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Since we know the following:

$\lim_{n\to\infty} 1/n=0$

Can we say that the this holds too:

$\lim_{n\to\infty} n*0=1$?

Thank you!

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No, you can't.

Writing $n\cdot 0$ is nothing more than a way to explain which numbers are in the sequence you're talking about. No matter whether you write $\lim_{n\to\infty} n\cdot 0$ or $\lim_{n\to\infty} 0$, the sequence is still the same, namely $$ (0,0,0,0,\ldots) $$ And the limit of that sequence is $0$ no matter how you write it down.


Note in particular that $$ \lim_{n\to\infty} 1/n = 0 \qquad\text{means}\qquad \Bigl( \lim_{n\to\infty} 1/n \Bigr) = 0 $$ In order for your rewriting to make sense, one would need to read it as $$ \lim_{n\to\infty}\; \underbrace{1/n = 0}_{\text{do algebra here}} $$ but that is not how the limit notation works. The $\lim$ sign applies only to an expression; it is nonsense to apply it to an entire equation.

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    $\begingroup$ Lol, I like the "do algebra here" part. Perfectly timed. $\endgroup$ – Simply Beautiful Art Jan 31 '17 at 19:09
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    $\begingroup$ Multiplying by ∞ is as undifined as dividing with 0? $\endgroup$ – Bende Jan 31 '17 at 19:50
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    $\begingroup$ $\infty$ is not even a number, so I'd say multiplication by $\infty$ is somewhat less defined than division by $0$. $\endgroup$ – Austin Mohr Jan 31 '17 at 20:11
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    $\begingroup$ @Bende: There's no arithmetic with $\infty$ going on here at all; the only appearance of $\infty$ is as part of the limit notation, but that refers to a particular definition of "$\lim\limits_{n\to\infty}\cdots$", and that definition does not depend on $\infty$ at all. $\endgroup$ – hmakholm left over Monica Jan 31 '17 at 20:19
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    $\begingroup$ @Henning: I did not question the limit notation at all. I was just trying to compare multiplying by infinity to something else which is clearly undefined that is dividing by 0. $\endgroup$ – Bende Jan 31 '17 at 20:28
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No, we cannot. Consider the graph of $f(n)=\frac{1}{n}$. As $n\rightarrow 0$, we can clearly see that $f(n)$ appears to approach infinity, approach being the keyword. We're not talking about what $f(n)$ actually is at $n=0$, but rather, what it approaches as $n$ approaches $0$. If you consider your second function then, $f(n)=0n$, we see that the graph of $f(n)$ is simply a straight line, $f(n)=0$. As we increase $n$, $n$ is never anything but $0$. For this reason, we say that the limit, $$\lim_{n\to\infty} 0n=0.$$ This does not mean that $f(\infty)=0$, but rather that it appears to approach $0$ as $n\rightarrow\infty$. This is never-minding the fact that $0\cdot\infty$ is undefined.

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    $\begingroup$ Thank you! I am starting to feel that multiplying by ∞ is as undifined as dividing with 0. $\endgroup$ – Bende Jan 31 '17 at 19:49
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Short answer: No.

What you described only works if you have two functions ($f(x)=1/n, g(x)=n$) and they grow at the same rate: $\lim_{x\rightarrow \infty }f(x)g(x)=1$.

This will always evaluate to 1 for any positive x because the point of intersection is where $f(x)=g(x)$ and this is only true for $x=1$.

If you instead use 2 separate limits as you indirectly did: $$\lim_{x\rightarrow \infty }\lim_{y\rightarrow \infty }f(x)g(y)\Leftrightarrow \infty\cdot0$$ you could evaluate to anything because the growth rate isn't synchronised.

You could imagine the functions being shifted left or right (due to the unsynchronised growth rates), thus moving the point of intersection left or right, thus making every result/solution possible.

2. Question "Multiplying by ∞ is as undefined as dividing by 0?"

If you think of them as limits then they are defined (see above), but if there is no insight to the "kind" of zero/infinity then you could evaluate to anything, thus making a "definition" impossible.

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  • $\begingroup$ Cheers! Makes good sense. $\endgroup$ – Bende Jan 31 '17 at 20:02
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Consider the sequence with terms $x_n = 5n/n$. It's limit is obviously 5 since it is actually the constant sequence! However, you pointed out that $1/n$ converges to 0 and $5n$ clearly converges to infinity. So we do not have '$0 \cdot \infty = 1$'. In general we can not say anything about $0 \cdot \infty$ since it is not defined.

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    $\begingroup$ Thanks! So '0⋅∞=1' is as undifined as 1/0. Right? $\endgroup$ – Bende Jan 31 '17 at 19:43
  • $\begingroup$ No: if we have a sequence $1/n$ and let $n$ approach to zero, we get that this tends to infinity, so we could define $1/0$ as infinity( 1/0 in this case means we divide one by something very small) whereas the outcome $0 \cdot \infty$ depends on the sequence, it does not have exactly one outcome... $\endgroup$ – Student Jan 31 '17 at 20:45

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