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Math Help - Limit[e*w*L/Sqrt[R^2 + (w*L - 1/(wC))^2], w -> Infinity]

  1. #1
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    Limit[e*w*L/Sqrt[R^2 + (w*L - 1/(w*C))^2], w -> Infinity]

    So I need to take the limit of

    e*w*L/Sqrt[R^2 + (w*L - 1/(w*C))^2] as w approaches infinity

    So in Mathematica it would look like

    Limit[e*w*L/Sqrt[R^2 + (w*L - 1/(w*C))^2], w -> Infinity]

    All variables are >0

    The answer is e, but I am curious as to how to do this. This is not a homework problem as this was something that was partially shown in our physics class and our professor took some serious shortcuts and basically just gave us the answer because he did not want to go through the proof, but I would like to see it.


    I understand L'Hopital's rule, but it does not seem to work here as you always have Sqrt[R^2 + (L w - 1/w*C)^2] in your answer no matter how many times you were to take the derivative.

    I am wondering if there is another formula to do something like this. If you have any suggestions that would be great.

    FYI: e=EMF, w=omega, L= inductance, R=resistance, C=capacitance. This equation is looking for the voltage across the inductor as the frequency approaches infinitely large

    Thanks,
    Robert
    Last edited by rlisac; April 21st 2010 at 05:43 PM.
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  2. #2
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    Quote Originally Posted by rlisac View Post
    So I need to take the limit of

    e*w*L/Sqrt[R^2 + (w*L - 1/(w*C))^2] as w approaches infinity

    So in Mathematica it would look like

    Limit[e*w*L/Sqrt[R^2 + (w*L - 1/(w*C))^2], w -> Infinity]

    All variables are >0

    The answer is e, but I am curious as to how to do this. This is not a homework problem as this was something that was partially shown in our physics class and our professor took some serious shortcuts and basically just gave us the answer because he did not want to go through the proof, but I would like to see it.


    I understand L'Hopital's rule, but it does not seem to work here as you always have Sqrt[R^2 + (L w - 1/w*C)^2] in your answer no matter how many times you were to take the derivative.

    I am wondering if there is another formula to do something like this. If you have any suggestions that would be great.

    FYI: e=EMF, w=omega, L= inductance, R=resistance, C=capacitance. This equation is looking for the voltage across the inductor as the frequency approaches infinitely large

    Thanks,
    Robert

    Let's see if I "guessed" correctly the expression in the limit: \lim_{w\to\infty}\frac{ewL}{\sqrt{R^2+\left(wL-\frac{1}{wC}\right)^2}} .

    Now multiply this expression by 1=\frac{1/w}{1/w} and remember: to take a positive number into the square root you square it inside and to take it into an squared

    expression you take its square root: w\sqrt{whatever}=\sqrt{w^2\cdot\,whatever}\,,\,\,w  (whatever)^2=(\sqrt{w}\cdot\,whatever)^2 ,and thus we get:

    \frac{eL}{\sqrt{\frac{R^2}{w^2}+\left(L-\frac{1}{w^2C}\right)^2}} and now some limits arithmetic: \frac{R^2}{w^2}\xrightarrow [w\to\infty]{}0\,,\,\,\frac{1}{w^2C}\xrightarrow [w\to\infty]{}0\Longrightarrow the denominator converges to \sqrt{0+(L-0)^2}=L , and since

    the numerator is constant wrt w the limit indeed is \frac{eL}{L}=e .

    Tonio
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  3. #3
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    Thanks so much. That was way simpler than I thought it was going to be. The professor started with L'Hopital's rule so I figured it would involve that or just be something really complex.
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