Then show that $ \ \sqrt[n]{n^3} \to 1 \ \text{as} \ n \to \infty \ $ We know that the sequence $ \ \sqrt[n]{n} \to 1 \ \text{as} \ n \to \infty \ $. 
Then show that $ \ \sqrt[n]{n^3} \to 1 \ \text{as} \ n \to \infty \ $
Answer:
$ \sqrt[n]{n^3}=(n^{\frac{1}{n}})^3=(\sqrt[n]{n})^3 \to 1 \ \text{as } \ n \to \infty \ $
Is not it quite correct?
 A: Let the limit be $L$, then 
$$\ln L = \frac{3\ln n }{n} \rightarrow 0$$
So $L \rightarrow e^0 = 1$
A: It's essentially correct. Maybe you still have to prove, that if a sequence $(a_n)$ converges to $1$, then the sequence $(a_n^3)$ also converges to one.
A: If
$a_n \to 1$
then,
for any $0 < c <1$,
$a_n \ge 1-c$
for all large enough $n$.
Since
$x^k-1
=(x-1)\sum_{j=0}^{k-1}x^j
$,
$a_n-1
=(a_n^{1/k})^k-1
=(a_n^{1/k}-1)\sum_{j=0}^{k-1}a_n^{j/k}
$
so
$\begin{array}\\
|a_n-1|
&=|a_n^{1/k}-1|\,|\sum_{j=0}^{k-1}a_n^{j/k}|\\
&\ge|a_n^{1/k}-1|\,|\sum_{j=0}^{k-1}(1-c)^{j/k}|\\
&\ge|a_n^{1/k}-1|k(1-c)
\quad\text{since } x^r \ge x \text{ for } 0 < x , r < 1\\
\text{so}\\
|a_n^{1/k}-1|
&\le \dfrac{|a_n-1|}{k(1-c)}
\end{array}
$
Therefore
$|a_n-1|\to 0$
implies that
$|a_n^{1/k}-1|\to 0$.
A: Notice that $$\lim_{n \to \infty} n^{\frac{1}{n}}=1.$$
Hence $$\lim_{n \to \infty} \sqrt[n]{n^3}=\lim_{n \to \infty} n^{\frac{3}{n}}=\lim_{n \to \infty} \left(n^{\frac{1}{n}}\right)^3=\left(\lim_{n \to \infty} n^{\frac{1}{n}}\right)^3=1^3=1.$$
