To prove this, we will use the definitions of limit superior (lim sup) and limit inferior (lim inf) of a sequence. 1. Let's start with lim sup. By definition, lim sup of a sequence (xn) is the smallest number 'L' such that for any positive number ε, there exists a natural number N such that for all n N, xn L + 1/k. This gives us a subsequence (xnk) of (xn) such that xnk L + 1/k for all k. On the other hand, since L is an upper bound of (xn), we have xnk ≤ L for all k. Therefore, by the squeeze theorem, we have limk→∞ xnk = L = lim sup xn. 2. The proof for lim inf is similar. By definition, lim inf of a sequence (xn) is the greatest number 'M' such that for any positive number ε, there exists a natural number N such that for all n N, xn M - ε. In other words, there are only finitely many terms of the sequence that are less than M - ε. Since (xn) is a bounded sequence, it has a lower bound. Let's denote the set of all lower bounds of (xn) as 'T'. Then, by definition, lim inf xn is the supremum of 'T', which we will denote as 'M'. Now, for each positive integer k, consider the number M - 1/k. Since M is the supremum of 'T', M - 1/k is not a lower bound of (xn), so there exists an element in the sequence, say xmk, such that xmk

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To prove this, we will use the definitions of limit superior (lim sup) and limit inferior (lim inf) of a sequence. 1. Let's start with lim sup. By definition, lim sup of a sequence (xn) is the smallest number 'L' such that for any positive number ε, there exists a natural number N such that for all n > N, xn < L + ε. In other words, there are only finitely many terms of the sequence that are greater than L + ε. Since (xn) is a bounded sequence, it has an upper bound. Let's denote the set of all upper bounds of (xn) as 'S'. Then, by definition, lim sup xn is the infimum of 'S', which we will denote as 'L'. Now, for each positive integer k, consider the number L + 1/k. Since L is the infimum of 'S', L + 1/k is not an upper bound of (xn), so there exists an element in the sequence, say xnk, such that xnk > L + 1/k. This gives us a subsequence (xnk) of (xn) such that xnk > L + 1/k for all k. On the other hand, since L is an upper bound of (xn), we have xnk ≤ L for all k. Therefore, by the squeeze theorem, we have limk→∞ xnk = L = lim sup xn. 2. The proof for lim inf is similar. By definition, lim inf of a sequence (xn) is the greatest number 'M' such that for any positive number ε, there exists a natural number N such that for all n > N, xn > M - ε. In other words, there are only finitely many terms of the sequence that are less than M - ε. Since (xn) is a bounded sequence, it has a lower bound. Let's denote the set of all lower bounds of (xn) as 'T'. Then, by definition, lim inf xn is the supremum of 'T', which we will denote as 'M'. Now, for each positive integer k, consider the number M - 1/k. Since M is the supremum of 'T', M - 1/k is not a lower bound of (xn), so there exists an element in the sequence, say xmk, such that xmk < M - 1/k. This gives us a subsequence (xmk) of (xn) such that xmk < M - 1/k for all k. On the other hand, since M is a lower bound of (xn), we have xmk ≥ M for all k. Therefore, by the squeeze theorem, we have limk→∞ xmk = M = lim inf xn. Therefore, we have shown that there exist subsequences (xnk) and (xmk) of (xn) such that limk→∞ xnk = lim sup xn and limk→∞ xmk = lim inf xn.

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To prove this, we will use the definitions of limit superior (lim sup) and limit inferior (lim inf) of a sequence. 1. Let's start with lim sup. By definition, lim sup of a sequence (xn) is the smallest number 'L' such that for any positive number ε, there exists a natural number N such that for all n > N, xn < L + ε. In other words, there are only finitely many terms of the sequence that are greater than L + ε. Since (xn) is a bounded sequence, it has an upper bound. Let's denote the set of all upper bounds of (xn) as 'S'. Then, by definition, lim sup xn is the infimum of 'S', which we will denote as 'L'. Now, for each positive integer k, consider the number L + 1/k. Since L is the infimum of 'S', L + 1/k is not an upper bound of (xn), so there exists an element in the sequence, say xnk, such that xnk > L + 1/k. This gives us a subsequence (xnk) of (xn) such that xnk > L + 1/k for all k. On the other hand, since L is an upper bound of (xn), we have xnk ≤ L for all k. Therefore, by the squeeze theorem, we have limk→∞ xnk = L = lim sup xn. 2. The proof for lim inf is similar. By definition, lim inf of a sequence (xn) is the greatest number 'M' such that for any positive number ε, there exists a natural number N such that for all n > N, xn > M - ε. In other words, there are only finitely many terms of the sequence that are less than M - ε. Since (xn) is a bounded sequence, it has a lower bound. Let's denote the set of all lower bounds of (xn) as 'T'. Then, by definition, lim inf xn is the supremum of 'T', which we will denote as 'M'. Now, for each positive integer k, consider the number M - 1/k. Since M is the supremum of 'T', M - 1/k is not a lower bound of (xn), so there exists an element in the sequence, say xmk, such that xmk < M - 1/k. This gives us a subsequence (xmk) of (xn) such that xmk < M - 1/k for all k. On the other hand, since M is a lower bound of (xn), we have xmk ≥ M for all k. Therefore, by the squeeze theorem, we have limk→∞ xmk = M = lim inf xn. Therefore, we have shown that there exist subsequences (xnk) and (xmk) of (xn) such that limk→∞ xnk = lim sup xn and limk→∞ xmk = lim inf xn.

Let (xn) be a bounded sequence in R. Show that there exist subsequences (xnk ) and(xmk ) of (xn) such thatlimk→∞ xnk = lim sup xn and limk→∞ xmk = lim inf xn.

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Let (xn) be a bounded sequence in R. Show that there exist subsequences (xnk ) and (xmk ) of (xn) such that

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