How can I find this limit:
$$\lim_{x\rightarrow\infty} \left(\frac{x+\ln x}{x-\ln x}\right)^{\frac{x}{\ln x}}$$
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I've tried to do ln on the two sides and than I get a multiple of x/lnx * ln(x+ln)/ln(x-lnx) and I can't continue.. I was looking for some way to get to infinity/infinity for lhopital but I couldnt – rab2004 Jan 25 '14 at 15:33
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Try to reduce to an expression involving $x/\log x$ only. – Siminore Jan 25 '14 at 15:34
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1How do you get $$ \frac{\ln(x+\ln(x))}{\ln(x-\ln(x))}$$ – Prahlad Vaidyanathan Jan 25 '14 at 15:34
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Hint: Look at the simpler problem $$\lim_{u\rightarrow\infty}\left(\frac{u+1}{u-1}\right)^u$$
and consider $u=x/ \log x$. Can you make this transformation and find this limit instead?
MPW
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mmm I'm trying but I can't.. is it the same way like on this post: http://math.stackexchange.com/questions/12307/how-to-calculate-displaystyle-lim-x-to-infty-left-fracx2x-right?rq=1 – rab2004 Jan 25 '14 at 15:41
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1Notice that $(\frac{u+1}{u-1})^u=(\frac{u-1+2}{u-1})^u=(1+\frac{2}{u-1})^u$. – Spock Jan 25 '14 at 15:56
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$\lim _{x \to \infty}\left(1+\frac{2\log x}{x(1-\log x/x)}\right)^{x/\log x}=e^{2}$
$\lim_{x \to \infty}\left(1+\frac{a}{x}\right)^{x}=\lim_{x \to \infty}\left(\left(1+\frac{a}{x}\right)^{\frac{x}{a}}\right)^a=e^a$
if $a(x) \to a \ne 0$, then
$\lim_{x \to \infty}\left(1+\frac{a(x)}{x}\right)^{x}=\lim_{x \to \infty}\left(\left(1+\frac{a(x)}{x}\right)^{\frac{x}{a(x)}}\right)^{a(x)}=e^a$
kmitov
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