0
$\begingroup$

Suppose that the arithmetic operation on extended complex plane are defined via arithmetic operations on the corresponding sequences, why $∞-∞，0*∞，∞/∞，0/0 $are not defined? Can anyone give me some example about that?

I know that ∞ + ∞ is undefined because if we have two sequences:

$1, 3, 3, 5, 5, 7, 7, 7, 9,....$ and

$-1,-2,-3,-4,-5,-6,-7,-8,....$

the sum of the above sequences will be: $0,1,0,1,0,1,0,1,...$

which is not converging or diverging. so ∞+∞ is meaningless.

Can I use it in $ ∞-∞，0*∞，∞/∞，0/0$?

Can anyone give me some example about that?

  • 1
    $\infty$ is not defined, because it converges, so I didn't understand your question. – 2017-02-01

1 Answers 1

1

  1. $[\infty-\infty]$

    • $\lim\limits_{n\to \infty} (n+1)^2-n^2=\infty$
    • $\lim\limits_{n\to \infty} (n^2+1)-n^2=1$

  2. $[0\cdot\infty]$, $\left[\frac{\infty}{\infty}\right]$, $\left[\frac{0}{0}\right]$

    • $\lim\limits_{n\to \infty} \frac{1}{n}\cdot n^2=\lim\limits_{n\to \infty} \frac{n^2}{n}=\lim\limits_{n\to \infty} \frac{\frac{1}{n}}{\frac{1}{n^2}}=\infty$
    • $\lim\limits_{n\to \infty} \frac{1}{n}\cdot n=\lim\limits_{n\to \infty} \frac{n}{n}=\lim\limits_{n\to \infty} \frac{\frac{1}{n}}{\frac{1}{n}}=1$
  • 1
    In the context of "arithmetic operation on extended complex plane" as stated by the OP, $\infty + \infty$ is considered undefined. – 2017-02-01
  • 0
    Yup, mea culpa. – 2017-02-01