# Is this a valid proof for a limit approaching infinity?

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 $\frac{1}{\sqrt{i}} <1$.

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When calculating a limit, note that if $N$ fits to some $\varepsilon$ then it necessarily fits any $\varepsilon'>\varepsilon$. – Asaf Karagila May 9 '11 at 6:43
You seem to prove that $\bigg |\frac{1}{\sqrt{i}}\bigg |\leq N$, but you need to prove it $\leq \epsilon$. – ShreevatsaR May 9 '11 at 6:44
can we not just choose N = int(1/E^2) + 1 or something simple and show it in one line? – The Chaz 2.0 May 9 '11 at 6:46
Yes we can.  – Did May 9 '11 at 7:45

As mentioned in the comments, you are confused about the definition of a limit. Suppose we have a sequence $\{u_n\}$, and want to prove that $u_n \rightarrow L$. You need to show that for any $\epsilon > 0$, there is a number $N$ such that for all natural numbers $n > N$ $$|u_n-L| < \epsilon.$$ The number $L$ is called the limit.
In this question, that is the same as showing $$\frac{1}{\sqrt{n}} < \epsilon.$$
To prove this, suppose $\epsilon>0$ and then let $n > N = \epsilon^{-2}$. (Note that you can't use $N$ as an index, it is not in general a natural number. This may differ with local usages.) Then $$\frac{1}{\sqrt{n}} < \frac{1}{\sqrt{\epsilon^{-2}}} = \epsilon,$$ as required.