0
$\begingroup$

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 $.

$\endgroup$
5
  • 1
    $\begingroup$ Infinity is not a real number. $\endgroup$ – Lucas Henrique Jan 28 '20 at 12:17
  • 3
    $\begingroup$ $\infty$ 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/… $\endgroup$ – Hans Lundmark Jan 28 '20 at 12:17
  • 1
    $\begingroup$ 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 $1$. $\endgroup$ – nicomezi Jan 28 '20 at 12:31
  • 1
    $\begingroup$ By the way, nice candidates for the function shape you're looking for could be $x/(x+1)$ or $\sqrt{x/(x+1)}$. $\endgroup$ – Magma Jan 28 '20 at 12:42
  • 1
    $\begingroup$ Warning, $\lim_{x\to\infty}((x+1)-x)\ne 0$. $\endgroup$ – Yves Daoust Jan 28 '20 at 12:44
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.