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 the statement that if 0 (≤) x (≤) 10, then 0 (≤) √(x+1) (≤) 10, we can break it down into two inequalities:

1. First, we need to prove that if 0 (≤) x (≤) 10, then 0 (≤) √(x+1).
To prove this, we will show that if 0 (≤) x (≤) 10, then x+1 (≥) 1, which implies √(x+1) (≥) 0.
Since the square root function is non-negative, we have proven the first part of the inequality.

2. Next, we need to prove that if 0 (≤) x (≤) 10, then √(x+1) (≤) 10.
To prove this, we will show that if 0 (≤) x (≤) 10, then x+1 (≤) 11, which implies √(x+1) (≤) √11.
Since 11 is a positive number, we have proven the second part of the inequality.

Therefore, combining the two parts, if 0 (≤) x (≤) 10, then 0 (≤) √(x+1) (≤) 10.

B) To prove the statement that if 0 (≤) u (≤) v (≤) 10, then 0 (≤) √(u+1) (≤) √(v+1) (≤) 10, we can use the transitive property of inequalities and the result from part A.

Using the transitive property, if 0 (≤) u (≤) v (≤) 10, then by part A, 0 (≤) √(u+1) (≤) √(v+1) (≤) 10.

Therefore, if 0 (≤) u (≤) v (≤) 10, then 0 (≤) √(u+1) (≤) √(v+1) (≤) 10.

C) To find the first five terms of the recursively defined sequence a_1 = 0.3 and a_(n+1) = √(a_n + 1) for n > 1, we can use the recursive formula to calculate each term step by step:

a_1 = 0.3
a_2 = √(0.3 + 1) = √1.3 ≈ 1.140175
a_3 = √(1.140175 + 1) ≈ 1.355249
a_4 = √(1.355249 + 1) ≈ 1.380277
a_5 = √(1.380277 + 1) ≈ 1.387040

To prove that this sequence converges, we can use the Monotone Convergence Theorem. The Monotone Convergence Theorem states that if a sequence is bounded and monotonic (either increasing or decreasing), then it must converge.

In this case, we can observe that the sequence is increasing because each term is the square root of the previous term plus 1. Additionally, we can see that the sequence is bounded by 0 and 10 because of the results from parts A and B.

Therefore, by the Monotone Convergence Theorem, the sequence defined by a_1 = 0.3 and a_(n+1) = √(a_n + 1) for n > 1 converges.

D) To find the exact limit of the sequence defined in part C, we can use the fact that the sequence converges.

Let's denote the limit as L. Taking the limit on both sides of the recursive equation a_(n+1) = √(a_n + 1), we get:

L = √(L + 1)

Squaring both sides, we have:

L^2 = L + 1

Rearranging the equation, we get:

L^2 - L - 1 = 0

This is a quadratic equation that can be solved using the quadratic formula:

L = [1 ± √(1 + 4(1)(1))] / 2
L = [1 ± √5] / 2

Since the elements of the sequence are positive, the only valid solution is:

L = (1 + √5) / 2 ≈ 1.618034

Therefore, the exact limit of the sequence defined in part C is approximately 1.618034.

The theorem that guarantees convergence in this case is the Monotone Convergence Theorem, mentioned in part C.

A) To prove that if 0 (≤) x (≤) 10, then 0 (≤) sqrt(x+1) (≤) 10, we need to show that the inequality holds for all values within the given range.

1. Start with the inequality assumption: 0 (≤) x (≤) 10.
2. Add 1 to both sides of each inequality: 1 (≤) x+1 (≤) 11.
3. Take the square root of each term: sqrt(1) (≤) sqrt(x+1) (≤) sqrt(11).
4. Simplify: 0 (≤) sqrt(x+1) (≤) sqrt(11).
5. Since the given range is 0 (≤) x (≤) 10, and sqrt(11) is less than or equal to 10, we have proven that if 0 (≤) x (≤) 10, then 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 use the result proven in part A.

1. Start with the inequality assumption: 0 (≤) u (≤) v (≤) 10.
2. Apply the result from part A to each pair of inequalities: 0 (≤) sqrt(u+1) (≤) 10 and 0 (≤) sqrt(v+1) (≤) 10.
3. Since u (≤) v, it follows that sqrt(u+1) (≤) sqrt(v+1) (≤) 10.
4. Therefore, if 0 (≤) u (≤) v (≤) 10, then 0 (≤) sqrt(u+1) (≤) sqrt(v+1) (≤) 10.

C) To find the first five terms of the recursively defined sequence and prove convergence, we need to apply the recursive formula to each term.

1. Start with the given initial term: a_1 = 0.3.
2. Use the recursive formula for each subsequent term: a_(n+1) = sqrt(a_n+1).
- a_2 = sqrt(a_1+1) = sqrt(0.3+1) ≈ 1.04881
- a_3 = sqrt(a_2+1) ≈ 1.39685
- a_4 ≈ 1.45535
- a_5 ≈ 1.46079
3. The first five terms of the sequence are approximately 0.3, 1.04881, 1.39685, 1.45535, 1.46079.

To prove convergence, we can observe that the sequence seems to approach a specific value as the terms increase. This suggests that the sequence is convergent.

The specific theorem that guarantees convergence for this type of recursively defined sequence is called the Monotone Convergence Theorem. It states that if a sequence is bounded and monotonically increasing or decreasing, then it converges to a limit.

In this case, we can see that the sequence is increasing and bounded above by 10 (as shown in part A), which satisfies the conditions of the Monotone Convergence Theorem.

D) To find the exact limit of the sequence defined in part C, we can use the property of convergence along with algebraic manipulation.

1. Assume that the sequence converges to a certain limit, denoted as L: lim (n -> infinity) a_n = L.
2. Apply the recursive formula to the sequence to express future terms in terms of the limit L: L = sqrt(L+1).
3. Square both sides of the equation: L^2 = L+1.
4. Rearrange the equation: L^2 - L - 1 = 0.
5. Solve the quadratic equation for L to find the possible limits of the sequence.

Using the quadratic formula, we get L = (1 ± sqrt(5))/2. Since the terms are positive, we take the positive root: L = (1 + sqrt(5))/2.

Therefore, the exact limit of the sequence defined in part C is (1 + sqrt(5))/2.