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Unformatted text preview: calculation: lim x  1 ( x + 1) 22 = lim x  1 x 2 + 2 x + 12 = lim x  1 ( 1 x 2 ) ( x 2 + 2 x + 1) 1 x 22 = lim x  1 x 2 1 + 2 x + 1 x 22 = 1 + 0 + 02 =2 6. lim u 3 u 2 + 1 u 2 + 3 = lim u (3 u 2 + 1) 1 u 2 ( u 2 + 3) 1 u 2 = lim u 3 + 1 u 2 1 + 3 u 2 = 3 + 0 1 + 0 = 3 7. The vertical asymptote occurs where the denominator is zero and the horizontal asymptote occurs where lim x 3 x +1 2x . Vertical: x = 2 Horizontal: y =3...
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This note was uploaded on 05/15/2008 for the course MATH 231 taught by Professor Mooney during the Spring '08 term at Wisconsin Milwaukee.
 Spring '08
 Mooney
 Limits

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