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The solution on the textbook is $\sqrt e$. I assume I have to use the notable limit $$\lim_{x\to0+} (1+x)^{1/x}=e$$ after transforming the function into $((1+\frac x2)^{\frac 2x})^{\frac12}$ but I am stumped on the intermediate passages.

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  • $\begingroup$ Just use taylor expansion of $e^x$ at $x=0$ $\endgroup$ Jul 8, 2018 at 5:43
  • $\begingroup$ Sorry, I should've specified in the title: no De L'Hopital and no Taylor either. $\endgroup$
    – Moss
    Jul 8, 2018 at 5:44
  • $\begingroup$ This is difficult without both $\endgroup$
    – Cloud JR K
    Jul 8, 2018 at 5:50
  • $\begingroup$ It is so simple with Taylor series. $\endgroup$ Jul 8, 2018 at 5:55
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    $\begingroup$ No Taylor series?! What are you allowed to use? What is your definition of $e^x$? $\endgroup$
    – Brian Tung
    Jul 8, 2018 at 5:57

2 Answers 2

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Given that $$ \lim_{x\to0}\left(\frac{e^x-1}{x}\right)^{1/x}=\lim_{x\to0}\exp\left[{\frac{1}{x}\;\log\left(\frac{e^x-1}{x}\right)}\right] $$ and that $$ \lim_{x\to0}\frac{1}{x}\;\log\left[1+\left(\frac{e^x-1}{x}-1\right)\right] =\lim_{x\to0}\frac{1}{x}\left(\frac{e^x-1}{x}-1\right)=\lim_{x\to0}\left(\frac{e^x-1-x}{x^2}\right) $$ this other question could be relevant:

How to find $\lim\limits_{x\to0}\frac{e^x-1-x}{x^2}$ without using l'Hopital's rule nor any series expansion?

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  • $\begingroup$ I knew there must have been an easier approach than mine. $\endgroup$
    – user123641
    Jul 8, 2018 at 8:24
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If you take $\log$, and use the inequality $$\frac{x-1}{x}\leq \log(x)\leq x-1$$ which is quite ideal around the value $x=1$ (which happens to be $\exp'(0)$), you can get the inequalities below: $$\frac{e^x-x-1}{x^2}\cdot \frac{x}{e^x-1}=\frac{\frac{e^x-1}{x}-1}{\frac{e^x-1}{x}}\cdot\frac{1}{x}\leq\frac{\ln\left(\frac{e^x-1}{x}\right)}{x}\leq \frac{\frac{e^x-1}{x}-1}{x}=\frac{e^x-x-1}{x^2}$$

Both functions on the left and right tend to $1/2$, if you can show those instead.

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