Continuity of f(x) = sin(x)/x at x = 0Date: 01/20/2004 at 11:52:00 From: Zaheer Subject: continuity Given f(x) = ((sin x)/x if x is not equal to 0) ( 1 if x is equal to 0) Please tell me how f(x) is continuous at 0? I think that we have to draw a graph of sinx/x and then see whether it is continuous at zero or not. Date: 01/21/2004 at 19:12:25 From: Doctor Fenton Subject: Re: continuity Hi Zaheer, Thanks for writing to Dr. Math. Drawing a graph can suggest whether the function is continuous, but it does not constitute a proof. A function f(x) is continuous at x = a if lim f(x) = f(a) , x->a so to show that f(x) = sin(x)/x is continuous at x = 0, when f(0) is defined as 1, you must show that sin(x) lim ------ = 1 . x->0 x This is usually "proved" by a geometric argument (one can argue about the rigor of the following argument): B D . | . : .| . : . . @ : . O ----------------------------+---- A C If @ (for the Greek letter theta) is angle AOB, and arc BC is an arc of the unit circle, so that OB and OA are unit radii, then (using radian measure) the area of sector COB is @/2. AB is sin(@), and OA is cos(@), so Area(triangle AOB) < area(sector COB) (1/2)*(OA)*(AB) < area(sector COB) (1/2)cos(@)sin(@) < (1/2)@ cos(@)sin(@) < @ , or sin(@) 1 ------ < ------ . @ cos(@) If OBD is a line, then CD = tan(@), and Area(sector COB) < Area(triangle COB) , sin(@) (1/2)@ < (1/2)tan(@) = -------- 2*cos(@) , so sin(@) cos(@) < ------ . @ Combining the inequalities, sin(@) 1 cos(@) < ------ < ------ @ cos(@) . Since lim cos(@) = 1 , @->0 then sin(@) lim ------ = 1 . @->0 @ If you have any questions, please write back and I will try to explain further. - Doctor Fenton, The Math Forum http://mathforum.org/dr.math/ |
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