Tham khảo tài liệu 'advanced mathematical methods for scientists and engineers episode 1 part 4', kỹ thuật - công nghệ, cơ khí - chế tạo máy phục vụ nhu cầu học tập, nghiên cứu và làm việc hiệu quả | 0 e 2 there is no value of 0 0 such that sin 1 x1 sin 1 x2 e for all x1 x2 e 0 1 and x1 x21 0. Thus sin 1 x is not uniformly continuous in the open interval 0 1 . Solution First consider the function x. Note that the function ựx 0 x is a decreasing function of x and an increasing function of 0 for positive x and 0. Thus for any fixed 0 the maximum value of Vx 0 x is bounded by ựõ. Therefore on the interval 0 1 a sufficient condition for l x L e is x e e2. The function x is uniformly continuous on the interval 0 1 . Consider any positive 0 and e. Note that 1 7 x x 0 for 1 x - 2 0 Thus there is no value of 0 such that 1 1 e x e for all x e 0. The function 1 is not uniformly continuous on the interval 0 1 . Solution Let the function f x be continuous on a closed interval. Consider the function e x 0 sup f e f x . í-x ổ Since f x is continuous e x 0 is a continuous function of x on the same closed interval. Since continuous functions on closed intervals are bounded there is a continuous increasing function e 0 satisfying e x 0 e 0 for all x in the closed interval. Since e 0 is continuous and increasing it has an inverse 0 e . Now note that f x f e e for aH x and e in the closed interval satisfying x e 0 e . Thus the function is uniformly continuous in the closed interval. 94 Solution 1. The statement lim an L n X is equivalent to V e 0 3 N . n N an L e. We want to show that V 5 0 3 M . m M a2 L2 5. Suppose that an L e. We obtain an upper bound on n L2 . a L2 an L an L C 2L e Now we choose a value of e such that n L2 5 e 2L e 5 e L2 5 L Consider any fixed 5 0. We see that since fore L2 5 L 3 N . n N an L e implies that n N a L2 5. Therefore V 5 0 3 M . m M a2 L2 5. We conclude that limn x an L2. 2. limn_ x an L2 does not imply that limn_ x an L. Consider an 1. In this case limn_ x an 1 and limn an 1. .
