Use the epsilon-delta definition to prove the limit?

Question

My attempt:

Let $\epsilon > 0$ be given. Define $N>0$: $N = \frac{1}{ε^2}$

It follows that for all $n > N$, $|a_n - 0| = |a_n| = |\frac{1}{\sqrt{n}}| < \epsilon$

And therefore $$\lim_{n\to\infty} \frac{1}{\sqrt{n}} = 0 $$

Answer 1

Let $\varepsilon>0$ be given, because of the archimedean principle, there is $N\in\mathbb{N}$ such that $\frac{1}{\varepsilon^{2}}<N$.

Therefore $\frac{1}{\sqrt{N}}<\varepsilon$.

So, for all $n>N$, $\frac{1}{\sqrt{n}}<\varepsilon$.