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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?

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  • $\begingroup$ Hmm, don't I want that? $\endgroup$
    – Analysis
    Commented May 26, 2019 at 20:06
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    $\begingroup$ My error, I thought the examples all referred to the first question. Your examples are all correct. $\endgroup$
    – lulu
    Commented May 26, 2019 at 20:07
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    $\begingroup$ It seems your examples are forgetting about the constant $c$. For instance, your example for (i) only considers $c=1$. $\endgroup$
    – Dave
    Commented May 26, 2019 at 20:12
  • $\begingroup$ ... which is an element of the real numbers, what's your point? $\endgroup$
    – Analysis
    Commented May 26, 2019 at 20:13
  • $\begingroup$ My interpretation of the questions is that you want the limit to equal the parameter $c$. $\endgroup$
    – Dave
    Commented May 26, 2019 at 20:14

2 Answers 2

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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.

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  • $\begingroup$ :-) Understood! $\endgroup$
    – Analysis
    Commented May 26, 2019 at 21:36
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Your examples are correct; you could generalize your $b_n$ to be $b_n=\frac{c}{n}$ for any fixed $c\in\mathbb R$. In this case, you'll satisfy all of the conditions you want.

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