I’m unsure and lack the tools to find out if $\lim_{x \to \infty} 0^{\frac{1}{x}}$ is 0 or 1 or is it an invalid limit. One way $0^0$ feels like it should give 1 but then again, the function is 0 for all $x$. So, how do we talk about such a limit? Also, does the limit remain the same when we replace $x$ by the natural number $n$. Please provide proper reasoning in your answers.
Note: Wolfram Alpha says that it is zero.
On
For any $x\in\Bbb R^+$, we have $0^{1/x}=0$. This means $\lim_{x\to\infty}0^{1/x}=0$. Now, one thing to point out about this is that the base of the exponential is literally $0$ (not just something that approaches $0$). If you were to ask something like
Given $f(x)\to0^+$ as $x\to\infty$, is $\lim_{x\to\infty}(f(x))^{1/x}=0$?
The answer to this question depends on the nature of $f(x)$ (specifically, how quickly it converges to $0$).
This function is identically zero in $(0,\infty)$. For any positive $x$ we have $0^{\frac{1}{x}}=0$. So of course the limit in that case is zero, you can even easily prove this by the definition of the limit.