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.)


2 Answers 2


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






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, note 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


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 $0$.


$$x^x=\exp(\log(x)x)=1+x\log(x)+o(x\log(x))$$ $$\begin{align} x^{x^x}&=\exp\left(\log(x)x^x\right)=\exp\left(\log(x)\big[1+x\log(x)+o(x\log(x))\big]\right)\\ &=\exp\left(\log(x)+x\log^2(x)+o\left(x\log^2(x)\right)\right) \\ &=x\exp\left(x\log^2(x)+o\left(x\log^2(x)\right)\right)=x(1+o(1)) \end{align}$$ 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 $(x^{x^{x}})^{-1}$

  • $\begingroup$ Sorry, what is the o function? $\endgroup$ Commented Dec 21, 2019 at 22:09
  • $\begingroup$ @顾泊洋 You're not familiar with little/big-oh notation? $\endgroup$
    – bjorn93
    Commented Dec 21, 2019 at 22:10
  • $\begingroup$ 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. $\endgroup$ Commented Dec 21, 2019 at 22:17
  • $\begingroup$ @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. $\endgroup$
    – bjorn93
    Commented Dec 21, 2019 at 22:35

Not the answer you're looking for? Browse other questions tagged .