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Unformatted text preview: Math 100, Section 105 Solutions to Problem Set 1 September 27, 2010 1. (Calculation of limits) (a) For x 6 = 1 we have x 2 x x 1 = x ( x 1) x 1 = x , and we know lim x → a x = a so lim x → 1 x 2 x x 1 = 1 . (b) lim x → 1 x 2 + x x 1 does not exist, since x 1 → while the numerator has a nonzero limit. Alternatively, if lim x → 1 x 2 + x x 1 existed and equalled L then the product rule would give 2 = lim x → 1 ( x 2 + x ) = lim x → 1 x 2 + x x 1 · ( x 1) = lim x → 1 x 2 + x x 1 lim x → 1 ( x 1) = L · 0 = 0 , a contradiction. (c) We have 1 h 1 (2+ h ) 2 1 4 = 4 (2+ h ) 2 h (2+ h ) 2 4 = 4 h + h 2 4 h (2+ h ) 2 = 4+ h 4(2+ h ) 2 . Now by the sum rule, lim h → (4+ h ) = 4 + 0 = 4 and by the product rule lim h → 4(2 + h ) 2 = 4 lim h → (2 + h ) 2 = 4 · 2 2 = 16 . Since the last number is nonzero we may apply the quotient rule and nd lim h → 1 h 1 (2+ h ) 2 1 4 = 4 16 = 1 4 ....
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This note was uploaded on 11/16/2010 for the course MATH 100 math 100 taught by Professor Lior during the Spring '10 term at The University of British Columbia.
 Spring '10
 LIOR

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