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Evaluating limits

  1. Oct 7, 2011 #1
    1. The problem statement, all variables and given/known data
    [tex]\lim_{x→1}(frac(x)+frac(-x))[/tex]

    frac() is the fractional part function.

    2. Relevant equations



    3. The attempt at a solution
    I tried it by graphing.
    Firstly i drew the graph of frac(x) and then frac(-x). I added them. The resulting graph was constant at 1 but discontinuous at integers. So the answer to the question is 1 but wolfram gave me the answer to be 0. :(
    I don't understand where i am wrong?
     
  2. jcsd
  3. Oct 7, 2011 #2

    SammyS

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    If 0 < x < 1 then -1 < -x < 0 and frac(x) = x and frac (-x) = 1 - x . So, what is frac(x) + frac(-x) ?

    Do similar for 1 < x < 2 .
     
  4. Oct 8, 2011 #3
    When 0<x<1, frac(x)+frac(-x)=1.
    When 1<x<2, then -2<-x<-1.
    frac(x)=x-1 and frac(-x)=2-x.
    Therefore, frac(x)+frac(-x)=1. :)

    But how would it help me to solve the problem?
     
  5. Oct 8, 2011 #4

    SammyS

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    When find the limit as x → 1, there is no need to be concerned about the value of the function at x=1, just be concerned about x approaching 1 from each side , and whether or not the function approaches the same value from either side.

    Although, I am having second thoughts about whether I have frac(x) correct for negative numbers.
     
  6. Oct 8, 2011 #5
    Thanks for your help Sammy. :)
    I have figured it out where i was wrong. I did a mistake in graphing and adding them.
    Now my answer comes to be correct. :)
     
  7. Oct 8, 2011 #6
    Do you think that this graph is correct? For frac(x)
    Wolfram gives me this:-
    [PLAIN]http://www3.wolframalpha.com/Calculate/MSP/MSP115319hcd32f187c18df000050fghaaf173h217b?MSPStoreType=image/gif&s=25&w=319&h=133&cdf=Coordinates&cdf=Tooltips [Broken]
     
    Last edited by a moderator: May 5, 2017
  8. Oct 8, 2011 #7

    SammyS

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    Last edited by a moderator: May 5, 2017
  9. Oct 8, 2011 #8
    I don't get you. frac(x) means the fractional part.
    One question:-
    If the number is negative, for instance, -1.30, then the fractional part is 0.70.
    Right..?

    So why wolfram gives me that type of graph? At the negative side too it should go upwards.

    Erm...what?
     
  10. Oct 9, 2011 #9
    Bump :(
     
  11. Oct 9, 2011 #10

    ehild

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  12. Oct 9, 2011 #11
    So is there no fractional part of -1.30?
     
  13. Oct 9, 2011 #12

    SammyS

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    There certainly is a fractional part of -1.30.

    Is it 0.30? ... or 0.70 ? ... or -0.30 ?

    I'm quite sure that for all of them, [itex]\lim_{x→1}(\text{frac}(x)+\text{frac}(-x))[/itex] exists, although the limit is not the same for them all.
     
    Last edited: Oct 9, 2011
  14. Oct 9, 2011 #13
    I think its 0.70. :)
     
  15. Oct 9, 2011 #14

    ehild

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    You could see that different people define the fractional part of negative numbers in different ways. Look at your notes what definition of fractional part was taught to you.

    ehild
     
  16. Oct 9, 2011 #15
    My notes says that if we have a negative number like -1.80, then the fractional part would be the (integer part +1)+(given negative number) i.e 0.20.
    Here the integer part is 1 and 1+1=2 and 2-1.80=0.20.
     
  17. Oct 9, 2011 #16

    ehild

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    With the definition you learnt, you get always a positive number for the integer part. It is then the same SammyS suggested.
    frac(A)=A-floor(A)
    where floor(A) is the highest integer which is less than A.
    You need to find the limit at x=1. Consider the cases x-->1+0 and when x-->1-0. It is enough to consider such x values for which |x-1|<1

    You were right, the result is 1 with this definition.

    ehild
     
  18. Oct 9, 2011 #17
    Thanks for the explanation ehild but how do you get this:- |x-1|<1 ?
     
  19. Oct 9, 2011 #18

    ehild

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    Finding the limit of a function at A, it is enough to investigate such x values which are close to A. A=1 in your problem, so it is enough to investigate the function for such x values which are at distance less than 1 from A=1, that is 0< x <2, that is -1<x-1<1, which is equivalent to |x-1|<1.

    ehild
     
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