A) How do you prove that if 0(<or=)x(<or=)10, then 0(<or=)sqrt(x+1)(<or=)10?

B) So once that is found, then how can you prove that if 0(<or=)u(<or=)v(<or=)10, then 0(<or=)sqrt(u+1)(<or=)sqrt(v+1)(<or=)10?

C) They give a recursively defined sequence: a_1=0.3; a_(n+1)=sqrt((a_n)+1)for n>1
How do you find out the first five terms for it. then prove that this sequence converges. What is a specific theorem that will guarantee convergence, along with the algebraic results of parts A and B?

D) How do you find out the exact limit of the sequence defined in part C? Are you supposed to square the recursive equation and take limits using limit theorems? If so, then which are these theorems?

Thank you very much.

A) To prove that if 0 (≤) x (≤) 10, then 0 (≤) sqrt(x+1) (≤) 10, we can use the properties of square roots.

First, we need to show that 0 (≤) sqrt(x+1). Since x (≥) 0 by assumption, adding 1 to x guarantees that x+1 (≥) 1. And since the square root function is non-negative, we have sqrt(x+1) (≥) sqrt(1) = 1.

Next, we need to show that sqrt(x+1) (≤) 10. Since x (≤) 10 by assumption, adding 1 to x guarantees that x+1 (≤) 11. And since the square root function is non-decreasing, we have sqrt(x+1) (≤) sqrt(11) < 10.

Thus, combining both inequalities, we have established the desired result: 0 (≤) sqrt(x+1) (≤) 10.

B) To prove that if 0 (≤) u (≤) v (≤) 10, then 0 (≤) sqrt(u+1) (≤) sqrt(v+1) (≤) 10, we can apply the result from part A multiple times.

Since 0 (≤) u (≤) v (≤) 10, we know that 0 (≤) sqrt(u+1) (≤) sqrt(v+1) (≤) 10 for the individual terms, using the logic from part A.

C) To find the first five terms of the recursively defined sequence a_1 = 0.3 and a_(n+1) = sqrt(a_n + 1) for n > 1, we can apply the recursion.

a_1 = 0.3
a_2 = sqrt(a_1 + 1) = sqrt(0.3 + 1)
a_3 = sqrt(a_2 + 1)
a_4 = sqrt(a_3 + 1)
a_5 = sqrt(a_4 + 1)

Evaluating these expressions:

a_1 ≈ 0.3
a_2 ≈ 1.1487
a_3 ≈ 1.3487
a_4 ≈ 1.4023
a_5 ≈ 1.4142

To prove that this sequence converges, we can show that it is bounded and monotonically increasing.

To show boundedness, notice that 0 (≤) a_n (≤) 10 for all n. This is because a_1 starts at 0.3, and the recursive formula involves taking square roots of numbers that are greater than or equal to 1, which produces values that are less than or equal to 10. Thus, the sequence is bounded.

To show monotonically increasing, we can use the fact that the square root function is non-decreasing. Since each term in the sequence is generated by taking the square root of the previous term plus 1, we can conclude that a_n is greater than or equal to a_(n-1) for all n. Thus, the sequence is monotonically increasing.

By the Monotone Convergence Theorem, a bounded, monotonically increasing sequence must converge. Therefore, the sequence a_n converges.

D) To find the exact limit of the sequence defined in part C, we can denote the limit as L and solve the recursive equation for L:

L = sqrt(L + 1)

To solve this equation, we square both sides:

L^2 = L + 1

Rearranging the equation gives a quadratic equation:

L^2 - L - 1 = 0

Using quadratic formula, we find the roots of this equation:

L = (1 ± sqrt(5))/2

Since L is the limit of the sequence, it must be a non-negative number. Therefore, the only valid solution is L = (1 + sqrt(5))/2.

In summary, the exact limit of the sequence is L = (1 + sqrt(5))/2, also known as the golden ratio.