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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.

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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: $$\lim_{x\rightarrow1}\frac{x^x-x^{x^x}}{(x-1)^2}=0.$$

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    $\begingroup$ +1. In other words, $\dfrac{x^x-x^{x^x}}{(1-x)^2} \approx 1-x$ near $x=1$ $\endgroup$ – Henry Mar 7 at 10:04
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$x^r=\bigg[1+(x-1)\bigg]^r\approx r(x-1)$ for $x\rightarrow 1$

$\displaystyle x^{x^{x}}-x^x\approx x^x(x-1)-x(x-1) \approx (x-1)(x^x-x)\approx (x-1)^2(x-1)$

$\displaystyle -\lim_{x\rightarrow 1}\frac{x^{x^{x}}-x^x}{(1-x)^2} =-\lim_{x\rightarrow 1}\frac{(x-1)^3}{(x-1)^2}=0$

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