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
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 Spring '08
 CASTLE
 Calculus

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