Euler Limit of $\frac{x^x-x}{1-x+\ln(x)}$ Without L'Hopital

Question

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.

Answer 1

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 $u = x-1$ $$=\lim_{u\to 0}\dfrac{(u+1)^{u+1}-u-1}{\ln(u+1)-u}$$ $$=\lim_{u\to 0}\dfrac{[1+u+u^2+O(u^3)]-u-1}{[u-\frac{u^2}{2}+O(u^3)]-u}$$ $$=\lim_{u\to 0}\dfrac{u^2+O(u^3)}{-\frac{u^2}{2}+O(u^3)}$$ $$=\lim_{u\to 0}\dfrac{u^2}{-\frac{u^2}{2}}$$ $$=\color{red}{-2}$$

Answer 2

$y=\ln(x)$ 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.

Answer 3

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 $-2$.

Answer 4

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 $0<a<1<b<\infty.$