Show that a sequence xn n1 converges to x if and only if for

Show that a sequence (xn) n=1 converges to x if and only if for all > 0, one can find N N such that |xn x| whenever n N. (Note the large inequality...)

Solution

Hi, I am Waqar. Below is the solution of above problem.

First we will assume that if sequence converges to x then we to show that  |xn x| 0 as n .

After that we will show the reverse inclusion

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x = limn xn, or xn x as n , if for every E > 0 there exists N N such that |xn x| < E for all n > N

If I give you an E > 0, you have to come up with an N. Also note that xn x as n means the same thing as |xn x| 0 as n .

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Suppose that (xn) is a sequence such that xn x and xn x 0 as n . Let E> 0. Then there exist N, N0 N such that |xn x| < E 2 for all n > N, |xn x\' | < E 2 for all n > N\'

Choose any n > max{N, N0}. Then, by the triangle inequality, |x x 0 | |x xn| + |xn x 0 | < E/ 2 + E/ 2 < E. Since this inequality holds for all E > 0, we must have |x x\' | = 0 (otherwise the inequality would be false for E = |x x\' |/2 > 0), so x = x\'

Happy to help you

Happy Chegging

Waqar