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samplequiz2

# samplequiz2 - SOLUTIONS TO QUIZ 2 Question 1[8 marks 1(a...

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SOLUTIONS TO QUIZ 2 Question 1 [8 marks] 1(a) Show that lim h 0 sin h h = 1. 1(b) Show that lim h 0 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 0 + 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 0 + sin h h = 1. 1(b) lim h 0 1 - cos h h = lim h 0 (1 - cos h )(1 + cos h ) h (1 + cos h ) = lim h 0 sin 2 h h (1 + cos h ) = lim h 0 sin h h !

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samplequiz2 - SOLUTIONS TO QUIZ 2 Question 1[8 marks 1(a...

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