# Did I compute this limit correctly?

I need to find $\lim_{x\rightarrow0^+} x^{x^x}$. Here is my solution.

Let $u=x^x$ $\lim_{x \rightarrow 0^+}x^u$

$\lim_{x \rightarrow 0^+}x^x$=$\lim_{x \rightarrow 0^+}e^{x\text{ln}(x)}$

Using L'Hopital's rule, I can show that $\lim_{x \rightarrow 0^+}x\text{ln}(x)=0$, thereore $\lim_{x \rightarrow 0^+}e^{x\text{ln}(x)}=1=\lim_{x \rightarrow 0^+}u$

Thus, the first limit is $\lim_{x \rightarrow 0^+}x^1=0$

So the limit is 0. Did I get this correct?

• SHWEET! Thanks a liot Commented Jun 13, 2014 at 17:51
• This depends on how you define $x^{x^x}$. $x^{(x^x)}$ gives your answer, but $(x^x)^x$ gives another answer. Commented Jun 13, 2014 at 17:52
• the question was simply $x^{x^x]$. No parentheses Commented Jun 13, 2014 at 17:58
• You can however get $\lim_{x\rightarrow 0^+} (x^x)^x = 1$ directly by looking at $u^x \rightarrow 1^0 = 1$ instead.
– mlk
Commented Jun 13, 2014 at 17:58
• I would assume that the expression represents $x^{(x^x)}$ since $(x^x)^x$ could be written more simply as $x^{x^2}$. Commented Jun 13, 2014 at 18:04

I'd say you got the right answer for the wrong reasons. You cannot treat the exponent as a separate variable $u$ which approaches $1$ independently of $x \to 0^+$.

In simple terms, every $x$ must approach $0$ at the same time. To illustrate this fact, consider

$$\lim_{x \to 0} \frac{x}{x}$$ This is (clearly!) equal to one, but if we let $u=x$, write $\lim_{u \to 0} u/x$ and let $u \to 0$ keeping $x$ fixed, then the limit is $0/x = 0$... What?

The problem here is that it is impossible to let $u \to 0$ while $x \not\to 0$. The variable $u$ depends entirely on $x$! Similarly, in your question to find that $u \to 1$ you are taking $x \to 0^+$, but then you still have $x$ approaching zero later. This does not, in general, produce valid results.

Try using your $a = e^{\ln(a)}$ trick twice. You should come up with a messy, but tractable limit which evaluates to $0$.

Here's a big hint:

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

• OK, thanks! I will try that. Commented Jun 13, 2014 at 17:52
• You actually can do the two limits seperate in this case, since $u\rightarrow 1$ and $x^y$ is continuous in the point $(0,1)$.
– mlk
Commented Jun 13, 2014 at 17:57
• so I basically have to compute the limit of $$e^{ln(x)}^{e^xln(x)}$$ Commented Jun 13, 2014 at 18:02
• The answer is 0, as per Maple and MATLAB. I got 0 as the answer after using your hint. Thanks Commented Jun 13, 2014 at 18:34

Your thinking is correct. I would be more careful to write some of the stuff there, though, like "$\lim_{x \to 0^+} u$", for example. Your new variable is $u$, so instead of that, it would be better to put "$\lim_{u \to 1} u = 1$", which is pretty much useless, since you already found out that $\lim_{x \to 0^+} x^x = 1$. Other thing that bothered me a little: "$\lim_{x \to 0^+} x^1$", in the end. See that $x^x$ is not $1$ in general. To justify it better, I would only write, informally: $\lim_{x \to 0^+} x^{x^x} = 0$, because $x \to 0$ and $x^x \to 1$, and $0^1 = 0$.

Well you seem to apply the rule that if $\lim_{x \to a}f(x) = A, \lim_{x \to a}g(x) = B$ then $\lim_{x \to a}\{f(x)\}^{g(x)} = A^{B}$. This is valid but only when $A > 0$ and $B$ is finite. Here $f(x) = x$ and $g(x) = x^{x}$ and clearly then $A = 0, B = 1$ so the rule can't be applied directly.

Rather we just proceed by taking logs. Consider the expression $v = x^{x}\log x$. As mentioned in the question $x^{x} \to 1$ as $x \to 0$ and we know that $\log x \to -\infty$ as $x \to 0$ so that $v = x^{x}\log x \to -\infty$ as $x \to 0$. It follows that $x^{x^{x}} = e^{v} \to 0$ as $v \to -\infty$.