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exam4_sol

exam4_sol - or ∞ ∞ form = lim x → 3 1 2 √ x 2 7 lim...

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4. (a) f ( x ) = x 3 + x - 1 , f 0 ( x ) = 3 x 2 + 1 , C 1 = 0 Use Newton’s method: C n +1 = C n - f ( C n ) f 0 ( C n ) . Then, C 2 = C 1 - f ( C 1 ) f 0 ( C 1 ) = 0 - f (0) f 0 (0) = 0 - - 1 1 = 1 C 3 = C 2 - f ( C 2 ) f 0 ( C 2 ) = 1 - f (1) f 0 (1) = 1 - 1 4 = 3 4 C 4 = C 3 - f ( C 3 ) f 0 ( C 3 ) = 3 4 - f ( 3 4 ) f 0 ( 3 4 ) 0 . 686 4. (b) f ( x ) = x 4 - 4 x 2 - 5 , f 0 ( x ) = 4 x 3 - 8 x, C 1 = 1 C 2 = C 1 - f ( C 1 ) f 0 ( C 1 ) = 1 - f (1) f 0 (1) = 1 - - 8 - 4 = - 1 C 3 = C 2 - f ( C 2 ) f 0 ( C 2 ) = - 1 - f ( - 1) f 0 ( - 1) = - 1 - - 8 4 = 1 C 4 = - 1 C 5 = 1 C 6 = - 1 . . . C 999 = 1 C 1000 = - 1 5. lim x 3 x 2 + 7 - 4 x 2 - 9 + 2 x - 8 x - 3 + x - 3 x 3 - 9 ! = lim x 3 x 2 + 7 - 4 x 2 - 9 + lim x 3 2 x - 8 x - 3 + lim x 3 x - 3 x 3 - 9 Use L’Hospital’s rule for first two terms. (Not for the third term since it’s not in
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Unformatted text preview: or ∞ ∞ form.) = lim x → 3 1 2 √ x 2 + 7 + lim x → 3 2 x ln 2 + lim x → 3 x-3 x 3-9 = 1 8 + 8 ln 2 + 0 = 1 8 + 8 ln 2 1...
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