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Suppose you have a sequence $(b_n)_{n\in\mathbb{N}}$ with $b_n\not=0$ for all $n\in\mathbb{N}$. I need to examine the following:

For all $(a_n)_{n\in\mathbb{N}}$, it holds that $\lim_{n\to\infty}(a_n-b_n)=0\Rightarrow \lim_{n\to\infty}\frac{a_n}{b_n}=1$.

I've come up with an argument, but I'm troubled with it:

Since $\lim_{n\to\infty}(a_n-b_n)=0$, we have $\lim_{n\to\infty}a_n -\lim_{n\to\infty}b_n=0$, i.e. $\lim_{n\to\infty}a_n=\lim_{n\to\infty}b_n$. Therefore we have by another limit-theorem that $\frac{\lim_{n\to\infty}a_n}{\lim_{n\to\infty}b_n}=\lim_{n\to\infty}\frac{a_n}{b_n}=1$.

The thing I'm troubled with is whether I can use the single limit-statements about $(a_n)$ and $(b_n)$ since I'm not sure whether they are convergent(as it is not required).

As a second point, how may I show that this statement is equivalent to $(\frac{1}{b_n})_{n\in\mathbb{N}}$ is bounded?

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You are absolutely right in doubting your own statement. It is true that $\lim_{n \to \infty} (n + \frac 1n)-n = 0$, however this is not equal to $\lim_{n \to \infty} (n + \frac 1n) = \lim_{n \to \infty} n$, since both quantities aren't even well defined! (the limits do not exist)

Hence, our argument must be more clever. We cannot split the limit across the subtraction, this much is known now.

First off, if $\lim_{n \to \infty} a_n$ does exist and is non-zero, then your argument is fine. This is because in that case, $\lim_{n \to \infty} a_n = \lim_{n \to \infty} b_n$, and both are non-zero, so the quotient rule applies.

The question is : what happens when both limits are zero, say, or both limits do not exist?

We now move to the second part : we claim that $\lim_{n \to \infty} (a_n - b_n) = 0 \implies \lim_{n \to \infty} \frac{a_n}{b_n} = 1$ if and only if $\frac 1{b_n}$ is bounded.

Now, suppose that $\frac 1{b_n}$ is bounded, say $\left|\frac{1}{b_n}\right| \leq M$ for all $n$. Then, $$0 \leq \left|\frac{a_n}{b_n} - 1\right| = \left|\frac{a_n-b_n}{b_n}\right| \leq M|a_n - b_n|$$ Note that we can make the right hand side as small as possible by increasing $n$, so it follows that $\lim_{n \to \infty} \frac{a_n}{b_n} = 1$.

I leave you to argue the other direction : given $\frac 1{b_n}$ not bounded, find $a_n$ so that $a_n - b_n \to 0$ but $\frac{a_n}{b_n} \not\to 1$. Think about $a_n = 2b_n$.

Of course, this completes the examination of the statement : we have shown that it is equivalent to another statement.

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  • $\begingroup$ How would I argue for the other direction, is finding a specific sequence (a_n) enough? $\endgroup$ – blub Nov 21 '17 at 16:19
  • $\begingroup$ Yes, finding any $a_n$ so that $a_n - b_n \to 0$ but $\frac{a_n}{b_n}$ does not go to $1$, is sufficient. $\endgroup$ – астон вілла олоф мэллбэрг Nov 21 '17 at 23:03

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