Originally Posted by

**santiagos11** For example, here is a partial proof:

Let $\displaystyle \epsilon>0$. Clearly $\displaystyle 1\leq n^{1/n}\leq 2 \ \forall n\in \mathbb{N}$

case 1: $\displaystyle \epsilon>1$

Let $\displaystyle N=1$. Suppose $\displaystyle n>N$. Then $\displaystyle |x_{n}-1|=|n^{1/n}-1|=n^{1/n}-1$. On the other hand $\displaystyle \epsilon + 1 > 2 \geq n^{1/n}$, so $\displaystyle n^{1/n}-1<\epsilon$. Therefore, $\displaystyle |x_{n}-1|<\epsilon$.

Do you see how this follows the definition that I gave in the top?

Now it remains to show the same, in the case $\displaystyle 0<\epsilon\leq1$