Epsilon-Delta proof for the $\lim \limits_{n \to \infty} \frac{i}{n}$ I have to figure out whether $z_n=\frac{i}{n}$ converges and if it does I have to find its limit as $n$ goes to infinity.
I know it converges and its limit is equal to zero when $n$ goes to infinity, however I want to prove this using an epsilon-delta proof.
We have $\lim \limits_{n \to \infty} \frac{i}{n}=0$ and from the epsilon-delta definition we get $|\frac{i}{n}-0| \lt \epsilon$ and $|n-n_0|=|n- \infty|$  
Now I don't know how to choose $\epsilon$ and $\delta$, I thought I had to manipulate $|\frac{i}{n}-0| \lt \epsilon$ and make it look like $|n- \infty| \lt \epsilon$ but I have no idea and that infinity is throwing me off.
 A: For each $n\in\Bbb N$, write $z_n=\frac{i}{n}$. We want to show that $$\lim_{n\to\infty}z_n=0.$$
Let $\epsilon>0$. Using the Archimedean Property, we can find $N\in\Bbb N$ such that $\frac{1}{N}<\epsilon$. Therefore, if $n>N$ then
$$|z_n-0|=\left|\frac{i}{n}\right|=|i|\cdot\frac{1}{n}=1\cdot\frac{1}{n}=\frac{1}{n}<\frac{1}{N}<\epsilon.$$
This proves our claim.
A: Hint: To show that the limit of a sequence $\{a_n\}$ as $n\to\infty$ is $L$, you need to show that for any $\epsilon>0$ the sequence gets within $\epsilon$ of $L$ and stays there. That is: for any $\epsilon>0$ there is some $N>0$ so that for $n>N$, $|a_n-L|<\epsilon$. That is the definition you need to apply to the given sequence.
A: By the Archimedean property, for any $x\in \mathbb{R}$, $x>0$ there is some 
$n\in \mathbb{N}$ with $n>x$. 
Then, taking reciprocals 
$$
1/x>1/n
$$
Since this works for any $x\in \mathbb{R}$, take $x=1/\epsilon$ for given $\epsilon$ and we will have
$$
\epsilon>1/n=|1/n|
$$
and we are done.
A: $x \rightarrow c$ and $x\rightarrow \infty$ aren't quite the same thing though they are very similar.
$x \rightarrow c$ means "$x$ gets very close to $c$" or more technically: For an arbitrarily small $\delta > 0$, $|x - c| < \delta$.
$x\rightarrow \infty$ doesn't mean "$x$ gets very close to infinity".  "infinity" isn't a value anything can get "close" to.  $x\rightarrow \infty$ means $x$ gets very large" of more technically: For an arbitrarily large $N$, $x > N$.
These differences are reflected in how we do $\lim_{x\rightarrow\{c, \infty} f(x) = \{k, \infty\}$.
1) $\lim_{x\rightarrow c} f(x) = k$ means:  For every $\epsilon > 0$ there exists a $\delta > 0$ so that $|x-c| < \delta \implies |f(x) - k |< \epsilon$.
2) $\lim_{x\rightarrow c} f(x) = \infty$ means:  For every $N$ there exists a $\delta > 0$ so that $|x-c| < \delta \implies f(x) > N$.
3) $\lim_{x\rightarrow \infty} f(x) = c$ (this is the one you want) means:  For every $\epsilon > 0$ there exists an $M$ so that $x > M \implies |f(x) - k |< \epsilon$.
4) $\lim_{x\rightarrow \infty} f(x) =\infty$ means:  For every $N$ there exists a an $M$ so that $x > M \implies f(x) > N$. 
So in your case you want to find an $M$ (in terms of $\epsilon$) so that $n > M \implies |\frac{i}n - 0| < \epsilon$.  Hint: use $M \ge \frac 1 \epsilon$.
