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## Main Question or Discussion Point

Hi!

Have I understood the concept of the limit of a sequence correctly?

Let's say we have a sequence {x_n} (dots for spacing):

----|------------------|-------------------|----------|------->

a-epsilon...................a.....................a+epsilon.........R............x_n

So "a" is the limit of {x_n} if there exists an n_0 = f(epsilon) so that for every n > n_0 the inequality |x_n - a| < epsilon is true.

How can |x_n - a| < epsilon be true for values of n which go beyond a+epsilon? For example, for R the inequality |R - a| < epsilon is not true, yet the definition of a limit of a sequence says that the inequality must hold for all n > n_0 (including that n which gives us R).

I guess they avoid that problem on the "left side" of "a" by saying that n_0 is a function of epsilon and that automatically cuts off all values of "n" before a-epsilon. Why is that not done on the "right side"? They allow "n" to increase towards positive infinity so that the inequality won't hold past the value of "n" which gives us a+epsilon.

- Kamataat

Have I understood the concept of the limit of a sequence correctly?

Let's say we have a sequence {x_n} (dots for spacing):

----|------------------|-------------------|----------|------->

a-epsilon...................a.....................a+epsilon.........R............x_n

So "a" is the limit of {x_n} if there exists an n_0 = f(epsilon) so that for every n > n_0 the inequality |x_n - a| < epsilon is true.

How can |x_n - a| < epsilon be true for values of n which go beyond a+epsilon? For example, for R the inequality |R - a| < epsilon is not true, yet the definition of a limit of a sequence says that the inequality must hold for all n > n_0 (including that n which gives us R).

I guess they avoid that problem on the "left side" of "a" by saying that n_0 is a function of epsilon and that automatically cuts off all values of "n" before a-epsilon. Why is that not done on the "right side"? They allow "n" to increase towards positive infinity so that the inequality won't hold past the value of "n" which gives us a+epsilon.

- Kamataat