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Limits that do not exist?

  1. Apr 12, 2005 #1
    Limits that do not exist??

    How do i show that

    lim h->0 (sqrt|(a*b)|*|h|)/h does not exist at 0

    so far i have
    lim h->0+ |h|/h =1
    and
    lim h->0- |h|/h =-1

    does sqrt|(a*b)| go to +/- inifinity?
    because as h gets smaller, sqrt|(a*b)| goes infinity.

    so
    lim h->0 (sqrt|(a*b)|*|h|)/h = +/- inifinity
    Callisto
     
  2. jcsd
  3. Apr 12, 2005 #2

    Hurkyl

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    Why doesn't √|a*b| go to √|a*b| as h goes to ∞?
     
  4. Apr 12, 2005 #3
    as h goes to infinity, sqrt|(a*b)| goes to zero
     
  5. Apr 12, 2005 #4

    dextercioby

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    Why?Is there something that u're hiding,like a dependence of "a" & "b" of "h"...?:bugeye:


    Daniel.
     
  6. Apr 12, 2005 #5
    OK my question is,

    does sqrt|(a*b)| go to +/- inifinity?
    because as h gets smaller, sqrt|(a*b)| goes to infinity?

    I thought that as h->0 then sqrt|(a*b)| -> infinity.
    am i not computing the limit correctly?
    please forgive my confusion.

    Callisto
     
  7. Apr 12, 2005 #6
    Are a and b constants?
     
  8. Apr 12, 2005 #7
    as long as a&b do not equal zero
     
  9. Apr 12, 2005 #8

    Hurkyl

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    What is [itex]\lim_{h \rightarrow \infty} 1[/itex]?
     
  10. Apr 12, 2005 #9
    I think perhaps you may not understand the terms you are using.
    When you say
    "as h->0 then sqrt|(a*b)| -> infinity"
    That means given any number N>0 there exists an d>0 such that if |h|<d then sqrt|(a*b)|>N.

    For example |1/h| -> infinity as h -> 0 because given an N>0 I can
    take d= 1/2N so if 0<|h|<d then |1/h| > 1/d =2N >N.

    So can you do that with sqrt|(a*b)|? If I gave you N=100 what would you give me for d? Why for that d that you've given me can you guarentee that sqrt|(a*b)| is greater than 100? What if I made N=1,000,000 instead then what would d be?
     
  11. Apr 12, 2005 #10
    You're confusing me.

    is this the answer to my question?

    callisto
     
  12. Apr 13, 2005 #11
    If f(x,y) = sqrt|a*b|, the partial derivatives at Dx(0,0) ,Dy(0,0) exist. i am trying to show that the directional derivatives in all other directions fail to exist.

    Callisto
     
  13. Apr 13, 2005 #12
    If f(x,y) = sqrt|a*b|, then the directional derivatives also exist. They are all 0. Because sqrt|a*b| is a constant. It doesn't change when x or y do.

    And if you take a limit as h -> 0, then sqrt|a*b| -> sqrt|a*b|. Because sqrt|a*b| is a constant. It doesn't change when h does.
     
  14. Apr 13, 2005 #13

    Zurtex

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    [tex]
    \lim_{\substack{a\rightarrow \infty\\b\rightarrow \infty\\c\rightarrow \infty\\x\rightarrow \infty\\y\rightarrow \infty\\z\rightarrow \infty\\\alpha\rightarrow \infty\\\beta\rightarrow \infty\\\gamma\rightarrow \infty\\\delta\rightarrow \infty}} \sqrt{4 \cdot 3} = \sqrt{4 \cdot 3}[/tex]

    :wink:
     
  15. Apr 13, 2005 #14
    You might want to check that.
     
  16. Apr 13, 2005 #15

    matt grime

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    ONly if you have failed to disclose what a and b are. you were asked but haven't bothered to say.
     
  17. Apr 13, 2005 #16
    Means a does not equal zero and b does not equal zero.

    From the definition of the directional derivative

    ie: The directional derivative of f at (x,y) in the direction of the unit vector
    u=<a,b> is

    Duf(x,y)=lim h->0 f(x+ha,y+hb)-f(x,y)/h

    IF THIS LIMIT EXISTS.
     
  18. Apr 13, 2005 #17

    matt grime

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    And that jsut confirms that you're confused abtou these things. partly a problem of the notation.

    for instance, writing f(x,y) = sqrt(ab) implies a and b are constants, so that taking partial differentials kills it.

    anyway, the poitn was jsut fix a and b. sqrt(ab) tends to sqrt(ab) as h tends to zero as it is independent of it.
     
  19. Apr 13, 2005 #18
    The confusion has only arisen as a result of the misinterpretation of my question on you and your colleagues behalf.

    since the sqrt|a*b| does not equal zero,

    and the limit of |h|/h = 1 as h tends to zero from the right and the limit of |h|/h=-1 as h tends to zero from the left.

    then the limit (sqrt|a*b|*|h|)/h as h tends to zero is +/- sqrt|a*b|.

    therefore the directional derivatives in all other directions fail to exist

    I must add i am not a mathematician so i dont know how to make it anymore clearer.

    callisto :!!)
     
  20. Apr 13, 2005 #19

    matt grime

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    We are correctly intepreting what you wrote in and of itself. However what you wrote was confused.

    Even the quote you use from yourself implies that f is a constant function, when it isn't.


    But the very first post contained some errors, that have been corrected. You do understand that the limt as h tends to zero of sqrt(|ab|) is sqrt(|ab|)?

    HOwever, you have got the gitst of why the derivatives do not exist.
     
  21. Apr 13, 2005 #20
    So am i correct?

    the limit as h tends to zero of sqrt(|ab|) is sqrt(|ab|) since sqrt(|ab|) is independent of h

    I am an undergraduate forgive me for the confusion.
     
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