4
$\begingroup$

Given a sequence $\{x_n\}_{n=1}^\infty$ of positive numbers, satisfying $\lim_{n \to \infty} x_n^{x_n} = 4$, prove $\lim_{n\to\infty}x_n=2$.
My attampt:
from $\lim_{n \to \infty} x_n^{x_n} = 4$ we know $\exists n_0\in\mathbb{N}:\forall n\ \geq n_0: |x_n^{x_n} -4| < \epsilon$ so
$|x_n-2| = \frac{|x_n^2-4|}{|x_n+2|} \leq |x_n^2-4| \leq |x_n^{x_n} -4| <\epsilon$
Which completes the proof if correct. Is it?

$\endgroup$
3
  • 5
    $\begingroup$ I don't think it is: why $|x_n^2 - 4| \leq |x_n^{x_n} - 4|$? $\endgroup$
    – Siméon
    Mar 6, 2015 at 12:18
  • 1
    $\begingroup$ Do you use the fact that $4$ has only one inverse image under the map $x \mapsto x^x$? Suppose you knew instead that $x_n^{x_n}$ tended to $1$, then you could have the sequence $1,\frac12,1,\frac13,1,\frac14,1,\frac15,\ldots$ which is divergent. $\endgroup$ Mar 6, 2015 at 12:19
  • $\begingroup$ So what are you suggesting? $\endgroup$
    – user114138
    Mar 6, 2015 at 12:19

4 Answers 4

4
$\begingroup$

First of all, there exists $N$ such that $x_n \geq 1$ for all $n \geq N$ since $x^x = e^{x\ln x} \leq 1$ for $x \leq 1$. This restriction allows us to use the following fact from which the result is obvious.

$|x^x - 4| \geq 3|x - 2|$ for all $x \geq 1$.

Proof. Let $\delta = x - 2$.

  • If $\delta \geq 0$, then $x^x - 4 \geq x^2 - 4 = 4\delta + \delta^2 \geq 3\delta = 3|x-2|$.
  • If $-1 \leq \delta \leq 0$, then $x^x - 4 \leq x^2 -4 = 4\delta + \delta^2 \leq 3\delta = - 3|x-2|$.
$\endgroup$
0
$\begingroup$

To simplify, We show that if $x_n>0$ and $y_n=x_n\log x_n\to 2\log 2$, then $x_n\to 2$. The study of the function $x\log x$ on $(0,+\infty)$ show that for $n$ large, we have $x_n\geq 1$. Now the function $g(x)\log x$: $[1,+\infty[\to [0,+\infty[$ is continuous, strictly increasing and onto (hence bijective). Let $h$ the reciprocal of $g$, ($h(g(x)=x)$; $h$ is a continuous function from $[0,+\infty[$ to $[1,+\infty[$, and we have $x_n=h(g(x_n))=h(y_n)$. As $y_n\to 2\log 2$, we get $x_n=h(y_n)\to h(2\log 2)=2$.

$\endgroup$
0
$\begingroup$

Here's part of one solution. You'll need to continue the argument to complete the proof. Proof by contrapositive.

Suppose that $\lim_{x\rightarrow\infty}x_n\not=2$. Then, there exists some $\epsilon$ such that for all $N\in\mathbb{N}$, there is some $n>N$ such that $|x_n-2|>\epsilon$.

Choose $\epsilon$ so that the definition of the limit fails. Now, assume that for infinitely many $n$, $x_n-2>\epsilon$ (you need to do the case where $x_n-2<-\epsilon$). Observe that $x_n>2+\epsilon$.

Since $f(y)=y$ and $g(z)=\ln(z)$ are increasing functions, consider $x_n\cdot\ln(x_n)>(2+\epsilon)\cdot\ln(2+\epsilon)$. Since $2+\epsilon>2$, it follows that $x_n\cdot\ln(x_n)>(2+\epsilon)\cdot\ln 2=2\cdot\ln 2+\epsilon\cdot\ln 2$.

Since $x_n\cdot \ln(x_n)>2\cdot\ln 2+\epsilon\cdot\ln2$, we can take the exponential of both sides to get $x_n^{x_n}>4\cdot2^\epsilon$. Therefore, $x_n^{x_n}-4>4(2^\epsilon-1)$, which shows that $x_n^{x_n}$ doesn't converge to $4$.

$\endgroup$
0
$\begingroup$

The function $f: \left[ 0,\infty \right) \to \left[ 0,\infty \right)$ given by $f(x)=x^x$ is continuous but not monotone on that interval. For example it attains the value $\frac57$ twice, see Wolfram Alpha's solution to $x^x=\frac57$ (they find $x\approx 0.530036$ and $x\approx 0.226742$).

You should restrict $f$ to an interval on which it is injective. Then argue that the inverse $f^{-1}$ of this restriction of $f$ is continuous. Then use the usual connection between continuous functions and convergent sequences. (You show that if necessary you can skip a finite number of terms from $\{ x_n \}$ to assume that $\{ x_n \}$ is within the required domain.)

$\endgroup$

You must log in to answer this question.

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