If lim┬(n→∞)⁡〖(x_n-x)/(x_n+x)〗=0, prove that lim┬(n→∞)⁡〖x_n 〗=x.

To prove that lim┬(n→∞)⁡〖x_n 〗=x, we can use the definition of a limit. In this case, we're given that lim┬(n→∞)⁡〖(x_n-x)/(x_n+x)〗=0.

We want to prove that for any positive number ε (epsilon), there exists a positive integer N such that for all n>N, |x_n - x| < ε.

To begin, let's manipulate the expression (x_n - x)/(x_n + x) using some algebraic simplifications:

(x_n - x)/(x_n + x) = [(x_n - x)/(x_n + x)] * [(x_n - x)/(x_n - x)]
= [(x_n - x)^2] / [(x_n^2 - x^2)]
= [x_n^2 - 2x_nx + x^2] / [x_n^2 - x^2]
= [x_n - x]^2 / [(x_n - x)(x_n + x)]
= [(x_n - x) / (x_n - x)] * [(x_n - x) / (x_n + x)]
= 1 * [(x_n - x) / (x_n + x)]
= (x_n - x) / (x_n + x)

Since the expression (x_n - x)/(x_n + x) approaches 0 as n approaches infinity, it means that for any positive number ε, there exists a positive integer N such that for all n>N, |(x_n - x) / (x_n + x)| < ε.

Now, we can continue the proof by cases:
1. If x ≠ 0, we can multiply both sides of the inequality by |x_n + x| and simplify:

|x_n - x| < ε * |x_n + x|
| x_n - x | / | x_n + x | < ε

Since the absolute value of a number divided by the absolute value of another number is always between 0 and 1, we can replace | x_n - x | / | x_n + x | with a positive constant K:

K < ε

So, for any positive number ε, there exists a positive integer N such that for all n>N, |(x_n - x) / (x_n + x)| < ε, which implies that lim┬(n→∞)⁡〖x_n 〗=x.

2. If x = 0, then the expression (x_n - x)/(x_n + x) becomes x_n / x_n, which is always 1. Therefore, the initial limit we were given is not valid for this case, and it remains unproven that lim┬(n→∞)⁡〖x_n 〗=x when x = 0.

In summary, when x ≠ 0, we have proven that lim┬(n→∞)⁡〖x_n 〗=x. However, when x=0, it remains unproven.