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Wolfram Alpha gives me this solution: $$\lim\limits_{x\to 0} \left(\frac {1+e^{x}}{2}\right) ^{1/x} = \sqrt{e}$$

But I have no idea how to get to that result.

I tried using L'Hopital but I found it impossible to get rid of the 1/x.

I also tried to use the fact that the result is $e^{1/2}$ and put an $e^{ln()}$ to use the logarithms properties, but I still couldn't simplify/rewrite the 1/x

I thought of using the squeeze theorem but I couldn't find functions that could work with the inequalities.

Átila Correia
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Atk
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2 Answers2

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According to the continuity of the $\exp$ function as well as the l'Hôpital's rule, one gets:

\begin{align*} \lim_{x\to 0}\left(\frac{1 + e^{x}}{2}\right)^{1/x} & = \lim_{x\to 0}\exp\left(\frac{1}{x}\ln\left(\frac{1 + e^{x}}{2}\right)\right) = \exp\left(\lim_{x\to 0}\frac{e^{x}}{1 + e^{x}}\right) = \exp\left(\frac{1}{2}\right) = \sqrt{e} \end{align*}

Hopefully this helps!

Átila Correia
  • 18,937
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The logarithm gives us $$ L := \lim\limits_{x\to 0} \left(\frac {1+e^{x}}{2}\right) ^{1/x} = \sqrt{e} \implies \ln(L) = \lim_{x \to 0} \frac {\ln \left( \frac{1+e^x}{2} \right)} x $$ As $x \to 0$, this is a $0/0$-type form. Differentiate $x$ and the logarithm to apply L'Hopital's: $$ \ln(L) = \lim_{x \to 0} \frac{1}{\left( \frac{1+e^x}{2} \right)} \cdot \frac 1 2 e^x = \lim_{x \to 0} \frac{e^x}{1+e^x} = \frac 1 2 $$ So $L = e^{1/2} = \sqrt e$.

PrincessEev
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  • Don't you need more details though to take the composition of functions? $ln$ needs to be continuous at x=0 for $((1 + e^x)/2)^{1/x}$, doesn't it? – bmitc Apr 06 '24 at 04:06
  • No, the limit need only exist there. – PrincessEev Apr 06 '24 at 04:15
  • That's what continuity means, but isn't that assuming the thing you're trying to show? – bmitc Apr 06 '24 at 05:29
  • Continuity does not mean that the limit exists. A function $f$ is continuous at $x_0$ if $$ \lim_{x \to x_0} f(x) = f(x_0) $$ The fact that limit not only exists, but equals $f(x_0)$, is what defines continuity. The limit existing is what we need, not continuity. – PrincessEev Apr 06 '24 at 07:08