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Unformatted text preview: versionMWA1.3.1,2009030223:28 Solutions to Midterm Examination Math V63.0121, sections 4/5/6 Calculus I October 15, 2008 versionMWA1.3.1,2009030223:28 1 Math V63.0121 Solutions to Midterm Examination October 15, 2008 1 1. (16 Points) Consider the two functions graphed below: y O 1 2 3 4 5 1 1 2 3 graph of f y O 1 2 3 4 5 1 1 2 3 graph of g Use the graphs to answer the following questions. No justification is necessary. No partial credit will be given. (i) At which points a in { 0, 1, 2, 3, 4, 5 } does lim x → a f ( x ) exist? (A) 1 , 2 , and 3 only (B) 3 only (C) , 4 , and 5 only (D) 1 , 2 , 3 , and 4 only (E) 2 and 3 only A B C D E Solution. The point 0 is the lefthand endpoint of the domain of f , so we can say nothing about lim x → f ( x ) . The limit (from the left) at 4 and 5 is ∞ and ∞ , but at the other three points the limit does exist. N (ii) At which points a in { 0, 1, 2, 3, 4, 5 } does lim x → a f ( x ) exist? (A) 3 only (B) , 2 , 3 only (C) 1 , 2 , and 3 only (D) 2 , 3 , and 4 only (E) 2 and 3 only A B C D E Solution. The limits at 0 from the right and 5 from the left do not exist. The limit at 1 does not exist because the left and right hand limits disagree. The other two do exist. N (iii) At which points a in { 0, 1, 2, 3, 4, 5 } does lim x → a ( f ( x ) + g ( x )) exist? (A) 1 and 2 only (B) 2 and 4 only (C) , 2 , 4 , 5 (D) 2 only (E) 1 , 2 , and 4 only A B C D E –1– versionMWA1.3.1,2009030223:28 1 Math V63.0121 Solutions to Midterm Examination October 15, 2008 1 Solution. Clearly lim x → 2 ( f ( x ) + g ( x )) exists because both lim x → 2 f ( x ) and lim x → 2 g ( x ) exist: the limit of the sum of two functions is the sum of the limit of the two functions when those limits exist. Contrapositively, when a function can be broken into a sum where one has a limit and the other does not, the sum cannot have a limit. Hence lim x → a ( f ( x ) + g ( x )) does not exist when a = 0, 3, 5. The sum of two positive infinite limits is ∞ , so lim x → 4 ( f ( x ) + g ( x )) does not exist. However, since lim x → 1 f ( x ) = 1 lim x → 1 g ( x ) = 1 and lim x → 1 + f ( x ) = lim x → 1 + g ( x ) = we have lim x → 1 ( f ( x ) + g ( x )) = 1 + ( 1 ) = lim x → 1 + ( f ( x ) + g ( x )) = + = and lim x → 1 ( f ( x ) + g ( x )) = 0. This is why “DNE + DNE = DNE” is a false rule. N (iv) Which of these limits must be + ∞ ? I. lim x → 5 ( g ( x ) f ( x )) II. lim x → 4 + ( f ( x ) + g ( x )) III. lim x → 4 ( f ( x ) g ( x )) (A) II and III only (B) I and II only (C) I, II, and III (D) II only (E) I only A B C D E Solution. The limit lim x → 5 ( g ( x ) f ( x )) is of the form 1 ( ∞ ) , so is +...
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This note was uploaded on 10/25/2010 for the course MATH V63.0121 taught by Professor Staff during the Fall '08 term at NYU.
 Fall '08
 staff
 Calculus

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