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Proving d/dx lnx = 1/x

  1. May 19, 2013 #1
    My question has to do with Euler's constant towards the end of the proof:

    d/dx ln/x = lim h → 0 1/h ln(x+h) - lnx
    = lim h→0 1/h ln[(x+h/x]
    = lim h→0 1/h ln(1 + h/x)
    = lim h→0 ln(1 + h/x)^(1/h)(1/x)(x/1)
    = lim h→0 1/x ln(1+h/x)^(x/h) ;

    let u = x/h

    = 1/x lim h→0 ln(1+1/u)^u
    = 1/x lne
    = 1/x

    Doesn't Euler's constant deal with the lim h→∞ instead of 0? Are they interchangeable? If so, why?
     
  2. jcsd
  3. May 19, 2013 #2
    If you let u→∞, you'll get e (which can be called Euler's number, Euler's constant usually means another number γ). Clearly, h→0 implies u→∞.
     
  4. May 19, 2013 #3
    EDIT: lim h-->0 should be lim u-->0
     
    Last edited: May 19, 2013
  5. May 19, 2013 #4
    Why? When you make a substitution or a change of variables, you have to check how the new variable (i.e. u) depends on the original one (i.e. h). Otherwise, you'd get ridiculous results like the following (let y=2x)
    [tex]2=\lim_{y\rightarrow 2} y = \lim_{x\rightarrow 2} 2x=4[/tex]

    EDIT: My point is that your final limit should be [itex]u\rightarrow \infty[/itex], not [itex]u\rightarrow 0[/itex].
     
    Last edited: May 19, 2013
  6. May 19, 2013 #5
    Aaaahhh. You're right. However if I allow h/x = u, then then lim u -->0 would be correct.
     
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