A simple question on limits Does the following equation hold:

$$ \lim_{x\to\infty} x^x = \left(\lim_{x\to\infty} x\right)^{\!x}?$$

I know that

$$ \lim_{x\to\infty} x^n = \left(\lim_{x\to\infty} x\right)^{\!n}$$

holds if $n$ is different from $x$, but I am not sure if it holds where $n=x$.
A notation like

$$\left(\lim_{x\to\infty}x\right)^x$$ is twice meanigless because


$x$ is used both as an independent variable (the exponent) and a dummy variable (argument of the limit), so which is which is ambiguous;
the limit diverges so that raising it to a power is not defined.
limit is similar to function, when you take limit, you let $x$ go to a specific point, or infinity, and you will get a exactly ONE answer if limit exist.( limit is unique if the space is Hausdorff, $R^n$ is Hausdorff).

I will slightly modify your question to :

$\lim_{x\to\infty} (\frac{1}{x}) ^x = (\lim_{x\to\infty} (\frac{1}{x}))^{x}$?

if we calculate the limit, we have   $0=0^x$.

for the right hand side of the equality, without any information on $x$, we cannot compute anything.