What is ∞ -(∞-1)?

Question

I have come up with two possible answers, but am unsure which is true. The first is that ∞-1 is still ∞,as ∞ is endless, and if taking away 1 made it not endless, then ∞ would not be endless either, it would just be one more then the number you got by subtracting 1 from ∞. Then, when you take ∞ away from ∞, you would get 0, as x-x=0, no matter the value of x. Therefore, ∞-(∞-1) would equal 0. My other possible answer is that x-(x-1) should equal one, because that x-1 equals 1 less than x, and a number minus one less than itself equals one more then zero. Therefore, ∞-(∞-1) would equal 1. I am not sure whether to base the answer off the fact that x-(x-1)=1, or that ∞-1=∞.

Answer 1

$\infty-(\infty-1)$ has the same status as $\infty-\infty$: an indeterminate form. Neither $\infty$ or $1$ is its resolution all the time, and the actual answer depends on where you got the form from.

Answer 2

It's still undefined. You would agree that

$${\lim_{x\rightarrow \infty}x=\infty}$$

and also that

$${\lim_{x\rightarrow \infty}(x-1)=\infty}$$

Now, if you take this limit:

$${\lim_{x\rightarrow \infty}x-(x-1)}$$

the answer, just by Algebraic simplification, is ${1}$. And the expression is approaching the form ${\infty-(\infty-1)}$. Equally, you can see that

$${\lim_{x\rightarrow \infty}x-((x-1)-1)}$$

is also approaching the form ${\infty-(\infty-1)}$ (since as we said, ${\lim_{x\rightarrow \infty}(x-1)}$ still approaches infinity). Yet the limit (again, through Algebraic simplification) is ${2}$. We have approached the expression in two different ways and got two different answers.