Prove that $ \lim_{x\to 0}x^x=1$.

Question

Prove that $\displaystyle \lim_{x\to 0}x^x=1$ for $\delta-\epsilon.$

We have to $|x|<\delta_{\epsilon}$ what will be chosen later.

$$\begin{align*}|x^x-1|<\epsilon &\Leftrightarrow -\epsilon<x^x-1<\epsilon\\ &\Leftrightarrow -\epsilon+1<x^x<\epsilon+1\\ &\Leftrightarrow \ln(-\epsilon+1)<x\ln x<\ln(\epsilon+1)\end{align*}$$

Hence I do not know how to continue, i.e. how to choose my $ \delta $. Someone help me, thanks.

Are you allowed to assume knowledge of the limit $\lim_{x\to -\infty}x^2e^x=0$ in your argument? — Alann Rosas, Sep 01 '21 at 07:36
If you can find a way to prove that $\left|\ln(x)\right|<\frac{1}{\sqrt{x}}$ for every sufficiently small $x>0$, a suitable $\delta$ will immediately follow from multiplying the inequality by $x$ and bounding the resulting right side (it's simply $\sqrt{x}$) by $\varepsilon$. — Alann Rosas, Sep 01 '21 at 07:44
@AlannRosas You can prove that, if I'm not wrong for $ x <2 $. So for $ 0 <x <2 $, that inequality is true. And it would be. — Andrew Madison, Sep 01 '21 at 07:49
That's correct. If you decide to go with this approach, the hard part of the proof will be proving this inequality. I'll spend some time on the problem to see if I get any clever ideas. — Alann Rosas, Sep 01 '21 at 07:53
Oh sure, I'll try to test that, with derivatives to see what I can get to. And if I already try it, then I fix the x <2 and I would have my proof complete. Thank you :). @AlannRosas — Andrew Madison, Sep 01 '21 at 07:58
For what it's worth, I tried using $\epsilon, \delta$ to show that $0 = \lim_{x \to 0^+} x\log x = ~$ [by Taylor Series on $~ \log (1 + u) ~] ~\lim_{y \to 1^-} (1-y)(-y - y^2/2 - y^3/3 - \cdots).$ Unfortunately this telescopes to $(-1) + [1 - 1/2] + [1/2 - 1/3] + \cdots = 0$, so I never get a chance to use $\epsilon, \delta$. — user2661923, Sep 01 '21 at 08:19
@AndrewMadison FYI, questions here which deal with solving your requested limit using techniques other than $\delta-\epsilon$ include What is the limit of $x^x$ as $x$ approaches $0$ ($\lim\limits_{x \to 0}x^x$)?, Different way solving limit $\lim \limits_{ x\rightarrow 0 }{ { x }^{ x } } $ and Tetration convergence: prove $\lim_{x\rightarrow0} {}^{n}x = \begin{cases} 1, & n \text{ even} \ 0, & n \text{ odd} \end{cases}$. — John Omielan, Sep 01 '21 at 08:37
Please take a look at this for an epsilon delta proof: https://math.stackexchange.com/questions/4113279/is-this-computation-of-lim-limits-x-to-0-xx-rigorous-enough — Koro, Sep 01 '21 at 08:46

JuanCarlos62 · Answer 1 · 2021-09-02T10:53:30.730

The solution involves the Lambert function. See https://en.wikipedia.org/wiki/Lambert_W_function

First of all, $x^x$ is defined for $x>0$ and has a minimum at $x=1/e$

Secondly the right inequality of the expression $1-\epsilon < x^x <1+\epsilon$ is true as long as if $\delta \leq 1$, and so it is true also for $\delta \leq 1/e < 1$

We focus on the left inequality and call $k=1-\epsilon \leq 1$ for simplicity.

We are looking for a value $\delta \leq 1/e$ such as $\delta^\delta >k$, and since $x^x$ decreases when $x \leq 1/e$ it suffices look for a $\delta$ that $\delta ^\delta = k$

The Lambert function is defined as $W(x)=f^{-1}(x)$ where $f(x)= x \cdot e^x$ and its domain is $[-1/e, \infty)$

Now we have some properties:

(1) $W(f(x))=f^{-1}(f(x))=x \hskip{1cm}$So $W(f(x))=W(xe^x)=x$

(2) $f(W(x))=f (f^{-1}(x))=x \hskip{1cm}$ So $x =f(W(x))=W(x)e^{W(x)} \Rightarrow e^{W(x)}=\frac{x}{W(x)}$

Now we focus in our search: $\delta^\delta = k \Leftrightarrow \delta \ln \delta = \ln k \Leftrightarrow e^{\ln \delta}\ln \delta =\ln k$

And applying Lambert: $W(e^{\ln \delta}\ln \delta) = W(\ln(k))$

If $z=\ln \delta$, the previous equation translates into $W(ze^z)=W(\ln(k))$

And since $W(ze^z)=z$ (prop 1), we have $\ln \delta = z=W(\ln(k)) \Rightarrow \delta = e^{W(\ln(k))}$.

But $e^{W(b)}=\frac{b}{W(b)}$ (prop 2)$ \Rightarrow \delta =\frac{\ln k}{W(\ln k)}$

So the solution is:

$\delta = \min \{1/e, \frac{\ln k}{W(\ln k)}\}$ where $k=1-\epsilon$

user2661923 · Answer 2 · 2021-09-02T05:50:28.557

As suggested, I scrapped my original answer in favor of an $\delta,\epsilon$ approach, where L'Hopital's rule is not used.

I am re-interpreting the problem to be :
To prove: $\lim_{x \to 0^+} x^x = 1.$
This is because, for any $\delta > 0$, there exists an $x$ such that $-\delta < x < 0$ and $x^x$ is undefined in Real Analysis.

For $~x > 0,~$ let $~f(x) = x\log x~$ and let $~g(x) = x^x.$
Then $~g(x) = e^{[f(x)]}.$

$\underline{\text{Lemma 1 :}}$
For $~0 < x < \dfrac{1}{e}~, f(x)~$ is a strictly decreasing function.
Proof:
$f'(x) = 1 + \log(x) < 1 + (-1) < 0.$
Thus, $f(x)$ is strictly decreasing throughout the interval.

$\underline{\text{Lemma 2 :}}$
Let $~\langle a_n\rangle~$ be defined by $a_n = \dfrac{3^n}{n} ~: ~n \in \Bbb{Z^+}.$
Then $~\langle a_n\rangle~$ is a strictly increasing sequence that grows unbounded.
Proof:
$\displaystyle \frac{a_{n+1}}{a_n} = \frac{3}{(n+1)/n} = \frac{3n}{n+1} > 1 ~: ~\forall n \in \Bbb{Z^+}.$
Therefore, $~\langle a_n\rangle~$ is a strictly increasing sequence.
Further, for $\displaystyle n \geq 3, \frac{3}{(n+1)/n} > \frac{3}{3/2} = 2.$
Therefore, for $\displaystyle n \geq 3, \frac{a_{n+1}}{a_n} > 2$.
Therefore, for $\displaystyle n \geq 4, a_n > 2^{(n-3)}a_3 = 2^{(n-3)} \times 9.$
Clearly, as $n \to \infty, 2^{(n-3)}$ grows unbounded.

$\underline{\text{Lemma 3 :}}$
$x < y \implies e^x < e^y.$
Proof:
$h(x) = e^x \implies h'(x) = e^x > 0.$
Thus, $h(x)$ is a strictly increasing function.

The challenge is that given any $\epsilon > 0,$
$\delta > 0$ must be found so that
$0 < x < \delta \implies 0 \leq |g(x) - 1| < \epsilon.$
This will be true $\iff 1 - \epsilon < g(x) < 1 + \epsilon.$

In constructing a $\delta, \epsilon$ proof, $\epsilon$ will be arbitrarily restricted to be less than $1$, so that $0 < (1 - \epsilon) < 1.$

This is justified, since if a suitable $\delta$ can be found for a specific $\epsilon_1 < 1$, such a $\delta$ would also be suitable for any $\epsilon > \epsilon_1$.

Given $\epsilon$ such that $0 < \epsilon < 1$, use the following procedure to construct $\delta$:

By Lemma 2, $n \in \Bbb{Z^+}$ may be chosen so that

$$\frac{n}{3^n} |\log(1/3)| < |\log (1-\epsilon)|.\tag 1$$

Choose $\displaystyle \delta = \frac{1}{3^n} \implies \delta \times |\log(\delta)| = \frac{1}{3^n} \times n |\log (1/3)|.$

Note that for any $x$ such that $0 < x \leq \delta$,
$\log (x) < 0 \implies x \log x < 0 \implies g(x) = x^x < 1 < (1 + \epsilon).$

Further, by (1) above, $\delta |\log (\delta)| < |\log(1-\epsilon)| \implies \delta \log (\delta) > \log(1 - \epsilon)$.

Also, since $\displaystyle \delta = \frac{1}{3^n} \leq \frac{1}{3} < \frac{1}{e}$,
Lemma 1 guarantees that for all $x$ such that $0 < x < \delta, ~~$ you will have that $~~x \log (x) > \log (1-\epsilon).$

Invoking Lemma 3, this implies that $ g(x) = x^x > 1 - \epsilon.$

I know how to do it for L'H. But, I would like to know some way how to do it for $ \delta- \epsilon $. And well, I can't find a way. Thank u :). — Andrew Madison, Sep 01 '21 at 07:42
It may be easier, but the OP explicitly asked for the use of the definition. — PierreCarre, Sep 01 '21 at 07:46

Prove that $ \lim_{x\to 0}x^x=1$.

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