Verified answer

You have

2 - x_{n+1} = 2 - √(2+x_n )

= (4 - (2 + x_n) )/(2 +√(2+x_n ) ) = (2 - x_n ) /(2 +√(2+x_n ) ).< (2 - x_n) / 2

Hence (2 - x_n) remains positive and decays faster than 1/2^n. So 2 is your limit.

(i) {x_n} is a positive sequence bounded above by 2:

Positivity of x_n is clear. Now, we'll show that x_n < 2 for all x_n.

This is clearly true for n = 1.

Assuming this is true for n = k,

x_(k+1) = √(2 + x_k) < √(2 + 2) = 2, as required.

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(ii) {x_n} is monotonically increasing:

x_(n+1) - x_n

= √(2 + x_n) - x_n

= [√(2 + x_n) - x_n] * [√(2 + x_n) + x_n] / [√(2 + x_n) + x_n]

= [(2 + x_n) - (x_n)^2] / [√(2 + x_n) + x_n]

= (2 - x_n)(1 + x_n) / [√(2 + x_n) + x_n]

> 0, by (i).

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By (i) and (ii), {x_n} converges by the Monotonic Convergence Theorem.

Let L denote the limit of the sequence.

So, letting n tend to infinity yields

L = √(2 + L)

==> L^2 = 2 + L

==> L^2 - L - 2 = 0

==> (L - 2)(L + 1) = 0

==> L = 2, since L > 0.

I hope this helps!

look at x ---> a million/(2+x). It decays on R+ and its by-product is often decrease than a million/4 in absolute fee. The fixed element L is given by technique of L(2+L) = a million so L = sqrt(2) - a million. hence (x_(n+a million)-L) = f'(c) (x_n - L) for some c between L and x_n; for this reason |x_(n+a million) - L| < a million/4 |x_n - L|. So the sequence (x_n) converges in direction of L and the version decays swifter than (a million/4)^n