Which of the following is of determinate form? Which of the following is of determinate form?
$a$. $\infty - \infty$
$b$. $\infty + \infty$
$c$. $\dfrac {\infty }{\infty}$
$d$. $\infty \times 0$.
My teacher said that $\infty + \infty $ is determinate and he proved it in the following way:
Let $\infty =1$ and $\infty =2,$
then $\infty + \infty = 1+2=3.$
Also, $\infty + \infty = 2+1=3.$
Is this correct? Or, what's the correct interpretation?
 A: It's “not wrong” (I dare not say “correct”) that $\infty+\infty$ is not an indeterminate form.
You can easily prove the following statement:

*Suppose $\lim_{x\to c}f(x)=\infty$ and $\lim_{x\to c}g(x)=\infty$; then
  $$
\lim_{x\to c}(f(x)+g(x))=\infty
$$

Here $c$ can also be $\infty$ or $-\infty$ and the limit can be one-sided.
I'll show the proof for the case where $c\in\mathbb{R}$, under the assumption that $f$ and $g$ are both defined on a set $D$ which has $c$ as a limit point (and $c\notin D$).
Fix $M>0$. Then, by assumption, there exist $\delta_1>0$ and $\delta_2>0$ such that


*

*for $x\in(c-\delta_1,c+\delta_1)\cap D$, $f(x)>M/2$,

*for $x\in(c-\delta_2,c+\delta_2)\cap D$, $g(x)>M/2$.


Take $\delta=\min\{\delta_1,\delta_2\}$; then,  for $x\in(c-\delta,c+\delta_1cap D$,
$$
f(x)+g(x)>\frac{M}{2}+\frac{M}{2}=M
$$
Therefore we are allowed to say
$$
\lim_{x\to c}(f(x)+g(x))=\infty
$$
according to the definition.
The other three cases are instead “indeterminate”, in the sense that there is no general theorem that can be applied.
For instance,


*

*for $f(x)=1/x^2$, $g(x)=1/x^2$, we have $\lim_{x\to0}(f(x)-g(x))=0$

*for $f(x)=1/x^4$, $g(x)=1/x^2$, we have $\lim_{x\to0}(f(x)-g(x))=\infty$


where $f$ and $g$ have both limit $\infty$ for $x\to0$. It's possible to modify the example in such a way that the limit of $f(x)-g(x)$ is any number or even the limit does not exist.
Similar examples can be given for cases c and d.
On the other hand, the “justification” with $\infty=1$ and $\infty=2$ is, in my opinion, at least misleading (choose a different word, if you like).
