By Tripathi M. M., Kim J., Kim S.

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5. Our proof of the Fundamental Theorem is complete (letting η = θ/(b − a)). The hypothesis says that if we can find a function f [x] so that lim ∆x→0 F [x + ∆x] − F [x] = f [x] ∆x uniformly for a ≤ x ≤ b then the conclusion is b f [x] dx = F [b] − F [a] a (Notice that the existence of the limit defining the integral is part of our proof. 2 Derivatives, Epsilons and Deltas The fundamental approximation defining the derivative of a real valued function can be formulated with or without infinitesimals as follows.

7. A Variation on Cauchy’s Problem Suppose an unknown differentiable function f [x] satisfies the (ExpSum) identity for all x and y, f [x + y] = f [x] · f [y] Does the function have to be f [x] = bx for some positive b? Since our unknown function f [x] satisfies the (ExpSum) identity and is differentiable, both of the following equations must hold: f [x + y] = f [x] · f [y] f [x + δx] = f [x] + f [x] · δx + ε · δx We let y = δx in the first identity to compare it with the increment approximation, f [x + δx] = f [x] · f [δx] f [x + δx] = f [x] + f [x] · δx + ε · δx so f [x] · f [δx] = f [x] + f [x] · δx + ε · δx f [x][f [δx] − 1] = f [x] · δx + ε · δx f [x] = f [x] f [δx] − 1 −ε δx or f [δx] − 1 f [x] = −ε f [x] δx with ε ≈ 0 when δx ≈ 0.

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. If x is nearly 1, then x + 1 is nearly 1 + 1 = 2. ” 31 32 3. 1 Plain Limits Technically, there are two equivalent ways to define the simple continuous variable limit as follows.