91Ó°ÊÓ

When we talk about the limit of a sequence, we're basically asking what happens to the terms of the sequence as we move further and further along it. Consider the sequence given in the problem: \(a_n = \frac{\sin^2 n}{\sqrt{n}}\). As \(n\), which represents the term number, moves towards infinity, we are interested in the behavior and ultimate fate of \(a_n\).
If a sequence reaches some specific value as \(n\) becomes infinitely large, then we say that this sequence converges, and the value it approaches is called the limit. In mathematical terms, a sequence \(\{a_n\}\) converges to a limit \(L\) if for every small positive number \(\epsilon\), there is a point in the sequence beyond which all the terms are within \(\epsilon\) of \(L\).
For the sequence \(a_n\), we observed through computations and bounding it using the Squeeze Theorem that its values approach zero as \(n\) becomes very large. Hence, \(\lim_{n \to \infty} a_n = 0\), confirming that the sequence converges to zero.

The Squeeze Theorem is a handy tool in calculus that can help determine the limit of a sequence when two other sequences 'squeeze' it. Imagine you have three sequences: one whose behavior you know very well and a second that bounds it tightly above and below.
For sequence \(a_n = \frac{\sin^2 n}{\sqrt{n}}\), knowing that the range of \(\sin^2 n\) is between 0 and 1 helps to formulate bounding sequences. This useful fact allows us to construct this inequality:

• \(0 \leq \frac{\sin^2 n}{\sqrt{n}} \leq \frac{1}{\sqrt{n}}\)

Although \(\sin^2 n\) oscillates value, the increasing size of \(\sqrt{n}\) pushes \(\frac{1}{\sqrt{n}}\) toward zero. By confirming that \(\lim_{n \to \infty} \frac{1}{\sqrt{n}} = 0\), the theorem lets us conclude that \(\lim_{n \to \infty} \frac{\sin^2 n}{\sqrt{n}} = 0\) as well.
Using the Squeeze Theorem effectively requires identifying these bounding sequences correctly—catching the 'sandwich' effect between them.

A sequence is considered bounded if there is a real number that serves as a maximum or minimum limit for all terms of the sequence. For example, with the given sequence \(a_n = \frac{\sin^2 n}{\sqrt{n}}\), knowledge about the bounded nature of \(\sin^2 n\) allows us to discuss its degrees of limitation.
Since \(\sin^2 n\) spans values from 0 to 1, this affects the sequence \(a_n\) making:\[0 \leq \sin^2 n \leq 1\] Subsequently, as \(n\) increases, \(\sqrt{n}\) increases, and thus \(\frac{1}{\sqrt{n}}\) diminishes, providing more boundary to \(a_n\).
If all values of a sequence fit within specific bounds, like here, each value of our sequence stays between those bounds. This bounded nature facilitated using Squeeze Theorem since starting boundaries were necessary.