2008Fall121Midterm-TR-A-sol

# 2008Fall121Midterm-TR-A-sol - versionTRA-1.3.1:21 Solutions...

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Unformatted text preview: versionTRA-1.3.1,2008-10-2414:21 Solutions to Midterm Examination Math V63.0121, sections 11/12/13 Calculus I October 16, 2008 versionTRA-1.3.1,2008-10-2414:21 1 Math V63.0121 Solutions to Midterm Examination October 16, 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 } does lim x → a + g ( x ) exist? (A) 4 only (B) 1 , 2 , 3 only (C) , 1 , 2 , and 3 only (D) , 1 , 2 , 3 , and 4 (E) 2 only A B C D E Solution. The limit (from the right) at 4 is ∞ , but the limit does exist at the other points. N (ii) At which points a in { 1, 2, 3, 4 } does lim x → a g ( x ) exist? (A) 1 , 3 , and 4 only (B) 1 , 2 , and 3 only (C) 2 only (D) 1 and 3 only (E) 2 and 4 only A B C D E Solution. We’re excluding the endpoints from the discussion. The limit at 1 does not exist because the left- and right-hand limits disagree. The same goes for 3, while the limit at 4 is ∞ . Among the given points, the limit exists only at 2. N (iii) At which points a in { 0, 1, 2, 3, 4, 5 } must lim x → a ( f ( x )- g ( x )) exist? (A) 1 , 2 , and 3 only (B) 1 and 4 only (C) 2 only (D) 2 and 3 only (E) 1 and 2 only A B C D E –1– versionTRA-1.3.1,2008-10-2414:21 1 Math V63.0121 Solutions to Midterm Examination October 16, 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 difference of two functions is the difference of the limit of the two functions when those limits exists. Contrapositively, when a function can be broken into a difference 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. Since both lim x → 4 f ( x ) = ∞ and lim x → 4 g ( x ) = ∞ , the limit lim x → 4 ( f ( x )- g ( x )) is indeterminate . It may exist or it may not, which means it is false to say that the limit must exist. This leaves only the point 1. Since lim x → 1- f ( x ) = 1 lim x → 1- g ( x ) = and lim x → 1 + f ( x ) = lim x → 1 + g ( x ) = 1 we have lim x → 1- ( f ( x )- g ( x )) = 1- = 1 lim x → 1 + ( f ( x )- g ( x )) =- 1 =- 1 So the limit lim x → 1 ( f ( x )- g ( x )) does not exist. N (iv) Which of these limits must be + ∞ ?...
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2008Fall121Midterm-TR-A-sol - versionTRA-1.3.1:21 Solutions...

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