4}. Then if 3 - d < x < 3, then, |x - 3| < (e - 2)/4

    |f(x) - L| = 2 |(x - 3) + 2|x - 3||/ |x - 3| < 2 + 4|x - 3| < 2 + (e - 2)/4 = e

    QED

    It appears to be very boring.
    I can explain it to you, but I cannot understand it for you

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    *Walks into thread*

    *Sees very very very very confusing lines and numbers*

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    Member deRusett's Avatar
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    Wantto write it out and post an image of it, I'm not making sence of it, because I have never seen Limits done typed on single lines, so its a bit confusing,


    side note.
    Want a good textbook for this and related items
    Pick up
    Essentials of Technical Mathematics with Calculus second Edition
    by Richard Paul/M.Leonard Shaevel
    ISBN 01-13-289091-7

    its an older book it was my First Year Math text book, but has become my book of referance when ever I need a refersher on this stuff
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  5. #5
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    Quote Originally Posted by cw823
    It appears to be very boring.
    You would appear to be right
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  6. #6
    Member deRusett's Avatar
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    thats why they discovered Differentiation

    because limits are very boring, but you need to understand them before you can understand what a Derivative actually is.

    then you get to forget how to take limits thats a happy day
    Success is..
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  7. #7
    Member shard's Avatar
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    *walks around lost and feeling stupid*
    *reads calculus book and head explodes*
    *whaaaaaaaaaaaaaaaaaaa he cries me stupid*
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    Inactive Pokémon Moderator JigPu's Avatar
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    Woah... Way too many parentathes and absolute value signs. My head is officially spinning now. I think the lack of knowing what's being divided by what is really throwing me off. Any way you could put up a copy that's not on one line?


    Quote Originally Posted by deRusett
    thats why they discovered Differentiation

    because limits are very boring, but you need to understand them before you can understand what a Derivative actually is.

    then you get to forget how to take limits thats a happy day
    The day I forgot how to take a limit (ok.. so it hasn't happened yet, but I've been close) was the worst day of my life! Reducing calculus to just sets of rules and regulations denies it of the mathematical beauty that lies underneath. While I think limits are evil (I prefer just plugging infinity/infintessimal straight into the equation ), the idea behind them is interesting and thought provoking.



    ...though the rules and regulations do make it a LOT easier to blaze through the zillion problems assigned!
    JigPu
    Last edited by JigPu; 07-15-04 at 12:24 AM.
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  9. #9
    Member Drec's Avatar
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    *pretends like he didn't see this thread*

    *walks away slowly*

  10. #10
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    wow do i got somthing to look forward to in a few years! Algebra 2 next year :P

  11. #11
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    i can't even read that thing at all, and my head hurts just by looking at it!!!!

    Shard, we are in the same boat. me is very stupid too.

    This hurts my head to much, i think i'm going to go look at some Pron now....less thinking required.
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  12. #12
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    Quote Originally Posted by deRusett
    Wantto write it out and post an image of it, I'm not making sence of it, because I have never seen Limits done typed on single lines, so its a bit confusing,
    Attached Images Attached Images

  13. #13
    Member germanjulian's Avatar
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    ***** whoo what*******
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  14. #14
    Just Freeze It
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    I don't have a clue, but I give anyone who understands this some mad props.

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    from what i remember from earlier this year in calc I, and some of calc II, that seems to be done more in depth than i remember doing epsilon and delta proofs when i had the unpleasant experience of doing them

    as i recall, you really don't need to know how to do these that much, just remeber the limits and how to take them and your set

    i get the limit to be 1.2 x 10^14th

    that is on an ti-86 doing it graphically and via table, and by hand getting something big

    i'd say your limit should be either infinity or something along the lines of the above
    Last edited by Roboman101; 07-15-04 at 12:19 PM. Reason: got the limit of f(x)

  16. #16
    Member deRusett's Avatar
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    question1, who said the Lim as X aproches 3 from the left is -2? I get it to be 12 from the equation


    questio2
    Why did you bother with this line
    "But if |x-3| <1, then -1 < x -3 < 1 and x+3 < 7

    X+3 can not equal 7 because we ignore the function after X=3

    as we aproche 3 from the left we get x=2.99999999999999
    which makes |x-3| = 0.0000000000001 ( I didn't count the 9's or zeros just held my finger down )


    I must be missing something because unless x is aproching -3 from the left there is no way it can be a negative value

    I'd go into more detail but I have 5 minutes to get ready for work, and I don't want to break out a pen and paper
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    At age 60 .... success is .... having sex.
    At age 70 .... success is .... having a driving licence.
    At age 75 .... success is .... having friends.
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    At Any Age ... success is ... Folding for Team 32

  17. #17
    Macaddict macky's Avatar
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    When did they start using letters in math?

  18. #18
    Member Docta_Z's Avatar
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    You are making life hard for yourself on the third line in as deRusett says.

    No need to rearrange it to test at 1.

    But looking back - is your whole first line given in the question?

    I don't think it equals 2...

    I did this stuff 2 years ago, now all my calculus is completely different. I'll think about it for a while though.

    Quote Originally Posted by macky
    When did they start using letters in math?
    Grade 8 for me...

  19. #19
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    I should probably know the answer to this.

    I should probably be able to give some input.

    I should have actually gone to more than half my maths lectures last year.

    Damn.
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  20. #20
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    Quote Originally Posted by deRusett
    question1, who said the Lim as X aproches 3 from the left is -2? I get it to be 12 from the equation


    questio2
    Why did you bother with this line
    "But if |x-3| <1, then -1 < x -3 < 1 and x+3 < 7

    X+3 can not equal 7 because we ignore the function after X=3

    as we aproche 3 from the left we get x=2.99999999999999
    which makes |x-3| = 0.0000000000001 ( I didn't count the 9's or zeros just held my finger down )


    I must be missing something because unless x is aproching -3 from the left there is no way it can be a negative value

    I'd go into more detail but I have 5 minutes to get ready for work, and I don't want to break out a pen and paper
    I didn't mean to waste your time. The equation is written wrong. I'll fix some stuff up

    questio2
    Why did you bother with this line
    "But if |x-3| <1, then -1 < x -3 < 1 and x+3 < 7

    X+3 can not equal 7 because we ignore the function after X=3
    It's a bit obsolete now that the equation is wrong. But I never said it was equal to 7.
    -1 < x - 3 < 1 => -1 + 6 < x + 3 < 1 + 6 => 5 < x + 3 < 7

    We don't worry about the lower bound since -1 < x - 3 < 1 => x + 3 < 7

    The reason why was to bound the factor in the expression by a number.

    Okay, another try. I've been feeling under the weather, but I hope I didn't commit any serious errors in this one either.
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