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I have been working on a proof which involves sums and products going to infinity. I am wondering whether the following proof of a limit is valid, and whether that result would allow me to come to another conclusion.

What is:

$$\lim \limits_{n \to \infty} f(n)\text {, where }f(n) = n-n$$

I have worked this out to be

$$\lim \limits_{n \to \infty} n-n = \lim \limits_{n \to \infty} n(1-1) = \lim \limits_{n \to \infty} n\cdot 0 = 0$$

I'm not sure whether this is the correct way of proving this limit, or whether the answer is correct. My math teacher had said that the whole limit raised a red flag in his mind, and he wasn't sure why.

If my limit is correct, though, I would like to know whether the following is also valid:

$$\lim \limits_{n \to \infty} f(n)\cdot n = 0$$

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    $\begingroup$ The red flag might be that $x_n\to\infty$ and $y_n\to0$ implies nothing about the asymptotics of $x_ny_n$. But here $f(n)=0$ for every $n$ hence $x_nf(n)=0$ for every $n$ hence $x_nf(n)\to0$ for every sequence $(x_n)$ (and to tell you the truth, this makes for a rather bizarre question). $\endgroup$ – Did Dec 17 '16 at 21:37
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    $\begingroup$ It's easier than that. f(n) = n-n = 0 is a constant function. $\endgroup$ – fleablood Dec 18 '16 at 2:37
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    $\begingroup$ $\lim n*0$ is ambiguous. It could mean $\lim (n*0)$ which is no easier or harder to solve than $\lim (n-n)$ or it could mean $(\lim n)*0$ which may not be a valid conclusion and may not be 0. BTW why do you think $\lim n*0 = 0$? $\endgroup$ – fleablood Dec 18 '16 at 2:41
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    $\begingroup$ $f(n)=n-n$ is just a roundabout way to say $f(n)=0\,$, so $\lim_{n \to \infty} f(n) = \lim_{n \to \infty} 0 = 0$. This should not be confused, however, with $\lim_{n \to \infty} (n-n) = \lim_{n \to \infty} n - \lim_{n \to \infty} n$ which is wrong and undefined. $\endgroup$ – dxiv Dec 18 '16 at 3:42
  • $\begingroup$ @dxiv AFAIK $\lim_{n \to \infty}{n-n}=0$. It's not completely clear from your comment that the error is in "splitting the limit". You can "split the limit" only: 1) when they are factors and 2) the individual limits exist and are finite (e.g. $\lim_{x \to \infty}{f(x) \cdot g(x)} = (\lim_{x \to \infty}{f(x)})\cdot(\lim_{x \to \infty}{g(x)})$ provided these last two limits exist and are finite). $\endgroup$ – Bakuriu Dec 18 '16 at 9:15
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If $f(n) = n-n$, then $f(n) = 0$ for all $n$. The limit you gave is true, i.e. $$\lim\limits_{n\to \infty} f(n) = 0$$ is correct.

Furthermore, we have that $n\cdot f(n) = n\cdot 0 = 0$ for all $n$, so the limit $$\lim\limits_{n\to\infty}n\cdot f(n) = 0$$ is also correct.


The red flag probably stems from the well known "indeterminate form" $\infty -\infty$. This is short hand notation for the fact that knowing $$\lim\limits_{n\to \infty} g(n) = \infty\quad\text{ and }\quad \lim\limits_{n\to\infty} h(n) = \infty $$ is not enough by itself to determine $$\lim\limits_{n\to\infty}(g(n)-h(n)).$$ Similarly, $\infty\cdot 0$ is also an indeterminate form, i.e. knowing $$\lim\limits_{n\to \infty} g(n) = \infty\quad\text{ and }\quad \lim\limits_{n\to\infty} h(n) = 0$$ is not enough to determine $$\lim\limits_{n\to\infty} g(n)\cdot h(n).$$

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Yes this is correct. If you define $f(n)=n-n$ it is sufficient to notice that $f(n) \equiv 0$

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When in doubt you can always go back to the definition of a limit. If $\lim_{n\to\infty} f(n)$ is going to have a value, $L$, it means that for any positive "deviation" $\epsilon$, no matter how small, there is a range $[N,\infty)$ such that, throughout that entire range, $f(n)$ never gets further away from $L$ than $\epsilon$.

In this case, $f(n) = 0$ for all $n$. So if we pick $L = 0$, then no matter what $\epsilon$ you pick, $f(n)$ is always that close to $L$ because it's exactly equal to $L$ everywhere. This reasoning also shows you that, for any function of a real number which always takes the same value (a constant function), its limit at any point or at $\pm\infty$ is equal to that value.

This is, of course, the "nontechnical" way to phrase it, but the idea is the same if you do a proper proof.

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