Thomson et al. provide a proof that $\lim_{n\rightarrow \infty} \sqrt[n]{n}=1$ in this book (page 73). It has to do with using an inequality that relies on the binomial theorem:
I have an alternative proof that I know (from elsewhere) as follows.
Proof.
\begin{align}
\lim_{n\rightarrow \infty} \frac{ \log n}{n} = 0
\end{align}
Then using this, I can instead prove:
\begin{align}
\lim_{n\rightarrow \infty} \sqrt[n]{n} &= \lim_{n\rightarrow \infty} \exp{\frac{ \log n}{n}} \newline
& = \exp{0} \newline
& = 1
\end{align}
On the one hand, it seems like a valid proof to me. On the other hand, I know I should be careful with infinite sequences. The step I'm most unsure of is:
\begin{align}
\lim_{n\rightarrow \infty} \sqrt[n]{n} = \lim_{n\rightarrow \infty} \exp{\frac{ \log n}{n}}
\end{align}
I know such an identity would hold for bounded $n$ but I'm not sure I can use this identity when $n\rightarrow \infty$.
Question:
If I am correct, then would there be any cases where I would be wrong? Specifically, given any sequence $x_n$, can I always assume:
\begin{align}
\lim_{n\rightarrow \infty} x_n = \lim_{n\rightarrow \infty} \exp(\log x_n)
\end{align}
Or are there sequences that invalidate that identity?
(Edited to expand the last question)
given any sequence $x_n$, can I always assume:
\begin{align}
\lim_{n\rightarrow \infty} x_n &= \exp(\log \lim_{n\rightarrow \infty} x_n) \newline
&= \exp(\lim_{n\rightarrow \infty} \log x_n) \newline
&= \lim_{n\rightarrow \infty} \exp( \log x_n)
\end{align}
Or are there sequences that invalidate any of the above identities?
(Edited to repurpose this question).
Please also feel free to add different proofs of $\lim_{n\rightarrow \infty} \sqrt[n]{n}=1$.
Best Answer
Here is one using $AM \ge GM$ to $1$ appearing $n-2$ times and $\sqrt{n}$ appearing twice.
$$\frac{1 + 1 + \dots + 1 + \sqrt{n} + \sqrt{n}}{n} \ge n^{1/n}$$
i.e
$$\frac{n - 2 + 2 \sqrt{n}}{n} \ge n^{1/n}$$
i.e.
$$ 1 - \frac{2}{n} + \frac{2}{\sqrt{n}} \ge n^{1/n} \ge 1$$
That the limit is $1$ follows.