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

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.

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 $n$.