Give examples showing why $0\cdot \infty$, $\infty/\infty$, and $0/0$ are meaningless Assuming arithmetic operations on $\overline{\mathbb{C}}$ (that's the extended complex plane) are defined via arithmetic operations on the corresponding sequences, I need to give examples showing why $0\cdot \infty$, $\infty/\infty$, and $0/0$ are meaningless.
I also had to show the same thing for $\infty - \infty$, but that was easy: Consider the sequences $\{1,3,3,5,5,7,7,9,\cdots \}$ and $\{1,2,3,4,5,6,7,8,\cdots \}$, both of which belong to the class $\infty$ (i.e., the equivalence class of all sequences converging to $\infty$). Then, subtracting the second sequence from the first, term-by-term, we obtain the resulting sequence $\{0,1,0,1,0,1,\cdots\}$, which does not belong to any convergence class, since it has no limit.
However, I can't approach the three cases I'm asking on here about in the same way. I must admit, I'm a little confused as to what is being asked. For reference, it is problems 2-4 on p. 79 of Markushevich's Theory of Functions of a Complex Variable, Vol I.
If someone could furnish me with an example, at least for $0 \cdot \infty$, and maybe either $0/0$ or $\infty/\infty$, perhaps I could use that to help me figure out the other case on my own.
Thank you.
 A: Here is an answer close to the proof you have for $\infty-\infty$.
Take $(0,\frac12,0,\frac14,0,\frac16,0,\ldots)$ as a representative of $0$ and let $(1,2,3,4,5,6,7,\ldots)$ represent $\infty$. If you multiply one by the other (term-by-term) you get $(0,1,0,1,\ldots)$, and this shows that $0\cdot \infty$ is nonsensical.
For $\frac00$, keep the first sequence above, and let the second sequence be $(1,\frac12,\frac13,\ldots)$. Again you get $(0,1,0,1,\ldots)$.
Lastly, for $\frac\infty\infty$, take, for instance, $(1,4,3,8,5,12,\ldots)$ divided by $(1,2,3,4,\ldots)$ to get $(1,2,1,2,1,\ldots)$.
A: {$1/n$} and {$1/n^2$} tend to 0.  {$n$} and {$n^2$} tend to $\infty$.
Let's compare:
{$(1/n)/(1/n^2) = n$}$\rightarrow \infty$ so $0/0 $~$ \infty$.
{$(1/n^2)/(1/n) = 1/n$}$\rightarrow 0$ so $0/0 $~$ 0$.
{$(n^2)*(1/n) = n$} so $\infty * 0 $~$\infty$
{$(n)*(1/n^2) = 1/n$} so $\infty * 0 $~$0$.
{$n^2/n = n$} so $\infty / \infty $~$\infty$
{$n/n^2 = 1/n$} so $\infty / \infty  $~$0$.
And we can choose other sequences (example $a/n, n$) to get any values we'd like.
{$(a/n)*n = a$} so $0*\infty$ ~ $a$.  etc.
A: The problem with $\dfrac 5 0$ is that the equation $0\cdot x=5$ has no solution for $x$.
The problem with $\dfrac 0 0$ is that the equation $0\cdot x=0$ has many solutions for $x$.
Things like $\dfrac 6 2$ are not problematic because the equation $2\cdot x = 6$ has just one solution for $x$.
Convergence:
As far as convergence goes, not that $\dfrac{6x} x$ converges to $6$ as $x$ approaches $0$ and the numerator and denominator both converge to $0$.  You can find functions $f$ and $g$ for which $\dfrac{f(x)}{g(x)}$ converges to $7$ as $x$ approaches something and $f(x)$ and $g(x)$ both approach $0$.
Next: $0\cdot\infty$
As $x$ approaches $0$ from above, look at the behavior of $6x\cdot\dfrac 1 x\vphantom{\dfrac 1{\displaystyle\sum}}$.  The first factor approaches $0$ and the second approaches $+\infty$, and the product approaches $6$, and you can do something similar with any number other than $6$.
Next: $\dfrac\infty\infty$
Notice that as $x\to\infty$ we have $\dfrac{2x+1}{x}\to2$ and $\dfrac{3x+1}x\to 3$, and in both cases the numerator and denominator both approach $\infty$.  The two fractions don't both approach the same thing.
A: Cheating a little, for $\dfrac\infty\infty$ use 
$$\frac{2^{S_k}}{2^{T_k}}=2^{S_k-T_k},$$ where $S$ and $T$ are the sequences as defined above.
For $0\cdot\infty$, use $$2^{-T_k}\cdot2^{S_k}=2^{S_k-T_k},$$
and for $\dfrac 00$, use
$$\frac{2^{-T_k}}{2^{-S_k}}=2^{S_k-T_k}.$$
A: The answer is much easier.
$\infty \notin \overline{\mathbb{C}}$
Yes, arithmetic operations are defined on $\overline{\mathbb{C}}$.
As $\infty$ is not part of $\overline{\mathbb{C}}$, the arithmetic operations defined on $\overline{\mathbb{C}}$ are meaningless.
