How can I approximate this equation? The approximation I'm having trouble with is this
$$V = a\ln\left(\frac{1+\frac{L}{a}}{1-\frac{L}{a}}\right)-2L,\space (a\gg L)$$
The hint was to use $\sqrt{1+x^2}=1+\frac{1}{2}x^2+\,...$ and $\log(1+x)=x-\frac{1}{2}x^2+\,...$
I couldn't find a way to use the first hint, so I tried to use the second hint by spreading the $\ln$ function into two different terms, and using the approximation on each. Then I approximated all the $\left(\frac{L}{a}\right)^m$ (m>1) terms into zero. As a result I obtained $V=0$, which I know isn't the wanted answer at all. Any help would be appreciated. Thanks!
 A: If $a\gg L$ then 
$$V = a\ln\left(\frac{1+\frac{L}{a}}{1-\frac{L}{a}}\right)-2L\approx \frac{2 L^3}{3 a^2}$$
Indeed if $a\gg L$ then $x=\dfrac{L}{a}\approx 0$ therefore plugging $L=ax$ in the formula we get
$$a \log \left(\frac{x+1}{1-x}\right)-2 a x$$
write MacLaurin Series up to 3rd power
$$\frac{2 a x^3}{3}+O(x^4)$$
plug back $x=\dfrac{L}{a}$ and you get the result.
I tried some values and it is a very good approximation
For $a=100;\;L=2$ the original formula gives $0.000533461$ and the approximation $0.000533333$
A: Since $a\gg L$
$\frac{1+\frac{L}{a}}{1-\frac{L}{a}} = \frac{1+2\frac{L}{a}+\left(\frac{L}{a}\right)^2}{1-\left(\frac{L}{a}\right)^2}\approx 1+2\frac{L}{a}$
$log\left(\frac{1+\frac{L}{a}}{1-\frac{L}{a}}\right) \approx log\left(1+2\frac{L}{a}\right)=\left(2\frac{L}{a}\right)-\frac{1}{2}\left(2\frac{L}{a}\right)^2+\,... \approx 2\frac{L}{a}-2\frac{L^2}{a^2}$
(above, we keep the quadratic term because we will multiply it by $a$ on the sequence)
$a\,log\left(\frac{1+\frac{L}{a}}{1-\frac{L}{a}}\right)-2\,L \approx 2\left(L-\frac{L^2}{a}\right)-2\,L=\boxed{-\frac{2\,L^2}{a}}$
Would that be the wanted answer?
