TITLE: Solve $\lim_{x\to +\infty}\frac{x^x}{(\lfloor x \rfloor)^{\lfloor x \rfloor }}$
QUESTION [13 upvotes]: 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

REPLY [2 votes]: 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 $n$.