Define x0, x1, x2, ... as follows (in Description). Find the limit of xn as n → ∞. (Assume that the limit exists.)

Question

Answer 1

If you just assume it has a limit, then √(2 + x) = x iff 2 + x = x^2 iff x^2 - 2 - x = (x - 2)(x + 1) = 0

As the sequence is positive, we can ignore the negative solution, so the solution is x = 2.

A way to show the limit exists and it is 2 follows.

The squeeze theorem is a great way to show that the limit is 2.

We will show that the sequence is always less than 2 and it is greater than or equal to x_k = (2 + x_k-1)/2, and that lim k -> ∞ xk, where x_k = (2 + x_k-1)/2, is 2

Consider x_k = (2 + x_k-1)/2, where x_o = 0

Let's show that xk = 2 - 2 * (1/2)^k

For n = 0, 2 - 2 * (1/2)ⁿ = 2 - 2 * (1/2)⁰ = 2 - 2 * 1 = 2 - 2 = 0

Assume for k

Then, for k + 1, (2 + x_k)/2 = (2 + 2 - 2 * (1/2)^k )/2 = (4 - 2 * (1/2)^k )/2 = 4/2 - 2 * (1/2)^k /2 =

2 - 2 * (1/2)^k * 1/2 = 2 - 2 * (1/2)^k+1

lim k -> ∞ 2 - 2 * (1/2)^k+1 = 2 - 2 * 0 = 2

Now, to show that this sequence is less than or equal to x_k = √(2 + x_k)

Note that the square root and square functions are monotone increasing for x > 0

For k = 0, both sequences are equal to 0. Assume it is true for k that x_k in x_k = √(2 + x_k-1) is greater than or equal to x_k in x_k = (2 + x_k-1)/2 = 1 + x_k-1/2

Then, consider xk-1 in the two sequences to be equal (if it is greater in the square root sequence, the next term will have an even higher value.

√(2 + x_k-1) >= (2 + x_k-1)/2 iff 2 + x_k-1 >= (1 + x_k-1/2)² = 1 + x_k-1 + x²_k-1 /4

Then, 2 + x_k-1 >= 1 + x_k-1 + x²_k-1 /4 iff x²_k-1 /4 <= 1 iff x²_k-1 <= 4 iff x_k-1 <= 2, which is the case.

Thus, we have shown that the square root sequence is greater than or equal to the sum sequence.

Then, for x_k-1 < 2, √(2 + x_k-1 ) < 2 iff 2 + x_k-1 < 2² iff x_k-1 < 2. As x_k-1 < 2, then the √(2 + x_k-1 ) < 2 for all k.

Now, we have shown that √(2 + x_k-1 ) is squeezed by (2 + x_k-1)/2 and 2, each of which has limit 2, so

the sequence √(2 + x_k-1) has limit 2.