# Limits of exponentials: $\lim_{x \to 1} {x^x - x^{x^x} \over {(1-x)^2}}$

It is a $$\frac00$$ limit, but I can't seem to figure it out. I tried writing $${x^x}$$ as $${e^{x\ln x}}$$ and going with L'Hospital from there but I got stuck. Any help is greatly appreciated.

I got the following expansions. $$x^x=1+(x-1)+(x-1)^2+\frac{1}{2}(x-1)^3+...$$ and $$x^{x^x}=1+(x-1)+(x-1)^2+\frac{3}{2}(x-1)^3+...,$$ which gives the answer: $$\lim_{x\rightarrow1}\frac{x^x-x^{x^x}}{(x-1)^2}=0.$$
• +1. In other words, $\dfrac{x^x-x^{x^x}}{(1-x)^2} \approx 1-x$ near $x=1$ – Henry Mar 7 at 10:04
$$x^r=\bigg[1+(x-1)\bigg]^r\approx r(x-1)$$ for $$x\rightarrow 1$$
$$\displaystyle x^{x^{x}}-x^x\approx x^x(x-1)-x(x-1) \approx (x-1)(x^x-x)\approx (x-1)^2(x-1)$$
$$\displaystyle -\lim_{x\rightarrow 1}\frac{x^{x^{x}}-x^x}{(1-x)^2} =-\lim_{x\rightarrow 1}\frac{(x-1)^3}{(x-1)^2}=0$$