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Determine if the following limits exist

$$\lim_{x\to +\infty}\dfrac{x^x}{(\lfloor x \rfloor)^{\lfloor x \rfloor }}$$

note that

$\lfloor x \rfloor \leq x < \lfloor x \rfloor + 1 \implies x-1 <\lfloor x \rfloor \leq x$

and $x^x=\exp(x\log x)$

$$\dfrac{1}{x^{\lfloor x \rfloor}} \le \dfrac{1}{\lfloor x \rfloor}< \dfrac{1}{(x-1)^{\lfloor x \rfloor}}$$

$$\dfrac{x^x}{x^{\lfloor x \rfloor}} \le \dfrac{x^x}{\lfloor x \rfloor}< \dfrac{x^x}{(x-1)^{\lfloor x \rfloor}}$$

Edit

Let $f(x)=\dfrac{x^x}{(\lfloor x \rfloor)^{\lfloor x \rfloor }}$, already $x^x$ is defined only for $x\geq 0$, then $f$ is defined on $\mathbb{R}^{*}_{+}$,

  • Case 1: $x\in \mathbb{N}^{*}$ $$\dfrac{x^x}{(\lfloor x \rfloor)^{\lfloor x \rfloor }}=1$$

    So a possible limit may be $1$

  • Case 2: otherswises

if we choose $x=n+0,5$, we have

\begin{align*} f(x)&=\exp((n+0,5)\log(n+0,5)-n\log(n)\\ &=\exp(n(ln(n+0,5)-ln(n))+0,5ln(n+05))\\ &\geq \exp(0,5\log(n+05)) \end{align*}

or $\exp(0,5\log(n+05))\to \infty$ when $x\to \infty$

  • $\forall\quad 0<\epsilon<1$, let $W_n=f(n+\epsilon)$

we have $$ \begin{align*} ln(W_n)&=(n+\epsilon)ln(n+\epsilon)-nln\\ &=n(ln(n+\epsilon)-ln(n))+\epsilon ln(n+\epsilon)\\ &\geq \epsilon ln(n+\epsilon) \end{align*} $$

then $$\lim_{x\to +\infty}W_n=+\infty $$

thus $$\lim_{x\to +\infty}f(x)=+\infty $$

therefore the function $f$ does not have a real limit as $x$ tends to infinity

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Let $f(x)=\frac{x^x}{\left ( \lfloor x \rfloor \right )^{\lfloor x \rfloor}}$. If $\lim_{x \to \infty} f(x)$ exists, it must be $1$, because we can go to $\infty$ along the sequence $x_n = n$ where $f(x_n)=1$. On each interval $[n,n+1)$, the furthest $f$ gets from $1$ will occur near $n+1$. More precisely, we can make $f$ be arbitrarily close to but less than $\frac{(n+1)^{n+1}}{n^n}$ by taking $x$ arbitrarily close to $n+1$. Now let's do some algebra:

$$\frac{(n+1)^{n+1}}{n^n} = (n+1) \frac{(n+1)^n}{n^n} = (n+1) \left ( 1 + \frac{1}{n} \right )^n.$$

So the graph of $f$ looks like this. $f(n)=1$ for each positive integer $n$. As $x$ approaches $n+1$, $f(x)$ increases to almost $(n+1)e$. Then the graph drops back to $1$ at $n+1$, etc. So the limit doesn't exist; in fact, $f$ is not even bounded.

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  • $\begingroup$ could you replace the part of graph with analytic part $\endgroup$ – Educ Jan 9 '15 at 8:31
  • $\begingroup$ @Educ It is pretty much analytic. The only non-analytic part is explaining what "almost $(n+1)e$" means. One precise answer would be that $f$ exceeds $n+1$ on each interval $[n,n+1)$ before dropping back to $1$. This works because $(1+1/n)^n>1$. Hence there is no $M$ such that $|f(x)-1|<1$ for $x>M$, so the limit isn't $1$. Then the first comment tells you the limit could only be $1$, so you're done. $\endgroup$ – Ian Jan 9 '15 at 15:48
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Let $x=n+a$, with $n\in\mathbb N$ and $a\in[0,1)$. If $a>0$ then

$$\dfrac{x^x}{\lfloor x\rfloor^{\lfloor x\rfloor}}=\biggl(1+\dfrac an\biggr)^n\biggl(1+\dfrac an\biggr)^a\,n^a\,,$$

whose limit when $n$ tends to $\infty$ is equal to $e^a\cdot1\cdot\infty=\infty$; on the other hand, if $a=0$ then the expression is constant equals to $1$ for all $n$.

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Let $\epsilon>0$ be fixed and tiny, and $x$ is an integer plus $\epsilon$.

$$\frac{\frac{x^x}{(x-\epsilon)^{(x-\epsilon)}}}{xe^{\epsilon}}=\frac{x}{xe^{\epsilon}(1-\frac{\epsilon}{x})^{\frac{x}{\epsilon}\frac{\epsilon(x-1)}{x}}}\rightarrow 1\text{ as }x\to\infty$$

Since $xe^{\epsilon}\to\infty$ as $x\to \infty$, we get that

$$\frac{x^x}{\lfloor x\rfloor^{\lfloor x\rfloor}}=\frac{x^x}{(x-\epsilon)^{(x-\epsilon)}}\to\infty$$

If $y$ is an integer then

$$\frac{y^y}{\lfloor y\rfloor^{\lfloor y\rfloor}}=1.$$

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