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Does L'Hospital's rule apply to complex functions?

  1. Sep 26, 2005 #1

    quasar987

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    I have to evaluate

    [tex]\lim_{x\rightarrow \infty} \frac{x-1}{e^{ipx/\hbar}}[/tex]

    is this equal to

    [tex]\lim_{x\rightarrow \infty} \frac{\hbar}{ip e^{ipx/\hbar}}[/tex]

    ??
     
  2. jcsd
  3. Sep 26, 2005 #2

    LeonhardEuler

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    L'Hospital's rule only applies when a limit approaches [itex]\frac{0}{0}[/itex] or [itex]\frac{\infty}{\infty}[/itex]. Niether is the case here because [itex]\lim_{x\rightarrow\infty}e^{ix}[/itex] does not exist. The real and imaginary parts oscilate between -1 and 1 as x approaches infinity. Your limit is equal to:
    [tex]\lim_{x\rightarrow \infty} \frac{x-1}{e^{ipx/\hbar}} =\lim_{x\rightarrow \infty} (x-1)\cos{\frac{ipx}{\hbar}}-i(x-1)\sin{\frac{ipx}{\hbar}[/tex]
    So the real and imaginary parts both oscillate between larger and larger positive and negative numbers as x gets larger, so the limit does not exist.
     
  4. Sep 26, 2005 #3

    quasar987

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    Dang. I guess I was trying a little too hard to make this QM problem work :tongue2:
     
  5. Jan 14, 2010 #4
    ok so the answer is infinity
    we have some thing bounded in the denominator
    and the numerator goes to infinity
     
  6. Jan 14, 2010 #5
    I suggest you choose different boundary conditions at infinity for your QM problem.
     
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