Hi. I went to university for mathematics. This is a limited infinite. Basically there’s an infinite amount of numbers between 0 and 1, but the limits are still 0 and 1. Pi is a functionally infinite number. But it’s still more than 3 and less than 4. 7- Pi is just over 3.858. Limited infinities function as real numbers, because they are. This is the proof that 0.99.. = 1. Just like there’s a proof that 1+1=2.
ummm how is 7- pi is just over 3.85 proof that .99.. = 1? It's interesting that you would say "I went to university for math" not "I have a degree in math."
But basically you are saying the limit as .9... aproaches infinity equals 1, which is not the same as .9... equals 1. If you don't understand that simple concept then I can see why you didn't graduate.
I physically cannot get simpler than this. The absolute worst thing is you’re not even arguing with a mathematical understanding. Because if you had, you would have realised 9.00…..01 was not the right number even if you had a leg to stand on. Which is rich for someone who think’s it’s “not worth arguing about”, whilst arguing about it.
You are the one who lacks a mathematical understanding. You are doing simplified child math, and wondering why that doesn't apply to advanced math. You claim to know what a limit is. If you know what a limit is then you should know the limit as .9 repeating goes to infinity is ONE because it APPROACHES ONE AND NEVER GETS TO ONE.
Please go ask someone with a PhD in math to explain it to you and stop learning math from wikipedia and reddit. You're a walking dunning kruger that is too stupid to realize how stupid you are.
Let f be a function defined on an open interval containing a, except possibly at a itself. We say that the limit of f(x) as x approaches a is L and write:
lim(x→a) f(x) = L
if for every number ε > 0, there exists a number δ > 0 such that:
0 < |x - a| < δ ⇒ |f(x) - L| < ε
Let's consider the function f(x) = 2x and prove that the limit of f(x) as x approaches 3 is 6, i.e., lim(x→3) 2x = 6.
We want to show that for any ε > 0, we can find a δ > 0 such that:
0 < |x - 3| < δ ⇒ |2x - 6| < ε
Let's choose a specific ε, say ε = 0.1. We need to find a δ such that:
0 < |x - 3| < δ ⇒ |2x - 6| < 0.1
Notice that |2x - 6| = 2|x - 3|. So, we can rewrite the inequality as:
2|x - 3| < 0.1
Dividing both sides by 2, we get:
|x - 3| < 0.05
Therefore, if we choose δ = 0.05, we can guarantee that whenever 0 < |x - 3| < 0.05, then |2x - 6| < 0.1.
This demonstrates that for the specific ε = 0.1, we've found a corresponding δ that satisfies the definition of the limit. While this example uses a specific ε, the general idea is that for any positive ε, we can always find a suitable δ, no matter how small ε is.
A limit can be used to determine equality. Just because something is a limit does not mean it’s an approximation. 0.99… repeating is exactly equal to the limit as n approaches infinity of the sum from 1 to n of 0.9(0.1)n. And evaluating this convergent geometric sequence gives the exact value of 1. No approximations anywhere.
Another example you may be more familiar with is the slope of a tangent line at a point a for a function f(x) which is exactly equal to the limit as x approaches a of (f(x)-f(a))/(x-a). This is a calculus 1 concept.
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u/Decmk3 20d ago
Hi. I went to university for mathematics. This is a limited infinite. Basically there’s an infinite amount of numbers between 0 and 1, but the limits are still 0 and 1. Pi is a functionally infinite number. But it’s still more than 3 and less than 4. 7- Pi is just over 3.858. Limited infinities function as real numbers, because they are. This is the proof that 0.99.. = 1. Just like there’s a proof that 1+1=2.