Test 1b solutions

Test 1b solutions - Solutions for Test E1 I: a) 7|h|3 <...

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Solutions for Test E1 I: a) 7 | h | 3 < ε δ = ( ε 7 ) 1 / 3 b) First statement is false. Take a = 1 / 100 then a = 1 / 10. Second statement is true. c) The expression is positive for all x < - 2 and all x > - 1 since x 2 + 3 x + 2 = ( x + 2)( x + 1) > 0 . d) ( x ) = - 2( x - 1) + 2 = - 2 x + 4 . e) ( x ) = ( x - 1) + 2 = x + 1 . II: a) Limit is 3 b) Limit is - 13 / 2 c) Limit is 1 / 2 d) Limit is 1 / 2 e) Limit does not exist, since the limit from the right yields +1 and the limit from the left yields - 1. III: a) For all values of x . b) g ( h ) = | 2 + h | ( 2 + p 1 + (1 + h ) 2 ) p 1 + (1 + h ) 2 c) One reasonable answer to this problem is as follows. For | h | < 1 g ( h ) < 3 ( 2 + 1) 1
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Hence for | h | < min(1 , (1 + 2) 3 ε ) we have that | f (1 + h ) - f (1) | < ε .
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Test 1b solutions - Solutions for Test E1 I: a) 7|h|3 &lt;...

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