A limit with $x^{x^x}$

Question

We need to find $$\lim_{ x\to 1^+}\frac{x^{x^x}-x^x}{x-1}$$ It seems to be not so hard to find , but I get stuck on this . Implicit I can find $\lim_{ x\to 1^+}x^{x^x}$ and $\lim_{ x\to 1^+}x^{x}$. also I think I can solve it by L'Hpoital . my question :Is it possible to find this limit with out using l'hopital rule ?

Answer 1

Since $\displaystyle x^x=1+(x-1)+(x-1)^2+\frac 12 (x-1)^3+ o((x-1)^3)$

and $\displaystyle x^{x^x}=1+(x-1)+(x-1)^2+\frac 32 (x-1)^3+ o((x-1)^3)$,

$$\frac{x^{x^x}-x^x}{x-1}=\frac{(x-1)^3+o((x-1)^3)}{x-1}=(x-1)^2+o((x-1)^2)$$

The limit is $0$.

Answer 2

Taking $x^{x} $ as a factor in numerator we can see that the limit in question is equal to $$\lim_{x\to 1^{+}}\frac{x^{x^{x}-x}-1}{x-1}$$ and this can be further expressed as $$\lim_{x\to 1^{+}}\frac{\exp((x^{x}-x)\log x) - 1}{(x^{x}-x)\log x} \cdot(x^{x} - x) \cdot\frac{\log x} {x-1}$$ and the desired limit is clearly equal to $1\cdot 0\cdot 1=0$.

Answer 3

On this very nice video you will see that:

$$ \frac{ d }{dx} x^x = x^x ( \ln x + 1 ) $$

and

$$ \frac{ d }{dx} x^{x^x} = x^{x^x} ~ ( ~ x^x (1+\ln x)\ln x + x^{x-1} ~ ) $$

thus, using L'Hopital, the limit is (1-1)/1 = 0.