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Finding f(x) given limit

  1. Sep 12, 2013 #1
    1. The problem statement, all variables and given/known data

    lin x-> 1
    f(x)-8 / x-1 = 10
    find f(x)

    2. Relevant equations



    3. The attempt at a solution
    i dont know where to start on finding f(x), i assume it includes (x-1) to eliminate the 0 denominator, i need some hints
     
  2. jcsd
  3. Sep 12, 2013 #2
    Please include parenthesis. Is it f(x)-8/x-1 or is it (f(x)-8)/(x-1). I'm assuming the latter. Also what is the given limit?

    Edit, I see the limit now. Your notation was confusing. To answer your question, you know you can multiply limits, right? So lim x-> 1 (f(x)-8)/(x-1)=10, then you can multiply by a limit that you KNOW exists.
     
  4. Sep 12, 2013 #3
    yeah sorry, it is indeed the latter
    lim x->1 (f(x)-8)/(x-1) = 10
     
    Last edited: Sep 12, 2013
  5. Sep 12, 2013 #4
    But even then, you can only know what the limit is at one point. You know nothing of the function. Is there more information?
     
  6. Sep 12, 2013 #5
    no, that was all that given; find f(x) at lim x-> 1
    any idea where i should start, i did trial and error on lim x->1, f(x) = (x-1)^n, but i dont think that worked
     
  7. Sep 12, 2013 #6
    Is the problem

    find f(x)

    or is the problem

    find lim x-> 1 f(x)
     
  8. Sep 12, 2013 #7
    its find lim x-> 1 f(x)
     
  9. Sep 12, 2013 #8
    So you can multiply limits, right? So you can multiply the given limit by something. What something would make it easier?
     
  10. Sep 12, 2013 #9
    i dont comprehend...
    do you mean bringing (x-1) to above by inversing it? i tried that.
    the whole f(x) within another function boggles me.
     
  11. Sep 12, 2013 #10
    That's not what I meant at all. You can create an entirely NEW limit, then multiply it with your given limit, so long as the limit you create exists. What would simplify things?
     
  12. Sep 12, 2013 #11
    so i.e.
    lim x->1 -8/(x-1) * f(x) where i can make up a new limit for f(x) which will also need (x-1)? do i need to get rid of lim x->1 (x-1) on the denominator with the f(x)?
     
    Last edited: Sep 12, 2013
  13. Sep 12, 2013 #12
    "Make up a new limit for f(x)". No, you cannot "make up" a limit for f(x). That limit just is. But you CAN create a new function, g(x), such that lim x-> 1 g(x)=c, some constant (you create g(x), and from this you know what the constant is)

    So then lim x-> 1 (f(x)-8)*g(x)/(x-1)=10*c

    What g(x) would simplify this?
     
  14. Sep 12, 2013 #13
    would this also address the 0 denominator given lim x->1 for (x-1)?
    f(x) * g(x) = 10x-2
    would this be on the right path?
     
  15. Sep 12, 2013 #14
    Huh? Would what address what? I think you might be overthinking this.
     
  16. Sep 12, 2013 #15
    I have no idea where that equation came from.
     
  17. Sep 12, 2013 #16
    does (f(x) - 8) / (x - 1) = 10 need to hold true for lim x->1? i thought id have to factor out the x-1 since it'd give a 0 denominator
     
  18. Sep 12, 2013 #17
    Isn't that the limit you're given?
     
  19. Sep 12, 2013 #18
    Also, you can't factor anything. x-1 is irreducible.
     
  20. Sep 12, 2013 #19
    i'm so lost.
    does that mean f(x) can just be (x-1) to factor out the denominator?
     
  21. Sep 12, 2013 #20
    There is no point in this problem where you will factor anything. Also, there is no point in the problem where you will ever know what f(x) is.

    Like I said, you need to multiply the limit you are given by another limit that you create.
     
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