140Bmid1csols

# 140Bmid1csols - 1 MATH 140B Midterm#1 Solutions Date Oct 9...

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Unformatted text preview: 1 MATH 140B Midterm#1 Solutions Date: Oct 9, 2007 1. (This is #3 on Version B) i) False. From the graph, the limit should be 0.1. ii) False. From the graph, the limit should be 0.2. iii) True. The function has no points of discontinuity on the open interval (2 , 3) . Therefore, the answer is C). 2. (This is #4 on Version B) Factoring the numerator gives: lim x → 3 parenleftbigg x 2 − 2 x − 3 x − 3 parenrightbigg = lim x → 3 ( x − 3)( x + 1) ( x − 3) = lim x → 3 ( x + 1) = 4 Therefore, the answer is D). 3. (This is #1 on Version B) i) True. If x → 1 − , then x − 1 is negative and close to zero. Since the numerator is negative, we conclude the limit is + ∞ . ii) False. If x → 1 + , then x − 1 is positive and close to zero. Since the numerator is negative, we conclude the limit is −∞ . iii) False. Since the left and right hand limits are unequal, the overall limit does not exist. Therefore, the answer is A). 4. (This is #2 on Version B) f ( x + h ) − f ( x ) h = 1 h bracketleftbigg 1 ( x + h ) 2 − 1 x 2 bracketrightbigg , by definition of f = 1 h bracketleftbigg 1 (1 + h ) 2 − 1 bracketrightbigg , put in x = 1 = 1 h bracketleftbigg 1 − (1 + 2 h + h 2 ) (1 + h ) 2 bracketrightbigg = − (2 + h ) (1 + h ) 2 Therefore, the answer is C)....
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140Bmid1csols - 1 MATH 140B Midterm#1 Solutions Date Oct 9...

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