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Math Help - limit when a>>x

  1. #1
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    limit when a>>x

    I'm not getting the algebric trick to do this limit, when a>>x
    \frac{1}{a}\ln(\frac{\sqrt{a^2+x^2} + a}{\sqrt{a^2+x^2} - a})
    In a similar question i did the taylor expansion with 2 terms: (1+b)^n=1+nb for b<<1, and it worked fine (i that case b=\frac{x^2}{a^2}). Probably i'll have to use the same in here.
    Thank's.
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  2. #2
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    "x>> a" is the same as saying that x goes to 0 (as compared to a).

    Start by rationalizing the numerator of the fraction inside the ln by multiplying both numerator and denominator by \sqrt{a^2+ x^2}- a.

    \frac{1}{a}ln\left(\frac{a^2+ x^2- a^2}{a^2+ x^2- 2a\sqrt{a^2+ x^2}+ a^2}\right)
    \frac{1}{a}ln(\left(\frac{x^2}{2a^2+ x^2- 2a\sqrt{a^2+ x^2}}\right

    Now, both numerator and denominator of the argument go to 0 as x goes to 0 so we can use L'hopital's rule:differentiate both, with respect to x. The derivative of the numerator is 2x. The derivative of the denominator is 2x+ 2ax(a^2+ x^2)^{-1/2}. The numerator and denominator still go to 0, as x goes to 0, so we use L'Hopital again.

    The derivative of 2x is 2. The derivative of [tex]2x+ 2ax(a^2+ x^2)^{-1/2}[tex] is 2+ 2a(a^2+ x^2)^{-1/2}- 2ax^2(a^2+ x^2)^{-3/2}. The first, of course, goes to 2 while the second goes to 2+ 2= 4 as x goes to 0.
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  3. #3
    Forum Admin topsquark's Avatar
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    Without using Calculus:

    Consider the numerator inside the logarithm:
    \displaystyle \sqrt{a^2 + x^2} = \frac{1}{a} \sqrt{1 + \left ( \frac{x}{a} \right ) ^2} \approx \frac{1}{a} \left ( 1 + \frac{1}{2} \left ( \frac{x}{a} \right )^2 + ~... \right )
    (by the binomial theorem. Since a >> x we can consider only the first two terms.)

    Thus
    \displaystyle ln \left [ \frac{\sqrt{a^2 + x^2} + a}{\sqrt{a^2 + x^2} - a} \right ]

    \displaystyle = ln \left [ \frac{\sqrt{1 + \left ( \frac{x}{a} \right )^2} + 1}{\sqrt{1 + \left ( \frac{x}{a} \right )^2} - 1} \right ]

    \displaystyle \approx ln \left [ \frac{1 + \frac{1}{2} \left ( \frac{x}{a} \right )^2 + 1}{1 + \frac{1}{2} \left ( \frac{x}{a} \right )^2 - 1} \right ]

    and after a little work:
    \displaystyle \approx ln \left [ 4 \left ( \frac{a}{x} \right )^2 + 1 \right ]

    So you need to calculate
    \displaystyle \frac{1}{a} \cdot ln \left [ 4 \left ( \frac{a}{x} \right )^2 + 1 \right ]
    for a>>x

    Can you take it from here?

    -Dan
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