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: 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.
A: No, they are not the same.  
We have two expressions here.$$\lim_{x\to\infty} x^x$$ 
and $$ \left(\lim_{x\to\infty} x\right)^{\!x}$$
The first expression makes perfect sense and the answer is $$\lim_{x\to\infty} x^x =  \infty $$
The second expression is ambiguous.  
You have   $$\lim_{x\to\infty} x =\infty$$ and you want to raise it to power of  $x$.
We have to know what is $x$ here to make an inference about $(\infty )^x.$
Of course we get a different answer for positive $x$ and for negative $x$.
Thus $$ \left(\lim_{x\to\infty} x\right)^{\!x}$$ depends on $x$.
A: 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.
