Geometric Proof of a Limit
Date: 12/06/2003 at 19:15:19 From: Jackos Subject: Limits Can you prove that lim[x->0](sinx)/x = 1 without using L'Hopital's rule?
Date: 12/06/2003 at 21:32:49 From: Doctor Rob Subject: Re: Limits Thanks for writing to Ask Dr. Math, Jackos! Here's a nice geometric proof. Draw a unit circle with center at the origin O. From the origin, draw a radius which makes an acute angle of t radians with the positive x-axis. Then 0 < t < Pi/2. It intersects the circle at the point P(cos[t],sin[t]). Draw the tangent line x = 1. Extend the radius to meet that tangent at the point R(1,tan[t]). Connect P to Q(1,0). Consider the areas of the triangle OPQ, the sector OPQ of the circle, and the triangle OQR. The area of triangle OPQ is (1/2)*1*1*sin[t] (half the product of two sides times the sine of the included angle). The area of sector OPQ is (1/2)*1^2*t (half the square of the radius times the central angle measure). The area of the triangle OQR is (1/2)*1*tan[t] (half the base times the altitude). Obviously by containment, those three areas satisfy A(triangle OPQ) < A(sector OPQ) < A(triangle OQR) (1/2)*sin[t] < (1/2)*t < (1/2)*tan[t] sin[t] < t < sin[t]/cos[t] Rearranging these inequalities, we have the general relationship cos[t] < sin[t]/t < 1, 0 < t < Pi/2. Also, if -Pi/2 < t < 0, then 0 < -t < Pi/2, so cos[-t] < sin[-t]/(-t) < 1, -Pi/2 < t < 0 cos[t] < sin[t]/t < 1, -Pi/2 < t < 0. Thus the inequality holds for all nonzero values of t between -Pi/2 and Pi/2. Now take the limit as t -> 0. 1 = lim cos[t] <= lim sin[t]/t <= lim 1 = 1, t->0 t->0 t->0 so lim sin[t]/t = 1. t->0 There's another proof which uses the infinite Maclaurin series sin(t) = t - t^3/3! + t^5/5! - t^7/7! + ... Feel free to write again if I can help further. - Doctor Rob, The Math Forum http://mathforum.org/dr.math/
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