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Unformatted text preview: 1f ( x ) = 3 vii. lim x 1 + f ( x ) = 3 viii. lim x 1 f ( x ) = 3 ix. lim x f ( x ) = (This is more precise than The limit does not exist, even the statement is true.) 2 2. Find constants a and b such that lim x ax + b2 x = 1 . Solution. Since the denominator is approaching 0, for this limit to exist the numerator must also approach 0. Thus 0 = lim x ( ax + b2) = b2 b = 2 b = 4 . Next 1 = lim x ax + 42 x = lim x ax + 42 x ax + 4 + 2 ax + 4 + 2 = lim x ax x ( ax + 4 + 2) = lim x a ax + 4 + 2 = a 4 Thus a = 4. 3...
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This note was uploaded on 06/27/2010 for the course PHYS phys120 taught by Professor Xx during the Fall '08 term at Simon Fraser.
 Fall '08
 xx

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