hw 3 soln - x . lim x 1 x 1 x 2 6 x + 5 = lim x 1 x 1 ( x...

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Calculus for the Life Science I MAT1330A , MAT1330B, MAT1330E Assignment 3 Due date: Oct. 7 Instructor (circle one): Jing Li , Catalin Rada , Frithjof Lutscher DGD (circle one): 1 , 2 , 3 , 4 Student Name (printed): Student ID Number: Question 1 To estimate the limit lim x 0 f ( x ), where f ( x ) = e x 1 x x 2 , one may use sequences of numerical values for x approaching 0. a ) Give two sequences to estimate the limit. A few terms for each sequence is enough. x n = 1 /n f ( x n ) 1 0.71828 1/2 0.594885 1/3 0.56051 1/4 0.5444 . . . . . . 1/1000 0.5001667 . . . . . . 1/10000 0.5000167 0 0.5 and x n = 1 /n f ( x n ) 1 0.367879 1 / 2 0.426122 1 / 3 0.448781 1 / 4 0.4608125 . . . . . . 1 / 100 0.499833 . . . . . . 1 / 1000 0.499983 0 0.5 b ) We may conclude that lim x 0 f ( x ) = 0 . 5 . Question 2 Does the limit lim x 2 | x 2 | x 2 exist ? Answer : no Justify your answer in one line. lim x 2 - | x 2 | x 2 = lim x 2 - 2 x x 2 = 1 n = lim x 2 + | x 2 | x 2 = lim x 2 + x 2 x 2 = 1 . Question 3 What is the value of the limit lim x 1 x 1 x 2 6 x + 5 ? Answer : 1 / 4 Justify your answer without using sequences of numerical values for
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Unformatted text preview: x . lim x 1 x 1 x 2 6 x + 5 = lim x 1 x 1 ( x 1)( x 5) = lim x 1 1 x 5 = 1 1 5 = 1 4 . Question 4 2 We consider the function f ( x ) = b if x Q x 2 if x n Q a ) Is the function f continuous at x = 0 ? Answer : Yes Justify your answer in one line. For x Q , f ( x ) = 0 0 = f (0) as x 0. For x n Q , f ( x ) = x 2 0 = f (0) as x b ) Does the derivative of f exist at x = 0 ? Answer : Yes and f (0) = 0 Justify your answer in one line. For x Q , f ( x ) = 0 and f ( x ) f (0) x = 0 0 as x 0. For x n Q , f ( x ) = x 2 and f ( x ) f (0) x = x 2 x = x 0 as x 0. . Note : Try to visualize the graph of f in your mind. Is it a nice continuous curve ?...
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This note was uploaded on 04/08/2010 for the course MATH MAT1330 taught by Professor Rad during the Spring '10 term at University of Ottawa.

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hw 3 soln - x . lim x 1 x 1 x 2 6 x + 5 = lim x 1 x 1 ( x...

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