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I want to calculate the limit: $$ \lim_{x\to +\infty}(1+e^{-x})^{2^x \log x}$$ The limit shows itself in an $1^\infty$ Indeterminate Form. I tried to elevate $e$ at the logarithm of the function:

$$\lim_{x\to +\infty} \log(e^{(1+e^{-x})^{2^x \log x}}) = e^{\lim_{x\to +\infty} \log((1+e^{-x})^{2^x \log x})} = e^{\lim_{x\to +\infty} 2^x \log x \cdot \log(1+e^{-x}) }$$

And then rewrite the exponent as a fraction, to get an $\frac{\infty}{\infty}$ form:

$$= \lim_{x\to +\infty} \frac{2^x \log x}{\frac{1}{\log(1+e^{-x})}} $$

But I don't know how to apply an infinite comparing technique here, and even applying de l'Hôpital seems to lead to nothing...

Could you guys give me some help?

Furthermore: is there a way to calculate this limit without using series expansions or other advanced mathematic instruments?

Thank you very much in advance.

P.S. Wolfram says this limit goes to 1, but I still really want to know how.

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  • $\begingroup$ If you are ok, you can set as solved. Thanks! $\endgroup$
    – user
    Dec 19 '17 at 15:55
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We have:

$$(1+e^{-x})^{2^x logx}=e^{2^x \log x\log(1+e^{-x})}\to e^0 = 1$$

indeed

$$2^x \log x\log(1+e^{-x})=\frac{2^x\log x}{e^{x}}\log\left[\left(1+\frac{1}{e^x}\right)^{e^x}\right]\to 0\cdot \log e=0$$

$$\frac{2^x\log x}{e^{x}}=\frac{\log x}{\left(\frac{e}{2}\right)^{x}}\stackrel{\text{l'Hospital}}=\frac{\frac{1}{x}}{\left(\frac{e}{2}\right)^{x}\log{\frac{e}{2}}}=\frac{1}{x\left(\frac{e}{2}\right)^{x}\log{\frac{e}{2}}}\to 0$$

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  • $\begingroup$ How could I tell that $\frac{2^x logx}{e^x} \to 0$ ? Could you explain me all the passages? Are you comparing infinites? $\endgroup$
    – ela
    Dec 18 '17 at 21:56
  • $\begingroup$ @AlessioMartorana it's a standard limit, I've added a derivation by l'Hospital's rule. $\endgroup$
    – user
    Dec 18 '17 at 22:10
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    $\begingroup$ You can avoid L'Hospital's Rule at the end by setting $e/2=a>1$ and then $(2^{x}\log x) /e^{x} =(x/a^{x}) \cdot ((\log x) /x) \to 0\cdot 0=0$ and both the limits used are pretty famous. $\endgroup$
    – Paramanand Singh
    Dec 19 '17 at 3:03
  • $\begingroup$ @Paramanandsingh that’s much better! thanks $\endgroup$
    – user
    Dec 19 '17 at 4:09
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If we call the expression $f(x)$ then $$\ln f(x)=2^x\log x\,\ln(1+e^{-x})=O\left(\frac{2^x\log x}{e^{x}}\right)\to0$$ as $x\to\infty$ as $2/e<1$.

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  • $\begingroup$ Could you give me a more detailed explanation of this solution? In particular: how did you passed from $\ln f(x)=2^x\log x\,\ln(1+e^{-x})$ to big O notation: $O\left(\frac{2^x\log x}{e^{x}}\right)$. Why this equality holds? $\endgroup$
    – ela
    Dec 18 '17 at 21:29
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Consider the more general case of $$y=(1+e^{-x})^{a^x\, \log(x)}$$ Tak logarithms of both sides $$\log(y)={a^x \log(x)}\log(1+e^{-x})$$ When $x$ is large $$\log(1+e^{-x})\sim e^{-x}$$ making $$\log(y)\sim \left(\frac a e\right)^x \log(x)$$ Now, consider the cases where $a<e$ and $a >e$.

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