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Finding a limit through algebra

  1. Jun 24, 2004 #1
    Hi, I'm having a bit of a problem with a certain problem assigned to me...
    I must find the limit of (lnx)/(x-1), as x approaches 1. However, I may not use L'Hopital's rule; I must stick to algebra methods to solve the problem, and apparently, the definition of limit is unnecessary.
    Thanks for the help.




    Note: as x approaches 0 has been changed to as x approaches 1
     
    Last edited: Jun 26, 2004
  2. jcsd
  3. Jun 24, 2004 #2
    This is actually easy if you forget everything about calc. for a sec. Looking at this equation simply, the numerator goes to negative infinity as x goes to zero and the denominator goes to a negative infitesimal number (-.00000000000...1). This fraction goes to positive infinity. Using l'Hôpital's Rule, f'(ln x) = 1/x and f'(x-1) = 1 therefore this limit is the same as 1/x/1= 1/x. I'm not sure if this is the ALGEBRAIC way to do this, but it certainly is a logical approach.
     
  4. Jun 24, 2004 #3
    why not lift the whole to e^(Inx/(x-1) = x^(1/(x-1)), which is roughly 1/x for small x. Thus In( infinity) equals infinity is a good guess for the answer.
     
  5. Jun 25, 2004 #4
    Here is a hint:
    Taylor series (from Schaum Mathematical Handbook):
    ln(x) = 2[(x-1)/(x+1) + 1/3*(x-1)^2/(x+1)^2 + ...)
     
  6. Jun 25, 2004 #5

    HallsofIvy

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    L'Hopital's rule doesn't apply anyway- this is not of the form 0/0 or inf/inf, etc. As napoleonmax pointed out, it is of the form -inf/(-0) which means that the limit is +infinity (which is just a way of saying that it has no limit).
     
  7. Jun 26, 2004 #6
    type

    The original limit problem was mistyped. Please note the change.
     
  8. Jun 26, 2004 #7

    HallsofIvy

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    I'm not sure what you mean by "algebra methods". I would write ln(x) as a Taylor's series about x= 1.
     
  9. Jun 27, 2004 #8

    Galileo

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    Here's another way:

    Recall the definition of a derivative: f'(a)=lim (x->a) (f(x)-f(a))/(x-a)
    You`ll see lim (x->1) ln(x)/(x-1) is the derivative of ln(x) at x=1.
     
  10. Jul 5, 2004 #9
    [tex]\frac{lnx}{x-1}=lnx^\frac{1}{x-1}[/tex] which is actually natural logarithm of x-1th root of x which converge to 1.
     
    Last edited: Jul 6, 2004
  11. Jul 5, 2004 #10

    turin

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    ln(10) = 1? I would think = 0.

    I'm very curious to see (the meaning of) the algebraic method of taking a limit. Until today, I had always thought that a limit was essentially a calculus concept.
     
  12. Jul 6, 2004 #11
    No it's lne because [tex]lnx^\frac{1}{x-1}=ln(1+x-1)^\frac{1}{x-1}[/tex]
    Make substitiution [tex]\frac{1}{x-1}=t[/tex]. Then t is going towards infinity when x is going to 1. Expression becomes [tex]\lim_{t\rightarrow\infty} ln(1+1/t)^t[/tex] which then converges to lne=1. If you don't beleive me check with mathematica or something else.
     
    Last edited: Jul 6, 2004
  13. Jul 6, 2004 #12
    ln x=a
    e^a=x
    lim a/(e^a-1)
    e^a->1
    now because e^a approaches 1 a approaches 0
    therefore we get 0/0.

    i dont know if this the way you were searching for but at least it's algebraic approach.
     
  14. Jul 6, 2004 #13

    turin

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    tomkeus,
    OK, yes, that makes sense. I misread your previous post. It helps to see the latex.

    LQG,
    0/0 is not defined in algebra, is it?
     
  15. Jul 7, 2004 #14
    that's my answer that it has no limit, doesnt converge.
     
  16. Jul 7, 2004 #15
    You cannot solve this entirely by "algebra means". You can only reduce it to [tex]\lim_{x\rightarrow\infty} (1+1/x)^x[/tex]. This is basic limit which cannot be proven with algebra only.
     
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