Limit of logarithm of square roots function

salparadise
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Homework Statement


I need to prove that the following limit holds when

\sqrt{s}>>m

\log\left(\frac{\sqrt{s}+\sqrt{s-4m^2}}{\sqrt{s}-\sqrt{s-4m^2}}\right) \rightarrow \log\left(\frac{s}{m^2}\right)

The Attempt at a Solution


I've tried several manipulations using logarithms properties but had no success.

Thanks
 
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You don't need to bother with the logarithm, just focus on the stuff inside. I'll give you a starting hint; you should provide the rest of your workings after that if you can't get it and want further help. To start, I would pull the factor of \sqrt{s} out from the top and bottom of the fraction inside the logarithm. Find the limiting form of the fraction in your desired limit.
 
Hello Mute, thanks for the help. I've managed to do it, after Taylor expansion of \sqrt{1-x}, where in my limit:

x=\frac{4m^2}{s}\rightarrow 0.

As I'm under some time pressure to finish this, very bigg, home exam I'm already not thinking so clearly..

Thanks again
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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