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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.

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  • $\begingroup$ What do you mean by $|n-\infty|$ $\endgroup$
    – user223391
    Feb 14, 2017 at 1:07
  • $\begingroup$ I am not sure, I was following an example from my textbook and it says when $\lim \limits_{z \to 3-i} (2z+3)=9-2i$ from this the book gets $|z-z_0|=|z-(3-i)|$, I tried to apply the same idea. $\endgroup$
    – idknuttin
    Feb 14, 2017 at 1:09
  • $\begingroup$ Knowing precisely what you want to prove is important. :) Check your book for the definition $\endgroup$
    – user223391
    Feb 14, 2017 at 1:11
  • $\begingroup$ Is n a natural number? i, even though it is a complex number, is constant and proving this is no different than findinng the limit of 1/n, 7/n, -3/n, or any other constant/n. $\endgroup$
    – fleablood
    Feb 14, 2017 at 1:34
  • $\begingroup$ @fleablood, yes n is a natural number $\endgroup$
    – idknuttin
    Feb 14, 2017 at 1:45

4 Answers 4

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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.

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$x \rightarrow c$ 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 $M \ge \frac 1 \epsilon$.

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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.

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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.

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