calculus - How to find the limit $limlimits_{ntoinfty}1/sqrt[n]{n}$ which is indeterminate on evaluation but is convergent?

When I evaluate the limit in the title above I get the following:

$$\begin{align} \lim\limits_{n\to\infty}\dfrac{1}{\sqrt[n]{n}} &= \lim\limits_{n\to\infty} \dfrac{1}{n^{\frac{1}{n}}} = \dfrac{1}{\infty^0} \quad\Rightarrow\quad Indeterminate\\ &= \lim\limits_{n\to\infty}\left(\dfrac{1}{n}\right)^\frac{1}{n} = 0^0 \quad\Rightarrow\quad Indeterminate \end{align}$$

But when I use a computer software (mathematica) to evaluate the same limit it says the limit is 1. What am I doing wrong?

Answer

Indeterminate forms can have values.

Note from L'Hospital's Rule that $\lim_{n\to \infty}\frac{\log(n)}{n}=\lim_{n\to \infty}\frac{1/n}{1}=0$. Hence, we have

$$\begin{align} \lim_{n\to \infty}\frac{1}{n^{1/n}}&=\lim_{n\to \infty}e^{-\frac1n \log(n)}\\\\ &e^{-\lim_{n\to \infty}\left(\frac1n \log(n)\right)}\\\\ &=e^0\\\\ &=1 \end{align}$$

as expected!