Let $c\in \mathbb{R}$. Find two sequences $(a_n)_n$, $(b_n)_n \subset \mathbb{R}$ with:
(i) $\lim\limits_{n\to\infty}a_n=\infty, \lim\limits_{n\to\infty}b_n=0 $ and $\lim\limits_{n\to\infty}a_nb_n=c$
My example:
$a_n:=n$ and $b_n:=\frac{1}{n}$
(ii) $a_n \neq 0 \neq b_n$ for all $n\in \mathbb{N}$, $\lim\limits_{n\to\infty}a_n=0=\lim\limits_{n\to\infty}b_n$ and $\lim\limits_{n\to\infty}\frac{a_n}{b_n}=c$
My example:
$a_n:=\frac{1}{n^2}$ and $b_n:=\frac{1}{n}$
(iii) $\lim\limits_{n\to\infty}a_n=\infty, \lim\limits_{n\to\infty}b_n=-\infty$ and $\lim\limits_{n\to\infty}(a_n+b_n)=c$
My example:
$a_n=n$ und $b_n=-n$
Are these valid examples?
These are so very close to being correct. The thing is, for example in part (i), the limit is
$$\lim_{n\to\infty}a_nb_n=\lim_{n\to\infty}1=1\neq c $$
How do you fix this? Well, you make $a_n=n$ as usual, but $b_n\frac{c}{n}$ so that their product is always $c$.
For part (ii), your answer isn't even correct -- the limit $$lim_{n\to\infty}\frac{a_n}{b_n}=0\neq c$$.
Basically the intuition here is we sort of want $a_n$ to be "c times" $b_n$ -- so if $a_n=\frac{c}{n}$ and $b_n=\frac{1}{n}$ then we're fine.
This is very similar to before -- your limit is $0$. You want $a_n=n+c,b_n=-n$ to ensure the limit is $c$ not $0$.
Pedantic note: When I write "$\neq c$" what I'm really saying is, there exists $c$ such that the two are unequal.