Proof that $\displaystyle \lim_{i \to \infty} \bigg | \frac{1}{\sqrt{i}}\bigg |=0 $
Let $\epsilon>0$ be arbitrary and let $N= \begin{cases} 1 & \text{ if } 1 \geq \epsilon\\ \epsilon & \text{otherwise.} \end{cases}$
Note: whether or not $1 \geq \epsilon$, it follows that $N \geq 1$ and $N \geq \epsilon$.
Assume $i>N$
Since $i>N\geq 1$, $\sqrt{i}>\sqrt{N}\geq \sqrt{1}$.
We then have $\bigg |\frac{1}{\sqrt{i}}\bigg |=\frac{1}{|\sqrt{i}|}=\frac{1}{\sqrt{i}}<1 \leq N$
My main worry about this proof is near $\frac{1}{\sqrt{i}}<1$. My justification here is that $i>N \geq 1$ and for any number $t>1$ we get $\sqrt{t}>\sqrt{1}=1$. Furthermore, for numbers $a$ and $b$, if $a>b>0$ then $\frac{b}{a} < 1$. Thus $\sqrt{i}>1>0$ so $\frac{1}{\sqrt{i}} <1$.