Show that $\lim_{n \rightarrow \infty} \sqrt[n]{a} =1$ This was a practice problem I was given. 

Show that $\lim_{n \rightarrow \infty} \sqrt[n]{a} =1$.

My argument is as follows:
\begin{align}
\lim_{n \rightarrow \infty} \sqrt[n]{a}
&= \lim_{n \rightarrow \infty} e^{\ln(\sqrt[n]{a})} \\
&= e^{\lim_{n \rightarrow \infty}(\frac{1}{n}) \ln(a)} \\
&= e^{\lim_{n \rightarrow \infty}(0 \cdot \ln(a))} \\
&= e^0 \\
&= 1.\end{align}
 A: I think that this one is due to Courant:
For all $n\in\mathbb{N}$ define $a_{n}$ such that 
$$
a_{n}=\sqrt[n]{n}-1
$$
Note that for every $n\in\mathbb{N}$, $a_{n}\geqslant0$. Now rewrite
as:
\begin{eqnarray*}
n & = & \left(a_{n}+1\right)^{n}\\
 & = & \sum_{k=0}^{n}\binom{n}{k}a_{n}^{k}\\
 & \geqslant & \binom{n}{2}a_{n}^{2}
\end{eqnarray*}
We get $n\geqslant\frac{n(n-1)}{2}a_{n}^{2}$. 
After rearranging:
$$
a_{n}\leqslant\sqrt{\frac{2}{n-1}}\xrightarrow[n\to\infty]{}0
$$
Concluding that 
$$
\lim_{n\to\infty}\left(\sqrt[n]{n}-1\right)=\lim_{n\to\infty}a_{n}=0\iff\lim_{n\to\infty}\sqrt[n]{n}=1
$$
A: Take any number $r>1$. Eventually $r^n>a$ and so eventually $\sqrt[n]a<r$. Now take $r<1$. Eventually $r^n<a$ and so eventually $\sqrt[n]a>r$.
Calculating $\lim\ r^n$ rigorously is a little tricky and depends on how deep you want to go, see this answer for one approach.
I did also use the fact that the $n$-th root function is increasing, but I'm not sure how to prove that without diving into the definition of the ordering in $\mathbb R$.
A: There is an essential piece in your reasoning, as you swap a limit and a function (the logarithm): you must state that the function is continuous.
You may have simply written
$$\log\left(\lim_{n\to\infty}\sqrt[n]a\right)=\lim_{n\to\infty}\log\left(\sqrt[n]a\right)=\lim_{n\to\infty}\left(\frac{\log a}n\right)=0.$$ 
