# Why is counting with infinity not indeterminable?

I was trying to find a function that looks like square root of x but I wanted to limit that function from above to 1. Finding solution was not as hard as I thought, but...

When I was thinking about solution I was thinking about infinity, because function of square root is not limited from above. So my first question was: How do I limit something that can be infinite?

Than I crossed to some page that gave me an idea: I can make two square root functions that will be offset and subtract one from the other.

$$f(x) = -\left(\sqrt{x+1}-\sqrt{x}\right)+1$$

Okay that worked. I got function that does exactly what I want, but I don't know why. I mean, it should not...? Main reason of my confusion is that $$\infty - \infty = \text{indeterminable}$$, but in my case is not...?

But that is not the only case. I also tried $$\lim \limits_{x \to \infty} (x+1)-x$$ and than also the most simple case $$x-x; x = \infty$$. It turns out that $$\infty - \infty = 0$$.

Both on Wolfram Alfa

Could you explain in little more detail why? or what am I doing or thinking wrong about?

But I guest I though about it all wrong. The theory or sentence $$\infty - \infty = \text{undefined}$$ is about two different infinity which cannot be subtracted because we are unable to know the outcome while in my case there is only one infinity and therefore $$\infty - \infty = 0$$.

• Infinity is not a real number. – Lucas Henrique Jan 28 '20 at 12:17
• $\infty$ is not a number, so you can't plug “$x=\infty$” into the formulas. The phrase “indeterminate form” refers to calculation of limits as $x \to \infty$, as explained in the answers to this question, for example: math.stackexchange.com/questions/554521/… – Hans Lundmark Jan 28 '20 at 12:17
• An indeterminate form does not mean the limit can never be computed. $\frac{x}x$ is surely an indeterminate form at $\infty$ if I work independently on the numetor and denominator, but, for $x>0$, this function is just $1$. – nicomezi Jan 28 '20 at 12:31
• By the way, nice candidates for the function shape you're looking for could be $x/(x+1)$ or $\sqrt{x/(x+1)}$. – Magma Jan 28 '20 at 12:42
• Warning, $\lim_{x\to\infty}((x+1)-x)\ne 0$. – Yves Daoust Jan 28 '20 at 12:44

The notation $$\infty-\infty$$ does not mean that you "subtract infinity from infinity", which is just meaningless.
If signifies that you are studying an expression which appears as the difference between two functions, $$f(x)-g(x)$$, when these are both unbounded (when $$x$$ approaches some predefined finite value, or goes to infinity).
$$\lim_{x\to\infty}(x^2-x)$$ and
$$\lim_{x\to3}\left(\frac1{x-3}-\frac1{\sin(x-3)}\right)$$ and do have this form.