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Extra info for Mathematical Background: Foundations of Infinitesimal Calculus
1 guarantees that ξ is defined and positive. √ √ Suppose ξ is not infinitesimal. Then there is a positive real number 0 < a with a < ξ. Squaring √ √ and using the Function Extension Axiom on the property 0 < b < c implies 0 < b < c, we see that 0 < a2 < ξ contradicting the assumption that ξ ≈ 0 is infinitesimal. 8. Infinite Limits We know that c + 7 = 0 because we are given that c = −7 and c ≈ −7 or c = −7 + ι with ι ≈ 0, but ι = 0. This means that c + 7 = ι = 0 and so c+7≈0 This, together with what we know about reciprocals of infinitesimals tells us that 1 is infinite c+7 We do not know if it is positive or negative; we simply weren’t told whether c < 7 or c > 7, but only that c ≈ 7.
Continuous Functions where f [x] is largest among points of the form a + k δx, even when δx ≈ 0 is infinitesimal, but it says this as follows: a ≤ x ≤ b and I x−a δx =1 ⇒ f [x] ≤ f [M [δx]] We interpret this as meaning that, ‘among the hyperreal numbers of the form a + k δx, f [x] is largest when x = M [δx],’ even when δx is a positive infinitesimal. 5. Bolzano’s Intermediate Value Theorem If y = f [x] is continuous on the interval a ≤ x ≤ b, then f [x] attains every value intermediate between the values f [a] and f [b].
The chapter also gives many computations of limits. A basic fact about the sine function is lim x→0 Sin[x] =1 x Notice that the limiting expression Sin[x] is defined for 0 < |x − 0| < 1, but not if x = 0. The x sine limit above is a difficult and interesting one. ” Here is a simpler limit where we can see what is being approached. x2 − 1 =2 x→1 x − 1 lim While this limit expression is also only defined for 0 < |x − 1|, or x = 1, the mystery is easily resolved with a little algebra, (x − 1)(x + 1) x2 − 1 = =x+1 x−1 (x − 1) So, x2 − 1 = lim (x + 1) = 2 x→1 x − 1 x→1 lim The limit limx→1 (x + 1) = 2 is so obvious as to not need a technical definition.