A sequence{an} is given by a1=sqrt(2), an+1=sqrt(2*an).

a) by induction or otherwise, show that {an} is increasing and bounded above by 3. Apply the Monotonic Sequence Theorem to show that lim n-->infinity an exists.
b) Find lim n-->infinity an.

To show that the sequence {an} is increasing and bounded above by 3, we will use induction.

a) Step 1: Proving the sequence is increasing
First, let's prove that {an} is increasing by showing that an+1 ≥ an for all n.

Base case: n = 1
We have a1 = √2, a2 = √(2*√2) = 2.
Since 2 ≥ √2, the base case holds.

Inductive step:
Assume an+1 ≥ an for some arbitrary value of n = k.

Now, we need to show that an+2 ≥ an+1.
an+1 = √(2*an) and an+2 = √(2*an+1).
Substituting the assumed inequality, we have:
an+2 = √(2*an+1) ≥ √(2*an) = an+1.

Therefore, an+2 ≥ an+1, which completes the induction.

Step 2: Proving the sequence is bounded above by 3
Now, let's prove that {an} is bounded above by 3, meaning that an ≤ 3 for all n.

Base case: n = 1
We have a1 = √2, which is less than 3. So, the base case holds.

Inductive step:
Assume an ≤ 3 for some arbitrary value of n = k.

Now, we need to show that an+1 ≤ 3.
an+1 = √(2*an).

Using the induction assumption, we have:
an+1 = √(2*an) ≤ √(2*3) = √6.
It can be shown that √6 is less than or equal to 3.

Therefore, an+1 ≤ 3, which completes the induction.

By the Monotonic Sequence Theorem, since {an} is increasing and bounded above by 3, the limit lim n→∞ an exists.

b) Finding lim n→∞ an
To find the limit, we can notice that as n approaches infinity, the sequence {an} is getting closer and closer to a single value. Let's denote the limit as L.

Then, we can write the recursive equation for the limit as:
L = √(2L).

Solving this equation, we square both sides:
L^2 = 2L.

Rearranging, we get:
L^2 - 2L = 0.

Factoring out L, we have:
L(L - 2) = 0.

From this equation, we get two possible solutions: L = 0 and L = 2.

However, since we know that {an} is increasing, it can be shown that L > √2 = a1.

Therefore, the limit of the sequence is L = 2.

Hence, lim n→∞ an = 2.