Answer
See the explanation
Work Step by Step
$\begin{array}
nn&2^{1/n}\\
1&2\\
2&\sqrt2=1.4\\
5&\sqrt[5] 2=1.15\\
10&\sqrt[10] 2=1.072\\
100&\sqrt[100] 2=1.01
\end{array}$,
As $n$ gets larger and larger, $1/n$ approaches $0$. Thus, $2^{1/n}$ approaches $1$.
$\begin{array}
nn&(\frac{1}{2})^{1/n}\\
1&\frac{1}{2}\\
2&\sqrt\frac{1}{2}=0.71\\
5&\sqrt[5] \frac{1}{2}=0.871\\
10&\sqrt[10] \frac{1}{2}=0.933\\
100&\sqrt[100] \frac{1}{2}=0.993
\end{array}$,
As $n$ gets larger and larger, $1/n$ approaches $0$. Thus, $(\frac{1}{2})^{1/n}$ approaches $1$.
$\begin{array}
nn&n^{1/n}\\
1&1\\
2&\sqrt2=1.4\\
5&\sqrt[5] 5=1.38\\
10&\sqrt[10] 10=1.26\\
100&\sqrt[100] 100=1.05
\end{array}$,
As $n$ gets larger and larger, $1/n$ approaches $0$. Thus, $n^{1/n}$ approaches $1$.
