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 $.
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).
For example,
$$\lim_{x\to\infty}(x^2-x)$$ and
$$\lim_{x\to3}\left(\frac1{x-3}-\frac1{\sin(x-3)}\right)$$ and do have this form.
Depending on the situation, the difference can be unbounded, tend to a particular finite value or even fail to tend to a single value.