sinx over x - To show: lim sin θ θ →0 θ = 1. The...

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Unformatted text preview: To show: lim sin θ θ →0 θ = 1. The following figure shows a sector OAC with angle θ of a unit circle and 2 right-angled triangles OBC and OAD. Observe that (1) area of sector OAC = ½ r 2 θ = θ/2, (2) area of ∆OBC = ½ OB · BC = ½ cos θ · sin θ, (3) area of ∆OAD = ½ OA · AD = ½ · 1 · tan θ = ½ tan θ. By comparing the above three areas, we have: ½ cos θ · sin θ ≤ θ/2 ≤ ½ tan θ. Multiplying these by 2/sin θ yields cos θ ≤ or θ →0 1 sin θ cos θ sin θ 1 ≤ . cos θ ≤ θ cos θ θ ≤ Since lim cos θ = 1, by Sandwich Theorem, lim sin θ θ →0 θ = 1. ...
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This note was uploaded on 05/10/2011 for the course MATH 1505 taught by Professor Yap during the Winter '11 term at National University of Singapore.

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