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On slide $10$ here: http://math.cmu.edu/~bwsulliv/basel-problem.pdf , it is stated that Euler may have used L'Hopital's Rule to show that the limit $$\lim_{x\to 1}\dfrac{x^x-x}{1-x+\ln(x)}=-2$$ I was able to reproduce this easily with L'Hopital's Rule Upon two applications of it (shown at the bottom of this post).

I've attempted and failed showing this limit using various methods including substitutions, Squeeze Theorem, splitting it up into $\dfrac{x^x}{1-x+\ln(x)}-\dfrac{x}{1-x+\ln(x)}$ (doesn't seem to work because it's just indeterminate), and series expansion representations (sort of).

Substitutions tried and problems with them:

  • Let $y=\ln(x)\implies x=e^y$ and $x^x=e^{ye^y}$, but I still get $\frac{0}{0}$ and couldn't find a way to transform it into a known limit.
  • Let $y=x^x\implies \ln(y)=x\ln(x)$, but don't know how to solve for $x$ here in terms of $y$.
  • Let $y=e^x\implies \ln(y)=x\implies x^x=(\ln(y))^{(\ln(y))}$. This limit did not seem to be easier.

For Squeeze Theorem, I simply couldn't find suitable comparisons mostly since I could not remove the $\ln(x)$ in those comparisons. Even removing the $x^x$ doesn't seem like it would help since $\dfrac{x^x-x}{1-x+\ln(x)}$ is strictly less than $\dfrac{-x}{1-x+\ln(x)}$. So, even if the new limit went to $-2$, which it doesn't, we wouldn't have strictly shown the original limit to be $-2$. As for series expansions, I have no idea how to get a good series expansion of that (tried Taylor at $x=0,1,2$, but very messy and still gave $\frac{0}{0}$ and didn't give any hints).

With L'Hopital $$\lim_{x\to 1}\dfrac{x^x-x}{1-x+\ln(x)}\to\dfrac{0}{0}\implies \lim_{x\to 1}\dfrac{x^x-x}{1-x+\ln(x)}=\lim_{x\to 1}\dfrac{x^x(\ln(x)+1)-1}{-1+\frac{1}{x}}=\dfrac{0}{0}$$ $$\implies \lim_{x\to 1}\dfrac{x^x(\ln(x)+1)-1}{-1+\frac{1}{x}}=\lim_{x\to 1}\dfrac{x^{x-1}+x^x(\ln(x)+1)\ln(x)+x^x(\ln(x)+1)}{-\frac{1}{x^2}}$$ $$=-(1^2)\cdot\left[1^{1-1}+1^1(\ln(1)+1)\ln(1)+1^1(\ln(1)+1)\right]=(-1)\cdot\left[1+0+1\right]=\boxed{-2}$$

I appreciate any helps or hints on changing this does to a bunch of known/simpler limits.

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    $\begingroup$ off topic, but i must say that the presentation is so nice $\endgroup$ – Domates Jan 27 '17 at 20:55
  • $\begingroup$ @ H. Ergul Why is this off-topic? Thank you. $\endgroup$ – user12345 Jan 27 '17 at 21:01
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    $\begingroup$ He was referring to his comment and not your question. $\endgroup$ – J. M. is a poor mathematician Jan 27 '17 at 21:40
  • $\begingroup$ @Domates Thank you so much! :-) $\endgroup$ – Brendan W. Sullivan Oct 27 '18 at 4:13
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You mention allowing series expansions. This problem is trivial with them. $$\lim_{x\to 1}\dfrac{x^x-x}{1-x+\ln(x)}$$ So that I can use McLaurin series I already know, I am going to enforce the substitution $u = x-1$ $$=\lim_{u\to 0}\dfrac{(u+1)^{u+1}-u-1}{\ln(u+1)-u}$$ $$=\lim_{u\to 0}\dfrac{[1+u+u^2+O(u^3)]-u-1}{[u-\frac{u^2}{2}+O(u^3)]-u}$$ $$=\lim_{u\to 0}\dfrac{u^2+O(u^3)}{-\frac{u^2}{2}+O(u^3)}$$ $$=\lim_{u\to 0}\dfrac{u^2}{-\frac{u^2}{2}}$$ $$=\color{red}{-2}$$

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  • $\begingroup$ Ahh I was not aware of how nice the series expansion for $(x+1)^{(x+1)}$ looked. Did you know that from experience or just something that happened to work with experimentation? Didn't think of this substitution. Thank you. $\endgroup$ – user12345 Jan 27 '17 at 21:00
  • $\begingroup$ @anonymaker000010001 Works with almost anything of the form $[P(x)]^{R(x)}$ for polynomials $P$ and $R$ $\endgroup$ – Simply Beautiful Art Jan 27 '17 at 21:10
  • $\begingroup$ @SimplyBeautifulArt It appears to be so after playing with them for a bit, but $x^x$ didn't give anything nice (many logs). Perhaps I just gave up too soon. Thanks! $\endgroup$ – user12345 Jan 27 '17 at 21:15
  • $\begingroup$ @anonymaker000010001 Well, you don't use Taylor expansions for these. Instead, use binomial expansion... $\endgroup$ – Simply Beautiful Art Jan 27 '17 at 21:16
  • $\begingroup$ @anonymaker000010001 this came down to personal experience :) I've worked a lot with $x^x$, and so I'm very familiar with the expansion in terms of $\log(x) $. How did I know there would be no logarithms when I made the limit approach zero? Pure intuition. The only thing that might have subconsciously influenced me is that I knew the limit was approaching one, and that logarithms dissappear at $1$. Therefore, when we approach zero we will just get power of $x$ (with some coefficients), because each of these terms will dissappear at zero. I hope this makes a bit of sense :/ $\endgroup$ – Brevan Ellefsen Jan 29 '17 at 1:22
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$y=\ln(x)$ actually works

$$\lim_{x\to 1}\dfrac{x^x-x}{1-x+\ln(x)}=\lim_{y \to 0}\dfrac{e^{ye^y}-e^y}{1-e^{y}+y}=\lim_{y \to 0}e^{y}\dfrac{e^{y(e^y-1)}-1}{1-e^{y}+y}=\lim_{y \to 0}\dfrac{e^{y(e^y-1)}-1}{1-e^{y}+y}$$

Now, by the definition of the derivative $$\lim_{z \to 0}\frac{e^z-1}{z}=1$$

Replacing $z=y(e^y-1)$ you get $$\lim_{y \to 0}\frac{e^{y(e^y-1)}-1}{y(e^y-1)}=1$$

Therefore $$\lim_{y \to 0}\frac{e^{y(e^y-1)}-1}{1+y-e^y}=\lim_{y \to 0}\frac{e^{y(e^y-1)}-1}{y(e^y-1)}\frac{y(e^y-1)}{1+y-e^y}=\lim_{y \to 0}\frac{y(e^y-1)}{1+y-e^y}$$

This last limit is much easier and can be calculated easily, either with power series, or by calculating $$ \lim_{y \to 0}\frac{y(e^y-1)}{1+y-e^y}=\lim_{y \to 0}\frac{y(e^y-1)}{y^2}\frac{y^2}{1+y-e^y}=\lim_{y \to 0}\frac{e^y-1}{y}\frac{y^2}{1+y-e^y}=\lim_{y \to 0}\frac{y^2}{1+y-e^y} $$ which is pretty standard.

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  • $\begingroup$ Nice. +1 Note that the last limit pretty much requires l'hospital. $\endgroup$ – Rene Schipperus Jan 27 '17 at 21:11
  • $\begingroup$ @ReneSchipperus Or Taylor series or Taylor approximation. $\endgroup$ – N. S. Jan 27 '17 at 21:13
  • $\begingroup$ Yeah I dont like those Taylor methods. $\endgroup$ – Rene Schipperus Jan 27 '17 at 21:14
  • $\begingroup$ @ReneSchipperus All you need is to show that $e^{y}=1+y+\frac{1}{2}y^2+o(y^2)$ which can be done by a standard error estimate. $\endgroup$ – N. S. Jan 27 '17 at 21:31
  • $\begingroup$ I dont like those type of arguments I find them very inelegant. L'Hospital accomplishes the same thing. $\endgroup$ – Rene Schipperus Jan 27 '17 at 21:33
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You can collect a factor $x$ from the numerator and then do $x=t+1$, so the limit becomes $$ \lim_{t\to0}(1+t)\frac{(t+1)^t-1}{\log(1+t)-t} $$ (surely Euler didn't use “ln”).

The factor $1+t$ contributes $1$, so we can concentrate on the fraction. The numerator is $e^{t\log(t+1)}-1$ and can be expanded like $$ 1+t\log(1+t)+o(t\log(1+t))-1=t^2+o(t^2) $$ whereas the denominator is $$ \log(1+t)-t=t-\frac{t^2}{2}+o(t^2)-t=-\frac{t^2}{2}+o(t^2) $$ so the limit is $-2$.

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This is a footnote about the method used in some of the answers already given.

The functions $f(x)=x^x-x$ and $g(x)=1-x+\ln x$ are infinitely differentiable for $x>0$. So $f'''$ and $g'''$ are continuous for $x>0.$

So $A=\sup \{|f'''(x)|:x\in [1/2,2]\}<\infty$ and $B=\sup \{|g'''(x)|:x\in [1/2,2]\}<\infty.$

Let $M=\max (A,B).$

For $1+y\in [1/2,2]$ we have $$f(1+y)=f(1)+yf'(1)+y^2f''(1)/2!+y^3f'''(c)/3!$$ for some $c$ such that $|c|\leq |y|.$ So $$|y^3f'''(c)/3!|\leq |y|^3M/3!.$$ Similarly for $g.$ The key is that $ M<\infty$ so we have $$f(1+y)=f(1)+yf'(1)+y^2f''(1)/2+O(y^3)$$ as $y\to 0.$

The choice of the interval $[1/2,2]$ is arbitrary. We could instead take any $[a,b]$ with $0<a<1<b<\infty.$

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