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Limits & continuity -solutions HW %

# Limits & continuity -solutions HW % - handa(nh5757...

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handa (nh5757) – Limits & continuity – bormashenko – (54880) 1 This print-out should have 16 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Below is the graph of a function f . 2 4 6 - 2 - 4 - 6 2 4 6 8 - 2 - 4 Use the graph to determine all the values of x on ( - 6 , 6) at which f fails to be continuous. 1. x = - 3 2. no values of x 3. x = 4 4. x = - 3 , 4 correct 5. none of the other answers Explanation: Since f ( x ) is defined for all values of x on ( - 6 , 6), the only values of x in ( - 6 , 6) at which the function f is discontinuous are those for which lim x x 0 f ( x ) = f ( x 0 ) or lim x x 0 - f ( x ) = lim x x 0 + f ( x ) . The only possible candidates here are x 0 = - 3 and x 0 = 4. But at x 0 = - 3 f ( - 3) = 8 = lim x → - 3 f ( x ) = 4 , while at x 0 = 4 lim x 4 - f ( x ) = 6 = lim x 4+ f ( x ) = 4 . Consequently, on ( - 6 , 6) the function f fails to be continuous only at at x = - 3 , 4 . 002 10.0 points Use continuity to evaluate lim x 3 π sin ( x + 4 sin x ) . 1. limit = 2. limit = 0 correct 3. limit = 3 π 4. limit = 1 5. limit = - 1 Explanation: Because both x are sin x is continuous on ( -∞ , ), the sum x + 4 sin x also is contin- uous everywhere on ( -∞ , ). But then the composition f ( x ) = sin( x + 4 sin x ) too is continuous everywhere on ( -∞ , ). Now by definition, lim x c f ( x ) = f ( c ) whenever f is continuous at x = c . For the given function f , therefore, lim x 3 π f ( x ) = sin(3 π + 4 sin(3 π )) . Consequently, limit = 0 .

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handa (nh5757) – Limits & continuity – bormashenko – (54880) 2 003 10.0 points Find all the points at which the function f ( x ) = x 2 - 2 x - 8 is not continuous. 1. x = 4 2. x = - 2 3. continuous everywhere correct 4. none of these 5. x = 4 , - 2 Explanation: Since f ( x ) = x 2 - 2 x - 8 is a polynomial function, it is continuous everywhere , including the points x = 4 , - 2 at which f (4) = f ( - 2) = 0 . 004 10.0 points Find all values of x at which the function f defined by f ( x ) = x - 6 x 2 - x - 30 is continuous, expressing your answer in in- terval notation. 1. ( -∞ , 6) (6 , ) 2. ( -∞ , - 5) ( - 5 , - 6) ( - 6 , ) 3. ( -∞ , - 6) ( - 6 , 5) (5 , ) 4. ( -∞ , - 5) ( - 5 , ) 5. ( -∞ , - 5) ( - 5 , 6) (6 , ) correct Explanation: After factorization the denominator be- comes x 2 - x - 30 = ( x - 6)( x + 5) , so f can be rewritten as f ( x ) = x - 6 ( x - 6)( x + 5) = 1 ( x + 5) whenever x = 6. At x = 6 both the numer- ator and denominator will be zero; thus f will not be defined, hence not continuous, at x = 6. Elsewhere f is a ratio of polynomial
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Limits & continuity -solutions HW % - handa(nh5757...

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