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Since we have $\frac00$,I Applied L'Hopital rule :

$$\lim_{x\to 0} (1+x)^{\tfrac1x}\times\left(\cfrac{-\ln(1+x)}{x^2}+\cfrac{1}{x(x+1)}\right)$$$$=\lim_{x\to 0}\cfrac{x^2(x+1)(1+x)^{\tfrac1x}-(x+1)\ln(1+x)+x}{x^2(x+1)}$$

But as you can see it is getting very ugly.

User
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    Hint: It is the derivative of $x \mapsto (1+x)^{1/x}$ at $x=0$. – Gary Nov 30 '20 at 07:46
  • Hint: you probably already know $\lim_{x\to0} (1+x)^{1/x}$, so concentrate on the term in parentheses at the end of the first line: combining them into a single fraction and using l'Hôpital's rule twice more should work. (Note also that there is an algebra error in going from the first line to the second.) – Greg Martin Nov 30 '20 at 07:46
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    @Gary: that's true but it leads to the exact same expression as l'Hôpital's rule does. – Greg Martin Nov 30 '20 at 07:47
  • @GregMartin I just saw that algebra error thanks. I had to calculate the fractions in parentheses then multiply first term to that. but now according to your hint I should replace the first term with $e$ . am I right? – User Nov 30 '20 at 07:51
  • Perhaps something like (using series expansions): $$\begin{align}L&=\lim_{x \rightarrow 0}\frac{\left(1+x\right)^{1/x}-e}{x}\&=e\cdot\lim_{x \rightarrow 0}\frac{e^{\ln\left(1+x\right)/x-1}-1}{x}\&=e\cdot\lim_{x \rightarrow 0}\frac{e^{\left(x-x^{2}/2\right)/x-1}-1}{x}\&=e\cdot\lim_{x \rightarrow 0}\frac{e^{-x/2}-1}{x}\&=e\cdot\lim_{x \rightarrow 0}\frac{\left(1-x/2\right)-1}{x}\&=-\frac{e}{2}\end{align}$$ – Mourad Nov 30 '20 at 08:26
  • Note that the use of L'Hopital's rule is somewhat circular here, or at the very least overkill, as the given limit is the definition of the derivative of $f(x)=(1+x)^{1/x}$ for $x\ne0$ and $f(0)=e$. See e.g. here for another such example. – Simply Beautiful Art Dec 01 '20 at 00:37

4 Answers4

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\begin{aligned} &\lim_{x\to 0}\frac{(1+x)^{\frac{1}{x}}-e}{x}\\ &=\lim_{x\to 0}\frac{e^{{\frac{1}{x}}\ln(1+x)}-e}{x}\\ &=e\lim_{x\to 0}\frac{e^{{\frac{1}{x}}\ln(1+x)-1}-1}{x}\\ &=e\lim_{x\to 0}\frac{{\frac{1}{x}}\ln(1+x)-1}{x}\\ &=e\lim_{x\to 0}\frac{\ln(1+x)-x}{x^2}\\ &=e\lim_{x\to 0}\frac{\frac{1}{1+x}-1}{2x}\\ &=e\lim_{x\to 0}\frac{-1}{2(1+x)}\\ &=\frac{-e}{2} \end{aligned}

fengbiqian
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Hint

Taylor series of $${(1+x)}^{1/x}=e(1-\frac{1}{2}x+\frac{11}{24}x^2+...)$$

If used you get $-e/2$

Hari Ramakrishnan Sudhakar
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Without series, only L'Hospital $$ \lim_{x\to 0}\dfrac{(1+x)^{\tfrac1x}-e}{x} $$ we get $$ \lim_{x\to 0}\left[\left(-\frac{\ln(1+x)}{x^2}+\frac{1}{x}(1+x)^{-1}\right)(1+x)^{\frac{1}{x}}\right] \\ =\lim_{x\to 0}\left(-\frac{\ln(1+x)}{x^2}+\frac{\frac{1}{1+x}}{x}\right)\lim_{x\to 0}(1+x)^{\frac{1}{x}} \\ =\lim_{x\to 0}\left(\frac{\frac{x}{1+x}-\ln(1+x)}{x^2}\right)e $$ another L'Hospital $$ = \lim_{x\to 0}\left(\frac{\frac{1}{1+x}+\frac{-x}{(1+x)^2}-\frac{1}{1+x}}{2x}\right)e = \lim_{x\to 0}\frac{-1}{2(1+x)^2}e = -\frac{e}{2}\\ $$

Physor
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  • Nice. how you recognized to write $-\frac{\ln(1+x)}{x^2}+\frac{\tfrac{1}{x+1}}{x}$? I normally write $x^2(x+1)$ at denominator – User Nov 30 '20 at 08:31
  • Yes, I tried later to write it in the other way but then one has to use the L'Hospital twice instead of once, and it works too,...honestly it was a coincidence! – Physor Nov 30 '20 at 08:33
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$$y=\dfrac{(1+x)^{\tfrac1x}-e}{x}$$ $$z=(1+x)^{\tfrac1x}\implies \log(z)=\tfrac1x\log(1+x)$$ Now Taylor $$\log(z)=\tfrac1x\left(x-\frac{x^2}{2}+\frac{x^3}{3}+O\left(x^4\right) \right)=1-\frac{x}{2}+\frac{x^2}{3}+O\left(x^3\right)$$ $$z=e^{\log(z)}=e-\frac{e x}{2}+\frac{11 e x^2}{24}+O\left(x^3\right)$$ $$y=-\frac{e}{2}+\frac{11 e x}{24}+O\left(x^2\right)$$

Claude Leibovici
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