Does $\lim_{n\rightarrow \infty }\sqrt[n]{a_{n}}=\sqrt[n]{\lim_{n\rightarrow \infty }a_{n}}$ hold?

Question

Is it possible to do the following step? Are there any restrictions? I would like to know to not make stupid mistakes using Cauchy's criterion in convergence test. $$\lim_{n\rightarrow \infty }\sqrt[n]{a_{n}}=\sqrt[n]{\lim_{n\rightarrow \infty }a_{n}}$$

Answer 1

No, it is not possible. $n$ is bound to the limit, with your operation, you make it a free variable on the right side.

To be more concrete, a known limit is $\lim_n\sqrt[n]n=1$, which obviously can not work the way you are thinking.

Answer 2

Consider $$ \lim_{n\to\infty}\sqrt[n]{a_n}=\sqrt[\large\color{#C00000}{n}]{\lim_{n\to\infty}a_n}\tag{1} $$ The index $\color{#C00000}{n}$ on the right is not bound by the limit. It is a free variable.

If $(1)$ is true and $\lim\limits_{n\to\infty}a_n$ exists and is not $0$, then $\lim\limits_{n\to\infty}\sqrt[n]{a_n}=1$ and thus, $$ \begin{align} \lim_{n\to\infty}a_n &=\left(\lim_{n\to\infty}\sqrt[n]{a_n}\right)^{\large\color{#C00000}{n}}\\ &=1\tag{2} \end{align} $$ no matter what $\color{#C00000}{n}$ is.