Proof verification that $\lim_{n\to \infty} n^\frac{1}{n}=1$

Question

We know that $n^\frac{1}{n}\geq1$

Let $\epsilon>0$ be arbitrary

Take $N=1+\epsilon$

Suppose $n>N$

$|n^\frac{1}{n}-1|=n^\frac{1}{n}-1<N^\frac{1}{N}-1<\epsilon \qquad \because \, 1+\epsilon<(1+\epsilon)^{1+\epsilon}$

But I feel like there is a problem somewhere because I can prove that the limit is $0$ if I take $N=\epsilon$

$|n^\frac{1}{n}|=n^\frac{1}{n}<N^\frac{1}{N}<\epsilon \qquad \because \, \epsilon<\epsilon^\epsilon$

So, where's the problem ?

Thanks in advance for your answers.

Answer 1

Well, the problem is that $f(x) = x^{\frac{1}{x}}$ is decreasing only on $(e, \infty)$. You can see this by considering $\log f(x) = \frac{1}{x}\log x$, taking derivative to get $\frac{f'}{f} = -\frac{1}{x^2}\log x + \frac{1}{x^2}$, and setting $f' = 0$. Therefore your statement $n > N \implies n^{\frac{1}{n}} < N^{\frac{1}{N}}$ does not hold for all $\epsilon > 0$. This means that both of your proofs are incorrect.

Graph: