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What's the limit?

  1. Jan 26, 2013 #1
    1. The problem statement, all variables and given/known data
    I don't know how to find a limit, and it's bothering me for a few hours now.
    Can someone help me?
    j - imaginary unit
    2. Relevant equations

    [itex]\lim_{\rho \to 0}{\frac{\frac{\sqrt{2}}{2}(-1+j)+\rho \exp(j\theta)}{(\frac{\sqrt{2}}{2}(-1+j)+\rho \exp(j\theta))^2+ \sqrt2(\frac{\sqrt{2}}{2}(-1+j)+\rho \exp(j\theta)))+1}}[/itex]

    3. The attempt at a solution
    Solution is:
    [itex]∞ exp( \frac{∏}{4} - \theta) [/itex]
     
  2. jcsd
  3. Jan 26, 2013 #2

    Dick

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    The denominator approaches 0 and the numerator doesn't. It doesn't have a limit.
     
  4. Jan 26, 2013 #3
    To be exact...
    [itex]\rho \rightarrow 0+[/itex]

    The solution I've written is correct for sure.:)
     
  5. Jan 26, 2013 #4

    Dick

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    Ok, lets write [itex]a=\frac{\sqrt{2}}{2} (-1+j)[/itex] and [itex]r=\rho exp( j \theta)[/itex] then your expression is [tex]\frac{a+r}{(a+r)^2+\sqrt{2} (a+r)+1}[/tex]
    If you expand the denominator, and putting in the value for a, you get [tex]\frac{a+r}{r^2+j \sqrt{2} r}[/tex]
    As ρ→0 you can ignore the r in the numerator and the r^2 in the denominator. Now you just have to express [tex]\frac{a}{j \sqrt{2} r}[/tex] as a magnitude and phase. Can you take it from there? It's not really a limit, it's a limiting behavior.
     
  6. Jan 26, 2013 #5
    Thanks...it helped me:)
     
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