Why does this limit exist $x^{x}$

Question

I was wondering what is the value of $\lim\limits_{x\to 0+} x^{x}$.

Assuming that the limit exists, I could show using the usual logarithm techniques that the limit is $1$. However, I am not able to show that the limit exists. Could some one help on that?

EDIT: It would be so nice if we could do it without l'Hopital...

Maple helps you with it by the $ Student[Calculus1][LimitTutor](); $ command. See the output in the screen. — user64494, Aug 22 '13 at 13:13
BTW, the l'Hopital rule also bases the existence of the limit $\lim_{x\to a+} \frac {f(x)} {g(x)}$ if the limit $\lim_{x\to a+} \frac {f'(x)} {g('x)}$ exists. — user64494, Aug 22 '13 at 15:23
Considering all the comments together, "ruler and compass" is a way of saying that the restriction to do it without L'Hopital is (in one user's opinion) unnatural, unnecessary, undesirable, or un-somethingelse. That is one answer to the question about this on the meta site. — zyx, Aug 28 '13 at 14:54
Does this answer your question? How do I prove that $\lim_{x\to0^+} x\cdot\ln x=0$ — Anne Bauval, Mar 05 '24 at 11:37

score 7 · Answer 1 · 2013-08-22T13:25:08.343

7

By the l'Hôpital theorem we have

$$\lim_{x\to0}x\log x=\lim_{x\to0}\frac{\log x}{\frac 1 x}=-\lim_{x\to0}\frac{\frac 1 x}{\frac{ 1 }{x^2}}=-\lim_{x\to0}x=0$$

and conclude with $$x^x=e^{x\log x}$$

Edited To find the limit without the l'Hôpital theorem: for $0<x<1$: $$ 0\leq|x\log x|=x\int_x^1\frac{dt}{t}\leq x\int_x^1\frac{dt}{t^{3/2}}=-2x\frac{1}{\sqrt{t}}\Big|_x^1\to0$$

edited Aug 22 '13 at 13:25

answered Aug 22 '13 at 13:06

1

That is a nice answer. But can we do it without recourse to l'Hopital? – Vishal Gupta Aug 22 '13 at 13:09
@ Vishal: What for? – user64494 Aug 22 '13 at 13:16
1

@Vishal I edited my answer. – Aug 22 '13 at 13:25

score 6 · Accepted Answer · edited Aug 22 '13 at 14:26

6

Hint: Using the change of variable $x=e^{-t}$ show that $$\lim_{x \to 0+} x \ln x = 0.$$

$$\lim_{x \to 0+} x \ln x = \lim_{t \to +\infty} -\frac{t}{e^t} = 0.$$

To prove the last equality you can use the Taylor expansion for $e^t$, for instance.

edited Aug 22 '13 at 14:26

Vishal Gupta

7,132

answered Aug 22 '13 at 13:02

njguliyev

14,543

1

Using Taylor expansion is basically L'Hopital. Which OP wants to avoid if possible. – Arthur Aug 22 '13 at 13:17
2

He added that condition after my answer. Then he can prove $\frac{t}{e^t} \to 0$ first for natural numbers (e.g., using binomial formula) then extend it to reals. – njguliyev Aug 22 '13 at 13:22
2

This makes perfect sense as for me, $e^{x}$ is defined as the series expansion. – Vishal Gupta Aug 22 '13 at 13:22

score 4 · Answer 3 · answered Aug 28 '13 at 23:15

Assume you know $$\lim_{n \rightarrow \infty}\sqrt[n]{n}=1$$ Proof:

$$1 \leq \sqrt[n]{n}<\frac{n-2+2\sqrt n}{n}, \,\,\,\,\,\,\,\,\,\, \mathrm{GM \leq AM}$$

Now for each $x$ near zero define $n(x)=[\frac{1}{x}]$ where $[\cdot]$ is the integral part function. Let's call $n(x)$ out of simplicity $n$.

Therefore \begin{align} &\frac{1}{n} \geq x \geq \frac{1}{n+1} \frac{}{} \Rightarrow\\ \left (\frac{1}{n} \right )^\frac{1}{2n}\geq &\left (\frac{1}{n} \right )^{\frac{1}{n+1}} \geq x ^{x}\geq \left (\frac{1}{n+1} \right )^\frac{1}{n} \geq \left (\frac{1}{2n} \right )^\frac{1}{n}& \end{align}

And because as $x$ tends to zero $n(x)$ tends to infinity taking limits at both sides you are done.

Why does this limit exist $x^{x}$

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