5333-Solns-Assignment6-Sp13 - Solutions Assignment#6 Section 21 x2 4x 5(x 1(x 5 = lim = lim(x 1 = 6 Set f(5 = 6 x5 x5 x5 x5 x5 3 lim 6(a True Set g(x k

5333-Solns-Assignment6-Sp13 - Solutions Assignment#6...

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Solutions, Assignment #6 Section 21. 3. lim x 5 x 2 - 4 x - 5 x - 5 = lim x 5 ( x + 1)( x - 5) x - 5 = lim x 5 ( x + 1) = 6. Set f (5) = 6. 6. (a) True. Set g ( x ) k , constant. Then g is continuous on R . By Theorem 21.6 (a) g · f = kf is continuous on D . (b) True. g = ( f + g ) - f . (c) False. f ( x ) = x 2 , g ( x ) = 1 /x x negationslash = 0 0 x = 0 on [ - 1 , 1]. f is continuous, g is not continuous at x = 0, f ( x ) g ( x ) = x is continuous. (d) False. Set f ( x ) = 1 if x is a rational number - 1 if x is an irrational number on [ - 1 , 1]. f 2 1 is continuous, f is everywhere discontinuous. (e) False. Set f ( x ) = 1 /x on D = (0 , 1]. (f) False. Let f be the example in (d) and let g = - f ; f + g 0 which is continuous on [ - 1 , 1]. (g) False. Let f be the example in (d) and let g = - f ; f · g 1 which is continuous on [ - 1 , 1]. (h) False. Let f and g be the function in (d). Then ( g f )( x ) = 1 for all x [ - 1 , 1]; f and g are not continuous, g f is continuous. 13. f is continuous at c and f ( c ) > 0. Let epsilon1 = f ( c ) / 2 > 0. Since lim x c f ( x ) = f ( c ), there is a neighborhood N ( c, δ ) = U such that | f ( x ) - f ( c ) | < epsilon1 for all x U D . Now | f ( x ) - f ( c ) | < epsilon1 implies - epsilon1 < f ( x ) - f ( c ) < epsilon1. It now follows that f ( c ) - epsilon1 < f ( x ) < epsilon1 + f ( c ) or f ( c ) 2 < f ( x ) < 3 f ( c ) 2 . for all x U D . In particular, f ( x ) > f ( c ) / 2 = α for all x U D . 14. Let epsilon1 = 1. There exists a neighborhood U of c such that | f ( x ) - f ( c ) | < 1 for all x U D.
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