Final answer:
In part (a), the limit lim(n->∞)an = 3 is proven using the ε-N definition of limit. In part (b), the limit lim(n->∞)bn = 0 is proven using the ε-N definition of limit. In part (c), the squeeze theorem is used to show that the limit of the sequence cn exists and is equal to 4.
Step-by-step explanation:
(a) To prove that limn→∞an = 3 using the ε-N definition of limit, we need to show that for any ε > 0, there exists an N such that |an - 3| < ε for all n > N. Let's start by setting |an - 3| < ε and solving for n.
3 + n² < ε
n² < ε - 3
n < √(ε - 3)
Since ε can be as small as we want, we can choose N = ceil(√(ε - 3)). Then for all n > N, we have |an - 3| < ε, which proves the limit is 3.
(b) To prove that limn→∞bn = 0 using the ε-N definition of limit, we need to show that for any ε > 0, there exists an N such that |bn - 0| < ε for all n > N. Since bn = 2/n, we can rewrite the inequality as |2/n - 0| < ε. This simplifies to 2/n < ε, which implies n > 2/ε.
Therefore, we can choose N = ceil(2/ε). Then for all n > N, we have |bn - 0| < ε, which proves the limit is 0.
(c) To use the squeeze theorem to show that limn→∞cn exists, we need to find two other sequences an and bn such that an ≤ cn ≤ bn and limn→∞an = limn→∞bn = L. Then, by the squeeze theorem, limn→∞cn = L.
In this case, we can let an = 4 and bn = 4 + 2/n. Since cos(n³ - 4n² + n - 3) is bounded between -1 and 1, we have -1 ≤ cos(n³ - 4n² + n - 3) ≤ 1. Therefore, an ≤ cn ≤ bn.
Now, we know limn→∞an = 4 and limn→∞bn = 4. By the squeeze theorem, limn→∞cn = 4.