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Solutions Homework 5

Solutions Homework 5 - MATH 106 CALCULUS I FOR BIO SOC SCI...

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MATH 106 CALCULUS I FOR BIO. & SOC. SCI. FALL 2011 INSTRUCTOR: JOS ´ E MANUEL G ´ OMEZ Solutions Homework 5. Please show all your work. (1) (5 Points) Consider the function f ( x ) = | x | . (a) (3 Points) Compute lim h 0 - f ( h ) - f (0) h and lim h 0 + f ( h ) - f (0) h . Solution: Let’s start by recalling that the function absolute value is defined by | h | = braceleftbigg h if h 0 , - h if h < 0 . Let’s compute first lim h 0 - f ( h ) f (0) h . In this case h 0 , thus h < 0 and therefore | h | = - h . With this in mind we have lim h 0 - f ( h ) - f (0) h = lim h 0 - | h | - | 0 | h = lim h 0 - | h | - | 0 | h = lim h 0 - | h | h = lim h 0 - - h h = lim h 0 - ( - 1) = - 1 . Similarly, to compute lim h 0 + f ( h ) f (0) h we have h 0 + , thus h > 0 and therefore | h | = h and lim h 0 + f ( h ) - f (0) h = lim h 0 + | h | - | 0 | h = lim h 0 + | h | - | 0 | h = lim h 0 + | h | h = lim h 0 + h h = lim h 0 - (1) = 1 . (b) (2 Points) Use the above to conclude that f (0) does not exist even though f is continuous everywhere. Solution: The quantity f (0) is defined by f (0) = lim h 0 f ( h ) - f (0) h . According to the previous part lim h 0 - f ( h ) f (0) h and lim h 0 + f ( h ) f (0) h are differ- ent and thus this limit does not exist. This means that f

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Solutions Homework 5 - MATH 106 CALCULUS I FOR BIO SOC SCI...

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