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The question is to evaluate $$\lim_{x\to\infty}\left(\sqrt{\ln(e^x+1)}-\sqrt{x}\right)^{1/x}$$

This is an indeterminate form of type $0^0$, so I've tried using the identity $a^b=e^{b\ln a}$ and somehow apply l'Hospital's, which leads to pretty complex derivatives and I'm getting nowhere. I've also tried multiplying by the conjugate and perhaps factorize, without success.

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    $\begingroup$ As always, limited expansions are the best way. But if, for some reason, one wants to avoid them, then multiplying by the conjugate quantity yields $$\sqrt{\ln(e^x+1)}-\sqrt{x}=\frac{\ln(e^x+1)-x}{\sqrt{\ln(e^x+1)}+\sqrt{x}}=\frac{\ln(1+e^{-x})}{\sqrt{x+\ln(1+e^{-x})}+\sqrt{x}}$$ Now, the limits of $$(\ln(1+e^{-x})^{1/x}$$ and $$(\sqrt{x+\ln(1+e^{-x})}+\sqrt{x})^{1/x}$$ should be clear. Are they to you? $\endgroup$ – Did Mar 11 '18 at 9:58
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Hint. Note that \begin{align} \bigg(\sqrt{\ln(e^x+1)}-\sqrt{x}\bigg)^{1/x}&= x^{1/(2x)}\bigg(\sqrt{1+\frac{\ln(1+e^{-x})}{x}}-1\bigg)^{1/x}\\ &=x^{1/(2x)}\bigg(\sqrt{1+\frac{e^{-x}+o(e^{-x})}{x}}-1\bigg)^{1/x}\\ &= x^{1/(2x)}\bigg(1+\frac{e^{-x}}{2 x}+o(e^{-x}/x)-1\bigg)^{1/x}\\ &= \frac{x^{1/(2x)}}{(2x)^{1/x}}\cdot e^{-1}\bigg(1+o(1)\bigg)^{1/x} \end{align} where we used the following expansions at $t=0$: $\ln(1+t)=t+o(t)$, $\sqrt{1+t}=1+\frac{t}{2}+o(t)$. So, what is the limit as $x\to +\infty$?

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Hint: $$ \begin{align} \sqrt{\log\left(e^x+1\right)}-\sqrt{x} &=\sqrt{x+\log\left(1+e^{-x}\right)}-\sqrt{x}\\ &=\frac{\log\left(1+e^{-x}\right)}{\sqrt{x+\log\left(1+e^{-x}\right)}+\sqrt{x}} \end{align} $$

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