Compute $\lim\limits_{x\to - \infty} x^x$

Question

How to compute $$ \lim\limits_{x\to - \infty} x^x? $$

My thoughts：

This is an $\infty^{\infty} $ indeterminate form, but I don't know how to approach it. I have thought about writing $x^x$ as $e^{x\ln x} $, but the problem is that $x \to - \infty$, so I have no clue how to compute it.

Updated context
Edit : the symbol $x$ is an integer. Thanks for pointing out that otherwise the function doesn't take real values.

Could we not view it as the limit of x to ∞ of $(-x)^{-x}$? Then it's just the limit of $e^{i \pi x} x^{-x}$, which would be $0$, since the first factor is bounded? — Ryan Goulden, Mar 07 '19 at 15:56
It doesn't exist, unless you consider the limit over the integers. — learner, Mar 07 '19 at 15:56
@maveric I know it is out of $\ln$'s domain, that' s why I said my approach doesn't work. Yet, I wanted to show the people here that I had actually thought about the question before posting. — MathEnthusiast, Mar 07 '19 at 15:57
See https://math.stackexchange.com/questions/394110/can-the-graph-of-xx-have-a-real-valued-plot-below-zero. — Michael Hoppe, Mar 07 '19 at 16:00
Because the function isn't even real for most $x$, except for the integers for which the function is real and approaches $0$ as $x\to\inf\Bbb Z$ — learner, Mar 07 '19 at 16:01
Thank you, I hadn't realised it didn't work! How about this limit over the integers? I will edit my question. — MathEnthusiast, Mar 07 '19 at 16:02
@learner thanks, I have just edited my question. Could you supply a proof for the result in your comment? — MathEnthusiast, Mar 07 '19 at 16:04
@MathEnthusiast Is $(-\pi)^\pi = \pi^\pi $ or $- \pi^\pi$? In case of odd/even integers you can clarify this, but here we have an issue. — Sarvesh Ravichandran Iyer, Mar 07 '19 at 16:04
Yes, indeed, now I see. Sorry for my initial mistake, we shall consider $x$ to be an integer from now on. — MathEnthusiast, Mar 07 '19 at 16:05
@MathEnthusiast It is good to clarify this! Note that $(-n)^{-n} = (-1)^{-n} n^{-n}$ so $|(-n)^{-n}| = |n^{-n}|$ for all $n > 0$. Now try to use the squeeze theorem. — Sarvesh Ravichandran Iyer, Mar 07 '19 at 16:11

score 2 · Answer 1 · answered Mar 07 '19 at 16:10

2

If $x\ll 0$ is an integer much lower than $0$, then $x^{-1}$ is very close to $0$, and $x^x$ is even closer to $0$. So the we get a determinate form, and the limit is $0$.

answered Mar 07 '19 at 16:10

Arthur

199,419

score 2 · Accepted Answer · answered Mar 07 '19 at 18:15

2

We need not restrict $x$ to be an integer. Note that for $x\in \mathbb{R}$, with $x<0$, we have

$$\begin{align} |x^x|&=\left| e^{x\log(x)}\right|\\\\ &=\left| e^{-|x|(\log(|x|)+i(2k+1)\pi)}\right|\\\\ &=\frac1{|x|^{|x|}} \end{align}$$

Inasmuch as the magnitude of $x^x$ approaches $0$ as $x\to -\infty$, we find that

$$\lim_{x\to-\infty}x^x=0$$

answered Mar 07 '19 at 18:15

Mark Viola

179,405

Can you please explain your 2nd step – M Desmond Mar 07 '19 at 18:34
1

@MDesmond Sure. The complex logarithm is multivalued and given by $$\log(z)=\log(|z|)+i(\arg(z)+2k\pi)$$with $\log(|z|)$ interpreted as the logarithm for real numbers and $k\in \mathbb{Z}$. Now for $x<0$, $x=-|x|$ and $\arg(-1)=e^{i(2k+1)\pi}$. – Mark Viola Mar 07 '19 at 21:44
@Mark Viola your answer is indeed very good. I actually forgot about this question, thank you for reminding me. – MathEnthusiast Mar 27 '19 at 15:06
You're welcome. My pleasure! – Mark Viola Mar 27 '19 at 15:06

score 1 · Answer 3 · answered Mar 07 '19 at 16:16

Let $x=-n$ where $n\in\Bbb N$.

$$x^x=(-n)^{-n}=\frac{(-1)^n}{n^n}$$

This is an alternating sequence with the denominator $n^n$ approaching $\infty$ as $n\to\sup\Bbb N=\infty$, so the reciprocal approaches $0$.

A more rigorous approach would be to use the $\epsilon$-$\delta$ definition though.

Compute $\lim\limits_{x\to - \infty} x^x$

3 Answers3