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Unformatted text preview: Section 4.1 #1) (a) Say  x 1  < 1 / 6 < 1. Then 0 < x < 2 and  x + 1  < 3.  x 2 1  =  x + 1  x 1  < 3 1 6 = 1 2 . (b) We need  x 2 1  < 1000. If  x 1  < 1, then 0 < x < 2 and  x +1  < 3.  x 2 1  =  x +1  x 1  < 3 < 1000. (c) Let  x 1  < 1 / (3 n ) < 1. Then 0 < x < 2 and  x + 1  < 3.  x 2 1  =  x + 1  x 1  < 3 1 3 n = 1 n . (d) Let  x 1  < 1 / (7 n ) < 1. Then 0 < x < 2 and  x 2 + x + 1   x  2 +  x  + 1 < 7.  x 3 1  =  x 2 + x + 1  x 1  < 7 1 7 n = 1 n . #3) lim x c f ( x ) = L ( > 0)( > 0)0 <  x c  < =  f ( x ) L  < ( > 0)( > 0)0 <  x c  < =  f ( x ) L    < lim x c  f ( x ) L  = 0 #8) Let > 0. Let = inf { c, c } > 0. Then if  x c  < , x > 0 and 1 x + sqrtc < 1 c . Therefore,  x c  =  x c   x + c  < c c = . So lim x c x = c . #9) (a) Let ( x n ) be any sequence tending to 2 that is never 2 or 1. We know that lim 1 1 x n = 1, so it follows from the Sequential Criterion that lim x 2 1 1 x = 1.1....
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This note was uploaded on 09/30/2009 for the course MATH 444 taught by Professor Junge during the Spring '08 term at University of Illinois at Urbana–Champaign.
 Spring '08
 JUNGE

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