The limit as x→0 of sinx/x

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.