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SCAN0027 - A Section 2.4 10 Using direct substitution...

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Unformatted text preview: A Section 2.4 10. Using direct substitution, 1111ng 1/7 x) 16 6 f(4) Therefore, 1’ is continuous at a = 4. . 1 . . . . 14. f (1) 2, but kn} ———1 does not eX1st. Therefore, f IS dlscontlnuous at 1. x0 x x2 x 1 16. f (x) 1, 1m} 2 1 3. Therefore, f is discontinuous at 1. x0 x 33- f(x) 3 x x 1, f(0) 1, f0) 1 There is a root of the equation {[32 l x in the interval (0,1) 40. f(x) lnx e x, f(1)[ 0.37, f(2)| 0.56 There is a root of the equation lnx e x in the interval (1,2) Sitiow 4. (a)2 (b)—2 (C) (b (d) ¢ (6) d> (1) Horizontal asymptotes: y 2, y 2 Vertical asymptotes: x 2, x 0, x 3 x2 40. x —————, and man others f() (x l)(x 3) y Se_<:ti9n_2§ 3 2 8. m 1imZx 5x 3 lim(2x 2x 3)(x 1) 1 y x 4 xo 1 x 1 x0 1 X 1 lel 0 2 10. m lim(x ) 2 2 y 2x x00 x 0 xoo(x 1) PDF created with pdfFactory trial version www.9dffactorycom ...
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