# Find $\lim_{x\rightarrow 0}x^{x^{x^x}}$ [duplicate]

I already known how to prove that $$\lim_{x\rightarrow 0}x^{x^x}=0$$ and $$\lim_{x\rightarrow 0}x^x=1$$. I also tried to use L'Hôpital's rule for this question but it didn't work. How to find the limit? (The limit should be $$1$$ from the graph sketching.)

$$x^x=\exp(\log(x)x)=1+x\log(x)+o(x\log(x))$$ \begin{align} x^{x^x}&=\exp\left(\log(x)x^x\right)=\exp\left(\log(x)\big[1+x\log(x)+o(x\log(x))\big]\right)\\ &=\exp\left(\log(x)+x\log^2(x)+o\left(x\log^2(x)\right)\right) \\ &=x\exp\left(x\log^2(x)+o\left(x\log^2(x)\right)\right)=x(1+o(1)) \end{align} Thus, \begin{align} x^{x^{x^x}}&=\exp\left(\log(x)x^{x^x}\right) \\ &=\exp\left(x\log(x)(1+o(1))\right)\to e^0=1 \end{align} using $$\lim_{x\to 0^+}x\log(x)=0$$.

Update: Since OP doesn't understand asymptotic arguments, I'm adding a solution with L'Hopital's rule. Let $$L=\lim_{x\to 0^+}x^{x^{x^x}}$$. Using the continuity of the logarithm: \begin{align} \log L&=\lim_{x\to 0^+}\log(x)x^{x^{x}}=\lim_{x\to 0^+}\frac{\log(x)}{1/{x^{x^{x}}}} \\ &=\lim_{x\to 0^+}\frac{1/x}{-x^{-x^x+x-1}(x\log^2(x)+x\log(x)+1)} \\ &=\lim_{x\to 0^+}-\frac{1}{x^{-x^x+x}(x\log^2(x)+x\log(x)+1)} \\ &=\lim_{x\to 0^+}-\frac{x^{x^x}}{x^x(x\log^2(x)+x\log(x)+1)} \end{align} and we know the limits of all expressions in the last line, so we can finish. To differentiate $$1/{x^{x^{x}}}$$, write it as $$(x^{x^{x}})^{-1}$$

• Sorry, what is the o function? – 顾泊洋 Dec 21 '19 at 22:09
• @顾泊洋 You're not familiar with little/big-oh notation? – bjorn93 Dec 21 '19 at 22:10
• No, I just started the college calculus course and proved the limit of x^x and x^x^x by L'Hopital's Rule. Can u tell me the formal name of the function so that I can wiki it. Thx. – 顾泊洋 Dec 21 '19 at 22:17
• @顾泊洋 In that case, I'll post another solution. You can read here: en.wikipedia.org/wiki/Big_O_notation#Little-o_notation but it takes a lot of practice to understand how these asymptotic arguments work. – bjorn93 Dec 21 '19 at 22:35

Let us consider the more general case of a power tower of $$x$$ with $$n$$ entries. Define

$$f_0(x)=x$$

$$f_1(x)=x^x$$

$$f_2(x)=x^{x^{x}}$$

$$f_3(x)=x^{x^{x^{x}}}$$

$$\vdots$$

and so on. So your question is what is

$$\lim_{x\to 0} f_3(x)=?$$

Now, note that the limit does not make sense for real numbers if we approach $$0$$ from the left. As such, we will only consider right sided limits from here on out. We shall show that

$$\lim_{x\to 0^{+}}f_n(x)=\left\{ \begin{array}{ll} 0 & \quad n\text{ even}\\ 1 & \quad n\text{ odd} \end{array} \right.$$

For the base cases, not that it is obviously true for $$n=0$$ and you have already proved it for $$n=1$$ (in fact, you have already proved it for $$n=2$$). Before continuing, we will note a useful recursion for $$f_n(x)$$. That is

$$f_{n+2}(x)=x^{f_{n+1}(x)}=x^{x^{f_n(x)}}$$

Then to prove the inductive step, assume the proposition is true for $$n-1\geq 1$$. For $$n$$ even we have

$$\lim_{x\to 0^{+}}f_n(x)=\lim_{x\to 0^{+}}x^{f_{n-1}(x)}$$

Now, by our assumption

$$\lim_{x\to 0^{+}}f_{n-1}(x)=1$$

as $$n-1$$ is odd. Thus, we can use the continuity of $$f_n(x)$$ to conclude

$$\lim_{x\to 0^{+}}x^{f_{n-1}(x)}=0^1=0$$

Consider the case where $$n$$ is odd. Since $$n-1\geq 1$$ we are assured $$n\geq 2$$. Thus

$$\lim_{x\to 0^{+}}f_n(x)=\lim_{x\to 0^{+}}x^{x^{f_{n-2}(x)}}=\lim_{x\to 0^{+}}\exp\left(x^{f_{n-2}(x)}\ln(x)\right)$$

Since the exponential is continuous, we can move the limit inside to get

$$=\exp\left(\lim_{x\to 0^{+}}x^{f_{n-2}(x)}\ln(x)\right)$$

So we now ask, what is

$$\lim_{x\to 0^{+}}x^{f_{n-2}(x)}\ln(x)=?$$

By our inductive assumption, we know $$f_{n-2}(x)$$ is eventually bounded between $$\frac{1}{2}$$ and $$\frac{3}{2}$$. Thus

$$x^{1/2}\ln(x)\leq x^{f_{n-2}(x)}\ln(x)\leq x^{3/2}\ln(x)$$

However, it is well known that

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

for all $$a>0$$. By the squeeze theorem, this implies

$$\lim_{x\to 0^{+}}x^{f_{n-2}(x)}\ln(x)=0$$

We may finally conclude that

$$\lim_{x\to 0^{+}}f_n(x)=\exp\left(\lim_{x\to 0^{+}}x^{f_{n-2}(x)}\ln(x)\right)=\exp\left(\lim_{x\to 0^{+}}x^{f_{n-2}(x)}\ln(x)\right)=\exp(0)=1$$

and we are done. We conclude $$f_3(x)$$ goes to $$1$$ as $$x$$ goes to $$0$$.