Limit by $\epsilon-N$ method Question is to prove that $\lim_{n\rightarrow \infty}n^{\frac{1}{n}}=1$.
This question is discussed in this site so many times and in all of the answers i have seen they have not considered the basic definition case :
Given $\epsilon>0$ there exists $N\in\mathbb{N}$ such that $\left|n^{\frac{1}{n}}-1\right|<\epsilon$ for all $n\geq N$.
We want to see if $(1+\epsilon)^n>n$
I have tried the following.
Let $\epsilon\geq 1$ then $(1+\epsilon)^n=1+n\epsilon+*\epsilon^2+\cdots+\epsilon^n>1+1+\cdots+1=n$.
So, we do have $(1+\epsilon)^n>n$ for all $n$ irrespective of $\epsilon$.
Suppose $\epsilon<1$ first of all i want one $N$ such that $(1+\epsilon)^N>N$. I was looking for $N$ such that $(1+\epsilon)^N>1+N\epsilon>N$ i.e., $\dfrac{1}{1-\epsilon}>N$.. As  $\epsilon<1$ it is possible to get such $N$.. But this argument does not make much sense.
Help me to solve this.

If you define $x_n=\sqrt[n]{n}-1$, then
$$
n = (1+x_n)^n \geq \frac{n(n-1)}{2}x_n^2.
$$
Hence
$$
0 \leq x_n \leq \sqrt{\frac{2}{n-1}}
$$
for all $n \geq 2$. Now you can play with $\epsilon$.
Bibliography
W. Rudin. Principles of mathematical analysis.
