Newton typed the following on his Apple IIe computer…
Quantities, and the ratios of quantities, which in any finite time converge continually to equality, and before the end of that time approach nearer the one to the other than by any given difference, become ultimately equal.
Any student of calculus should spend some time mulling over (or reckoning) how the above paragraph-like sentence explains the notion of a limit.
In this blog we wish to talk about two sets that are open sets, open at least on the end containing the answer. We will deviate in one respect. We are going to talk about f(x) approaching 5 but Newton does one more step and talks about f(x) – 5 approaching 0. This would be one more column of calculations on your spreadsheet. Newton will speak of making the difference “as small as you want” and this corresponds to get getting “as close as you want” to the target answer (for us, 5).
We know that if f(x)=x+3 then if x is approaching 2 from the less-than side (1.999, 1.99999, 1.99999999, etc.) then we will get answers like 4.9999 for our approximations and the actual answer (and the limit) is 5.
One open set is an interval approaching 2. This is where x pulls all those 1.99999 values.
The second open set is an interval approaching 5. This has all those 4.99999 values.
We argue that we can hit any value on the pink interval (any 4.9999….) using a value found on the yellow interval (all those 1.99999….).
It is fair for you to ask “where do those two intervals start? That was determined by you. You started this process with your first guess. Possibly you tried x=1.9 and you got f(x) = 4.9. When you did that you started the pink interval at 4.9 and in this example we have the two open intervals as:
[1.9, 2)
[4.9, 5)
Newton takes everything that we have above and determines that we have “5” when approaching 2 because if we suppose a different value instead of 5, we create a contradiction.
Suppose that we were guessing a value greater than 5. Suppose we got confused and thought it was 6.
We get out 4.999 and 4.999999 values but no choice for x on [1.9, 2) can give us a value like 5.1 and the above carries the promise that if the answer is 6 then we can hit any value on the open interval [4.9, 6).
Some work remains. We need to show that the language guides us in a way that the answer can’t be something inside the yellow interval.
This part is made complicated somewhat because a function is allowed to change direction and there can be a function where f(x)=5 for an x that is not 2. We have to argue that we made our first guess so close that this problem wouldn’t happen.
If we are approaching the limit over an interval where the function is monotonic then the value 5 won’t exist within the yellow interval.
Vectoreverything 