Practice Quiz 1c solutions

Practice Quiz 1c solutions - Practice Quiz IC for Calculus...

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Unformatted text preview: Practice Quiz IC for Calculus I, MATH 1501 (I) Let f ( x ) be given by f ( x ) = 1- x 1 + x for x 6 = 0. (a) Find all x such that | f ( x )- f (1) | < 1. (b) Find a ( ) > 0 such that | x- 1 | < ( ) | f ( x )- f (1) | < . and justify your answer. ANSWERS For (a), x > 0. for (b), here are two ways: | f (1 + h )- f (1) | = | h | | 2 + h | First Way Apply | a + b | | a | - | b | with a = 2, b = h to get | 2 + h | 2- | h | . Hence | f (1 + h )- f (1) | | h | 2- | h | Now for | h | < 2, 2- | h | > 0, so | h | 2- | h | < | h | < 2- | h | | h | < 2 1 + This gives ( ) = min 2 , 2 1 + . Second Way Inpose the condition | h | < 1, under which 1 < 2 + h < 3 and hence | 2 + h | > 1. Decreasing the denominator gives | f (1 + h )- f (1) | | h | 2- | h | < | h | and so if | h | < , we have what we want. Therefore, ( ) = min { 1 , } . (II) Which of the following limits exist? If they dont exist, explain why not. If they do, compute the value of the limit.compute the value of the limit....
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This note was uploaded on 02/07/2011 for the course MATH 1501 taught by Professor N/a during the Fall '08 term at Georgia Institute of Technology.

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Practice Quiz 1c solutions - Practice Quiz IC for Calculus...

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