Understanding the Limit Point of a Sequence

The question asks us to find the limit point of the sequence given as < f > = 1. A sequence is an ordered list of numbers. In this case, the sequence is denoted by < f >, and the rule for generating its terms is simply 1. This means every term in the sequence is the number 1.

Let's write out the first few terms of this sequence:

• First term: 1
• Second term: 1
• Third term: 1
• ...and so on.

So, the sequence can be written as < f > = (1, 1, 1, 1, ...).

What is a Limit Point?

A limit point (also sometimes called an accumulation point or cluster point) of a sequence is a value that the terms of the sequence get arbitrarily close to as the index goes to infinity. More formally, a number $L$ is a limit point of a sequence $(a_n)$ if for every neighborhood around $L$, there are infinitely many terms of the sequence within that neighborhood.

For a sequence to converge to a limit $L$, for every $\epsilon > 0$, there must exist a positive integer $N$ such that for all $n > N$, $|a_n - L| < \epsilon$. If a sequence converges to a limit $L$, then $L$ is the only limit point of the sequence.

Finding the Limit Point for < f > = 1

Our sequence is $a_n = 1$ for all $n \ge 1$. Let's consider the definition of a limit point. Does this sequence approach any specific value?

Since every term of the sequence is exactly 1, the terms are already at the value 1. They don't need to "get close" to 1; they are always equal to 1. Let's check if 1 satisfies the definition of a limit point. Consider any small neighborhood around 1. A neighborhood around 1 of radius $\epsilon > 0$ is the interval $(1 - \epsilon, 1 + \epsilon)$.

For the sequence < f > = (1, 1, 1, ...), every term $a_n = 1$. Is $a_n$ inside the interval $(1 - \epsilon, 1 + \epsilon)$ for any $\epsilon > 0$? Yes, because $1 - \epsilon < 1 < 1 + \epsilon$ for any $\epsilon > 0$.

Since every term of the sequence is 1, and there are infinitely many terms in the sequence, there are infinitely many terms of the sequence within any neighborhood around 1. This confirms that 1 is a limit point of the sequence.

Now, let's consider if there could be any other limit point, say $L \ne 1$. If $L \ne 1$, then the distance between $L$ and 1 is $|L - 1| > 0$. Let's choose $\epsilon = |L - 1| / 2$. The neighborhood around $L$ with radius $\epsilon$ is $(L - \epsilon, L + \epsilon)$. The number 1 is not in this neighborhood because the distance between 1 and $L$ is $|L - 1|$, which is greater than $\epsilon$. Since every term of the sequence is 1, no term of the sequence is in the neighborhood $(L - \epsilon, L + \epsilon)$. Therefore, no value other than 1 can be a limit point of this sequence.

Analyzing the Options

Based on our analysis, the only limit point of the sequence < f > = 1 is 1. Let's look at the given options:

1. 2: This is not the value of the sequence and not a limit point.
2. infinity: The sequence terms are always finite (equal to 1), so infinity is not a limit point.
3. 0: This is not the value of the sequence and not a limit point.
4. 1: This is the value that every term of the sequence equals. It satisfies the definition of a limit point.

Thus, the correct limit point is 1.

Conclusion on Sequence Limit Point

The sequence < f > = 1 is a constant sequence where every term is 1. For a constant sequence $(c, c, c, ...)$, the terms do not change and remain at $c$. As the index goes to infinity, the terms are still $c$. Therefore, a constant sequence always converges to the constant value, and this value is its unique limit point.

For the sequence < f > = 1, the constant value is 1. Hence, the limit point is 1.

Revision Table: Sequence and Limit Point

Additional Information on Sequences and Limits

Constant Sequence: A sequence $(a_n)$ is called a constant sequence if $a_n = c$ for some fixed number $c$ and for all $n \ge 1$. For any constant sequence, the limit exists and is equal to the constant $c$. This means the constant value is the unique limit point.

Convergence: A sequence $(a_n)$ converges to a limit $L$ if for every $\epsilon > 0$, there exists an integer $N$ such that for all $n > N$, $|a_n - L| < \epsilon$. For a constant sequence $a_n = c$, we can choose $L = c$. Then $|a_n - c| = |c - c| = 0$. Since $0 < \epsilon$ for any $\epsilon > 0$, the condition is satisfied for any $N$ (we can even take $N=0$). Thus, a constant sequence converges to the constant value.

Limit Superior and Limit Inferior: For any bounded sequence, the limit superior (limsup) and limit inferior (liminf) always exist. The limit points of a sequence are the values between the liminf and limsup (inclusive). For a convergent sequence, limsup = liminf = limit = unique limit point.

Bolzano-Weierstrass Theorem: This theorem states that every bounded sequence in $\mathbb{R}^n$ (real numbers) has at least one limit point. The sequence < f > = 1 is bounded (e.g., between 0 and 2), so it must have at least one limit point, which we found to be 1.