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