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Unformatted text preview: Problems graded: 20.16 (10 points), 28.3 (15 points); the others were graded for completeness and worth five points each (though 28.7, which was longer, was worth ten) and scores were then doubled to make the total value 100. 20.12. (a) (Graph should start out just below the origin for large negative x ; gradually decreasing as x increases until x = 0, where f ( x ) = − . 25. From here on out, f should be rapidly decreasing until x = 1, where f is undefined, but the limit from the left is −∞ . The limit from the right is ∞ ; from here, the function should swoop down to a minimum value for x somewhere between 1 and 2; technically, the coordinates are (4 / 3 , 6 . 75). Then the function rapidly increases to another vertical asymptote at x = 2; past x = 2 the function decreases, staying positive for x > 2 although the function should approach the x axis as x gets large). (b) lim x → 2 + f ( x ): Consider x = 2 + h for h small and positive. f ( x ) = (1 + h ) − 1 ∗ h − 2 ; h − 2 approaches ∞ as h approaches zero from the right and 1 + h approaches 1; this gives that f ( x ) approaches ∞ as h approaches 0 from the right by the limit laws from Section 9. lim x → 2 − f ( x ): Consider x = 2 + h for x small and negative. f ( x ) = (1 + h ) − 1 ∗ h − 2 ; h − 2 approaches ∞ as h approaches zero from the left and 1 + h approaches 1; this gives that f ( x ) approaches ∞ as h approaches 0 from the left by the limit laws from Section 9. lim x → 1 + f ( x ): Consider x = 1 + h for h small and positive. f ( x ) = h − 1 ∗ (1 + h ) − 2 ; (1 + h ) − 2 approaches 1 as h approaches zero from the right and h − 1 approaches ∞ ; this gives that f ( x ) approaches ∞ as h approaches 0 from the right by the limit laws from Section 9. lim x → 1 − f ( x ): Consider x = 1 + h for h small and negative. f ( x ) = h − 1 ∗ (1 + h ) − 2 ; (1 + h ) − 2 approaches 1 as h approaches zero from the left and h − 1 approaches −∞ ; this gives that f ( x ) approaches −∞ as h approaches 0 from the right by the limit laws from Section 9. (c) By Theorem 20.10, we know from the preceding part that since the right hand limit and the left hand limit of f ( x ) agree (both are ∞ ) as x approaches two, this is indeed lim x → 2 f ( x ). However, since these limits are different (one is ∞ and the other is −∞ ) for x approaching one, lim x → 1 f ( x ) does not exist....
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 Spring '08
 hitrik
 Limit, Mathematical analysis, Limit of a function, Lefthandedness, Multiplicative inverse

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