On page 112 of Tenenbaum and Pollard's [i]Ordinary Differential Equations[/i] I am confronted with the a diagram like this:

The value Δ controls the accuracy of the displayed approximations ds, A and dA. Tenenbaum goes to the heart of the matter. dA is the area enclosed by OP, OQ, and the arc [i]ds[/i]. I will use a physical definition of equality: "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." (Newton) As the evanescent quantity vanishes, quantities whose ultimate ratios are those of equality can be used interchangeably. [b]The first obstacle is that, for (a) dA/dθ = r²/2 to hold, dθ must be indistinguishable from sin(dθ).[/b] Next: _______________________ To measure the ratio of areas, I would like very much to have, as [i]dθ[/i] vanishes, the ultimate ratios *rdθ/u = 1, *ds/h = 1 *h sinϕ/dr = 1 Then I could write interchangeably, in the limit, u² + dr² = h² = (rdθ)² + dr² = ds² I find the geometric details more instructive than the abstract answers which I know and am happy to take for granted.

Two limits: [b]1) PQ approaches the arc s:[/b] The arc s is caught between PQ and PC +CQ < PB. Drive θ to zero. The differences between the three lengths can be made as small as we wish (smaller than any finite quantity). Result 1: In the limit, the tangent PB, arc [i]s[/i], and subtended chord PQ are equal. We may use them interchangeably. [b]2) lim sinθ/θ = 1[/b]: [list] [*]Put sinθ and θ in a box together. Pick θ or sinθ. [i]I choose θ.[/i] Create two inequalities: θ > A sinθ θ < B sinθ A, B expressions in θ and x. This is the box: A sin[i]θ[/i] < [i]θ[/i] < B sin[i]θ[/i] Put sinθ/θ in the middle. [/list] [list] [*]Shut the box: drive x to 0 (the limiting value). [/list] [list] [*] The left and right sides are defined and positive for all values of x < 1. How far apart are they at x=0? They are the same. [/list] It is always possible to take a limit in this way (place the ratio in question at the center of an inequality, and let the bounds approach the limit). The upside and downside are the same: [i]I will get the limiting interval I asked for.[/i] I could choose badly and get an answer like: -∞ <sinθ/θ < ∞. Here, I found two expressions which are equal at θ=0.