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Unformatted text preview: SOLUTIONS TO QUIZ 2 Question 1 [8 marks] 1(a) Show that lim h → sin h h = 1. 1(b) Show that lim h → 1- cos h h = 0 . Hint: multiply 1- cos h h top and bottom by 1 + cos h and then use a trig identity and properties of limits. Solution to Question 1: 1(a) We need only show that lim h → + sin h h = 1 since sin h h = sin(- h )- h . Thus suppose 0 < h < π/ 2. Note that cos h > 0 for this range of values of h . Now consider the diagram below of the unit circle centered at the origin. By trig we have area( 4 OAB ) = cos h sin h 2 , area(sector OCB)= h 2 π π = h 2 and area( 4 OCD ) = tan h 2 . Because of the way the triangles and sector are nested within one another we have cos h sin h 2 < h 2 < tan h 2 = sin h 2 cos h . This is equivalent to cos h < sin h h < 1 cos h , for 0 < h < π/ 2. Using the continuity of cos h at h = 0, the squeeze principle and the fact that cos 0 = 1 we get lim h → + sin h h = 1....
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This note was uploaded on 01/12/2010 for the course STAT 100 taught by Professor Denissjerve during the Winter '06 term at San Jose State University .
- Winter '06