1873_solutions

Positive integers n we conclude that x n is not

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Unformatted text preview: the set R 0, 1 then, since R 0, 1 is a neighborhood of x and x n is not frequently (or, indeed, ever) in the set R 0, 1 , the number x must fail to be a partial limit of x n . m k Exercises on the Elementary Properties of Limits 1. The purpose of this exercise is to use Scientific Notebook to gain an intuitive feel for the limit behaviour of a rather difficult sequence. a. Point at the equation xn  and then click on the button nn n n! e n to supply the definition to Scientific Notebook. When you see the screen make the selection “A function argument” so that Scientific Notebook knows that you are defining a sequence. b. Point at the expression x n and click on the button to display the sequence graphically. Revise your graph and set the domain interval as 1, 500 . Double click into your graph to make the buttons appear in the top right corner and click on the bottom button to select it. Trace your graph with the mouse and show graphically that nn n lim  0. 3989. n Ý n! e n c. Point at the expression nn n n! e n and ask Scientific Notebook to evaluate it numerically. Compare the result with the limit value that you lim nÝ 137 found graphically. d. Point at the expression and ask Scientific Notebook to evaluate it exactly to show that the limit is 1/ 2 . 2. Prove that 5 n /n! 0 as n Ý. Solution: Whenever n 5 we have 5n  5 5 5 5 0 1 2 3 4 n! Now, to prove that 5 n /n! 0 as n Ý, suppose that Choose an integer N such that 54 4! 5 55 n 5 6  0. 5 n 5 N 5 . 4! Then whenever n N we have 5n n! 0 3. Prove that n!/n n 0 as n 54 4! 54 4! 5 n 5 N . Ý. Hint: Make use of the fact that, for each n we have 0 n!  nn 1 n 2n n n 1. n 4. Given that x n is a sequence of real numbers, that x  0 and that x n x as n integer N such that the inequality x n  0 holds for all integers n N. Hint: The interval Ý, prove that there exists an 0, Ý is a neighborhood of the number x. 5. Given that x n 0 for every positive integer n and that x is a partial limit of the sequence x n , prove that x 0. We need to show that no negative number can be a partial limit of x n . Suppose that y  0. Since the interval Ý, 0 is a neighborhood of y and x n is not frequently (or, indeed, ever) in the interval Ý, 0 , we deduce that y is not a partial limit of x n . 6. Suppose that x n is a sequence of real numbers and that x equivalent: a. x n x as n R. Prove that the following conditions are Ý. b. |x n x | 0 as n Ý. Condition a says that for every  0 there exists an integer N such that whenever n have |x n x |  . Condition b says that for every  0 there exists an integer N such that whenever n have ||x n x | 0 |  . These two conditions clearly say the same thing. 7. Suppose that x n is a sequence of real numbers, that x n Ý. R and that x n x as n N we N we Ý. Prove that |x n | |x | as Hint: Make use of the fact that for each n we have 0 ||x n | |x || |x n x |. 8. Suppose that x n is a sequence of real numbers, that x R and that x n x as n Ý. Suppose that p is an integ...
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