How do I Show that, $\lim\limits_{x\to 0^+}x^{x^{x^x}} = 1$?

Question

$$\lim_{x\to 0^+}x^{(x^{(x^{x})})}$$

I tried to solve them by:

$$\lim_{x\to 0^+}\ln(x^{(x^{(x^{x})})}) = \lim_{x\to 0^+}x^{(x^{x})} \ln(x)$= \lim_{x\to 0^+}x^{(x^{x})} \cdot \lim_{x\to 0^+}\ln(x)$$

and then ends up with $0 \cdot -\infty $ (undefined)

I tried to render the graph in a web application and it shows that the result should be $1$, not undefined

What is the correct strategy to solve this limits?

Answer 1

Hin

$ \lim_{x\to 0^+}x\ln x=0$ that is ${x\ln(x)}$ stay ner zero as $x$ is close to zero

so by Taylor expansion of first oder near zero give

$$ x^x= e^{x\ln(x)} \sim 1+x\ln x \implies x^{x^x} = e^{x^x\ln(x)} \sim e^{(1+x\ln x)\ln(x)} = e^{\ln x+x\ln^2 x} =x e^{x\ln^2 x}$$

That is near zero we have $$\bbox[yellow]{x^{x^x} \sim x e^{x\ln^2 x}} $$

But $$ x\ln^2x = [\frac{1}{2}\sqrt{x}\ln\sqrt{x}]^2\to 0 \qquad as\qquad x\to 0^+$$

Therefore, near zero we have $$\bbox[yellow]{x^{x^x} \sim x e^{x\ln^2 x}\sim x \qquad as \qquad x\to 0^+}$$ Hence,

$$x^{x^x} \sim x\implies \lim_{x\to 0^+}x^{x^{x^{x}}} =\lim_{x\to 0^+}e^{x^{x^{x}}\ln(x)} \sim \lim_{x\to 0^+}e^{x\ln(x)} = 1$$

Answer 2

Here is a detailed expansion in $0^+$ and justification of the chain $x^{x^{x^x}}\sim x^{x^1}\sim x^x\sim 1$

• $\lim\limits_{x\to 0+}x^\alpha\ln^{\beta}x=0$ for $\alpha,\beta>0$
• $\exp(1+u)=1+u+o(u)$ when $u\to 0$

$\begin{array}{ll} f(x) &= x^x \\ &=\exp(x\ln x)\\ &=1+O(x\ln x)&\sim 1\\ \end{array}$

$\begin{array}{ll} g(x) &= x^{x^x}=x^{f(x)} \\ &=\exp(f(x)\ln x)\\ &=\exp(\ln x+O(x\ln ^2x))\\ &=x\exp(O(x\ln^2 x))\\ &=x(1+O(x\ln ^2x))\\ &=x+O(x^2\ln^2x)&\sim x\\ \end{array}$

$\begin{array}{ll} h(x) &= x^{x^{x^x}}=x^{g(x)}\\ &=\exp(g(x)\ln x)\\ &=\exp(x\ln x+O(x^2\ln^3 x))\\ &=1+x\ln x+O(x^2\ln^3 x)&\sim 1\end{array}$

Note: $\frac 12x^2\ln^2 x=o(x^2\ln^3 x)$ so this term does not appear in the last line of $h(x)$ expansion.