They both get closer to but never reaching zero. Is the reason simply that one gets 1/x^2 gets closer to zero faster? So whats the threshold for something to be considered finite or infinite?
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> So whats the threshold for something to be considered finite or infinite?
This was a big problem for pre-17th-century mathematicians; how to deal with things that appear finite or not finite.
The trick turns out to be to use “limits.” The idea of saying that even if we can’t get to a finite number, we can get arbitrarily close to it.
Let’s take these integrals/areas.
Suppose we’re finding the area under these curves, “to the right” of x = 1.
For 1/x^2 our text books tell us we get an area of 1. For 1/x we get an undefined answer.
Let’s see how we could approach this using limits. We could change the “upper bound” of our area. Instead of going “all the way to the right” (whatever that means, given infinities are involved), what if we only go from x = 1 to x = 2?
For 1/x^2 we get an area of 0.5, for 1/x we get an area of log(2) = 0.69315…
We could do the same going up to x = 3, x = 4 and so on:
| Upper Bound | Area under 1/x^2 | Area under 1/x
| :— | –: | –:
| 2 | 0.5 | 0.69…
| 3 | 0.66… | 1.09…
| 5 | 0.75… | 1.38…
| 10 | 0.9 | 2.30…
| 20 | 0.95 | 2.99…
| 100 | .99 | 4.60…
| 1,000 | 0.999 | 6.90..
and so on.
From what we can see so far it looks like the 1/x^2 area is heading towards 1, but the 1/x one is getting bigger and bigger.
If we put in “infinity” as our upper limit… we can’t, because infinity isn’t really a number. But what if we sneak up on it using limits? We can say we think the limit of this sequence of numbers (for 1/x^(2)) is 1, while there isn’t a limit for the 1/x sequence of numbers.
The way we prove limits usually is by a challenge or game. You give me some number (bigger than 0), as small as you like. I then have to find some point in our sequence so that all numbers from that point onwards are closer to my proposed limit that the number you’ve given me.
So for example, if you give me the number 0.0001, I have to find some value for the upper bound for which all subsequent areas will be within 0.0001 of 1. Which I can do for the 1/x^2 area. The formula we get for those areas is 1 – 1/x, so I just need to find an x for which:
> 1 – (1 – 1/x) < 0.0001
If we solve this we get:
> 1/x < 0.0001
> x > 1 / 0.0001
> x > 10,000
So I can pick x = 10,001 and I win.
No matter how small a target you give me I can always find a point in our sequence where that number, and every subsequent number, will give me an area closer to 1 than your target. Our sequence never actually gets to 1 but it can get *arbitrarily close to 1.*
So we say that this sequence (1 – 1/x) converges to 1 as x tends to infinity. It has a finite value.
We cannot do that with the 1/x areas, as that gives us a sequence of log(x). log(x) does not converge to a value as x goes to infinity; playing a similar game, if you give me a target, no matter how high, I can find a value for x for which log(x) is bigger than your target.
> log (x) > n
> x > e^n
So try x = e^n + 1, and I win.
——————-
While in both cases the areas increase as we go further to the right, the amount with which they increase gets smaller.
For 1/x^2 the amount we are “adding on” as we increase the upper bound gets closer and closer to zero (if we are increasing our upper bound from A to B, the new area is (B-A)/AB, which will get smaller as B gets bigger). But with 1/x the amount we are “adding on” as we increase the upper bound doesn’t get closer to zero (we get log(B/A) which will increase as B increases).
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