math1501 tests - Georgia Tech FALL 2009 Heinrich Matzinger...

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Unformatted text preview: Georgia Tech FALL 2009 Heinrich Matzinger SOLUTION FOR QUIZ 1 FOR MATH 1501: CALCULUS I 1) a) Draw the graph of the function f ( x ) := | x- 1 | x- 1 with domain [0 , 2]. At x = 1 we define the function f to be equal to f (1) = 0. b) Is the function continuous at x = 1. Explain. Answer: Discontinuity at the point x = 1. The reason is that when we approach x = 1 from right, then the value f ( x ) tends to 1. (It is always equal to 1 to the right of x = 1). When we approach from the left, then f ( x ) tends to- 1. Hence, there is a jump of size two at x = 1, so there the function is not continuous. 2) What is the limit lim x + ( sin ( x )) 2 x equal to? Why? Answer: we have that the above limit is equal to lim x + x ( sin ( x )) 2 x 2 = lim x + x ( sin ( x )) x 2 . Now, x goes to 0 as x goes to 0. Also, sin( x ) /x goes to 1 as x goes to 0. Hence, our limit is equal to lim x + x lim x + ( sin ( x )) x 2 (1) 2 = 0 ....
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This note was uploaded on 11/12/2009 for the course MATH 1501 taught by Professor N/a during the Fall '08 term at Georgia Institute of Technology.

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math1501 tests - Georgia Tech FALL 2009 Heinrich Matzinger...

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