Final answer:
A sequence with a unique fixed point and boundedness will have a limit because the Bolzano-Weierstrass Theorem guarantees convergent subsequences to the fixed point, and uniqueness prevents oscillation or divergence to infinity, thus the whole sequence converges.
Step-by-step explanation:
To show that a sequence with a unique fixed point and boundedness has a limit, let's explore the concept of a fixed point iteration. If a sequence of numbers produced by successive iterations of a function converges, the limit it approaches is called the fixed point. In this case, given a unique fixed point means there is only one point to which the sequence can converge.
In mathematical analysis, bounded sequences are sequences with a finite upper and lower bound. According to the Bolzano-Weierstrass Theorem, every bounded sequence in ℝ (the set of real numbers) has a convergent subsequence. Since the sequence is bounded and has a unique fixed point, it implies that every convergent subsequence must converge to the same point, the unique fixed point.
To illustrate further, consider the sequence as a repeated application of a function f on an initial value x_0, producing the elements of the sequence x_1, x_2, x_3, .... Since the sequence is bounded, it doesn't diverge to infinity. This characteristic coupled with the fact that there is a unique fixed point implies that the sequence cannot oscillate indefinitely; otherwise, it would imply the existence of more than one fixed point, conflicting with the 'unique' condition.
Therefore, as we take the limit of the sequence as the number of steps approaches infinity, every subsequence converges to the same unique fixed point. And due to the boundedness of the whole sequence, this result extends to the entire sequence, not just to the subsequences. Hence, the entire sequence has a limit.