Why does $\lim_{x\rightarrow\infty} x-x^{\frac{1}{x}^{\frac{1}{x}}}-\log^2x=0?$

Question

Why does $$\lim_{x\rightarrow\infty} x-x^{\frac{1}{x}^{\frac{1}{x}}}-\log^2x=0?$$

Moreover, why is $$x-x^{\frac{1}{x}^{\frac{1}{x}}}\approx\log^2 x?$$

Answer 1

By a Taylor expansion, $$ \left( \frac{1}{x} \right)^{1/x} = \exp\left( -\frac{1}{x}\ln x \right) = 1-\frac{\ln x}{x} + O\left(\frac{\ln^2 x}{x^2}\right) . $$ Thus the second term equals $$ (1+\delta)\exp\left( \ln x \left( 1 - \frac{\ln x}{x} \right) \right) = x \left( 1-\frac{\ln^2 x}{x} + O\left(\frac{\ln^4 x}{x^2}\right) \right)(1+\delta) $$ with $\delta = O(\ln^3 x/x^2)$, so this gives the claim.

Answer 2

I give a completely rigorous proof. Let $y=\frac{1}{x}$ then the expression reads

$$\frac{1}{y}-y^{-y^y}-\ln^2 y$$

Now $$y^{-y^y}=e^{-\ln y e^{y\ln y}}.$$

Note that $$\lim\limits_{x \rightarrow 0}\frac{e^x-1-x}{x}=0$$ therefore

$$\lim\limits_{y \rightarrow 0}\frac{e^{y\ln y}-1-y\ln y}{y\ln y}=0$$

And since $$\lim\limits_{y \rightarrow 0} y \ln^2 y=0$$ we have

$$\lim\limits_{y \rightarrow 0}\ln y(e^{y\ln y}-1-y\ln y)=0$$

and so

$$\lim\limits_{y \rightarrow 0}e^{\ln y(e^{y\ln y}-1-y\ln y)}=1$$

Now we have

$$e^{-\ln y e^{y\ln y}}=e^{-\ln y (1+y\ln y)}e^{-\ln y (e^{y\ln y}-1-y\ln y)} $$

We shall want to multiply the entire limit by $e^{\ln y (e^{y\ln y}-1-y\ln y)} $

This gives us

$$\left(\frac{1}{y}-\ln^2 y\right)e^{\ln y (e^{y\ln y}-1-y\ln y)}- e^{-\ln y (1+y\ln y)}$$

And we want to know that

$$\left(\frac{1}{y}-\ln^2 y\right)(e^{\ln y (e^{y\ln y}-1-y\ln y)}-1)\rightarrow 0$$ which will reduce the question to finding the limit of $$\left(\frac{1}{y}-\ln^2 y\right)-e^{-\ln y (1+y\ln y)}.$$

The proof of this last limit is actually not so difficult, $\frac{e^x-1}{x}\rightarrow 1$ as $x\rightarrow 0$ so $$\frac{e^{\ln y (e^{y\ln y}-1-y\ln y)}-1}{\ln y (e^{y\ln y}-1-y\ln y)}\rightarrow 1$$

Further since $\frac{e^x-1-x}{x^2}\rightarrow \frac{1}{2}$ we get $$\frac{e^{y\ln y}-1-y\ln y}{y^2\ln^2 y}\rightarrow \frac{1}{2}$$ and we are reduced to looking at the limit $$y^2\ln^2 y\left(\frac{1}{y}-\ln^2 y\right)$$ as this limit is clearly zero. Thus we have shown the limit above and now the problem is completely reduced to the limit of $$\left(\frac{1}{y}-\ln^2 y\right)-e^{-\ln y (1+y\ln y)}.$$

Here we have

$$e^{-\ln y (1+y\ln y)}=\frac{1}{y}e^{-y\ln^2 y}$$

Now note that $$\lim\limits_{x \rightarrow 0}\frac{e^x-1-x}{x^2}=\frac{1}{2}$$ and therefore

$$\lim\limits_{y \rightarrow 0}\frac{e^{-y\ln^2 y}-1+y\ln^2 y}{y^2\ln^4 y}=-\frac{1}{2}$$ and since $$\lim\limits_{y \rightarrow 0} y\ln^4 y=0$$ we have $$\lim\limits_{y \rightarrow 0}\frac{e^{-y\ln^2 y}-1+y\ln^2 y}{y}=0$$

Now $$\frac{1}{y}e^{-y\ln^2 y}-\frac{1}{y}+\ln^2 y=\frac{e^{-y\ln^2 y}-1+y\ln^2 y}{y}$$ so we are finished. Its so easy....