Limit Question Solution | Simplifying Infinite Limit with e^2 Result

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Homework Statement


I've done thsi limit, i think I'm half right. I'm sure I have done something wrong, but I'm not sure what.

lim_{x-> \infty} ( \frac {x+lnx}{x-lnx})^(x/lnx)


The Attempt at a Solution


lim_{x-> \infty} ( \frac {x+lnx}{x-lnx})^{x/lnx} = lim_{x-> \infty} ( \frac {x(1 +\frac {lnx}{x})}{x(1- \frac {lnx}{x})})^{x/lnx}=
<br /> t= \frac {x}{lnx}; lim_{t-&gt; \infty} (\frac {1 + \frac {1}{t}} {1 - \frac{1}{t}})^t=
\frac {e}{e^{-1}}=e^2
 
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That's right.
 
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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