Question

define x0, x1, x2, as follows: xk = 2 xk − 1 for each integer k ≥ 1 x0 = 0 find lim n → [infinity] xn. (assume that the limit exists.)

Answer 1

The limit of the sequence xn as n approaches infinity is 0.

Step-by-step explanation:

To find the limit of xn as n approaches infinity, we need to understand the pattern of the sequence x0, x1, x2, ...

Given that xk = 2 * xk-1 for each integer k ≥ 1 and x0 = 0, we can compute the first few terms of the sequence to see the pattern:

x0 = 0

x1 = 2 * x0 = 2 * 0 = 0

x2 = 2 * x1 = 2 * 0 = 0

x3 = 2 * x2 = 2 * 0 = 0

From these calculations, we can observe that the sequence xn is a constant sequence of zeroes. Therefore, the limit of xn as n approaches infinity is 0.

Answer 2

The sequence xk = 2x(k-1) with x0 = 0 grows without bound as k approaches infinity, as each term is double the previous term, leading to a limit of infinity.

Step-by-step explanation:

The sequence defined by xk = 2xk-1 for each integer k ≥ 1 with an initial condition of x0 = 0 is a geometric sequence where each term is double the previous term.

As we proceed with this sequence, each subsequent term becomes larger by a factor of 2. Considering the limit of xn as n approaches infinity, we observe that since every term is multiplied by 2, the terms will increase without bound, growing towards infinity.