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

"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 $\displaystyle \sqrt{a^2+ x^2} a$.
$\displaystyle \frac{1}{a}ln\left(\frac{a^2+ x^2 a^2}{a^2+ x^2 2a\sqrt{a^2+ x^2}+ a^2}\right)$
$\displaystyle \frac{1}{a}ln(\left(\frac{x^2}{2a^2+ x^2 2a\sqrt{a^2+ x^2}}\right$
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 $\displaystyle 2x+ 2ax(a^2+ x^2)^{1/2}$. 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 $\displaystyle 2+ 2a(a^2+ x^2)^{1/2} 2ax^2(a^2+ x^2)^{3/2}$. The first, of course, goes to 2 while the second goes to $\displaystyle 2+ 2= 4$ as x goes to 0.

Without using Calculus:
Consider the numerator inside the logarithm:
$\displaystyle \displaystyle \sqrt{a^2 + x^2} = \frac{1}{a} \sqrt{1 + \left ( \frac{x}{a} \right ) ^2} \approx \frac{1}{a} \left ( 1 + \frac{1}{2} \left ( \frac{x}{a} \right )^2 + ~... \right )$
(by the binomial theorem. Since a >> x we can consider only the first two terms.)
Thus
$\displaystyle \displaystyle ln \left [ \frac{\sqrt{a^2 + x^2} + a}{\sqrt{a^2 + x^2}  a} \right ]$
$\displaystyle \displaystyle = ln \left [ \frac{\sqrt{1 + \left ( \frac{x}{a} \right )^2} + 1}{\sqrt{1 + \left ( \frac{x}{a} \right )^2}  1} \right ]$
$\displaystyle \displaystyle \approx ln \left [ \frac{1 + \frac{1}{2} \left ( \frac{x}{a} \right )^2 + 1}{1 + \frac{1}{2} \left ( \frac{x}{a} \right )^2  1} \right ]$
and after a little work:
$\displaystyle \displaystyle \approx ln \left [ 4 \left ( \frac{a}{x} \right )^2 + 1 \right ]$
So you need to calculate
$\displaystyle \displaystyle \frac{1}{a} \cdot ln \left [ 4 \left ( \frac{a}{x} \right )^2 + 1 \right ]$
for a>>x
Can you take it from here?
Dan