For each positive real number lambda,let A_la | vedclass

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(C) Using the Mean Value Theorem,$|\sin(\sqrt{n+1}) - \sin(\sqrt{n})| = |\cos(c) \cdot (\sqrt{n+1} - \sqrt{n})|$ for some $c \in (\sqrt{n}, \sqrt{n+1}

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$A_{1/2}^c, A_{1/3}^c, A_{2/5}^c$ are all finite sets

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(C) Using the Mean Value Theorem,$|\sin(\sqrt{n+1}) - \sin(\sqrt{n})| = |\cos(c) \cdot (\sqrt{n+1} - \sqrt{n})|$ for some $c \in (\sqrt{n}, \sqrt{n+1})$.Since $\sqrt{n+1} - \sqrt{n} = \frac{1}{\sqrt{n+1} + \sqrt{n}}$,we have $|\sin(\sqrt{n+1}) - \sin(\sqrt{n})| = |\cos(c)| \cdot \frac{1}{\sqrt{n+1} + \sqrt{n}}$.As $n 	o \infty$,$\frac{1}{\sqrt{n+1} + \sqrt{n}} 	o 0$.Since $|\cos(c)| \le 1$,the expression $|\sin(\sqrt{n+1}) - \sin(\sqrt{n})|$ approaches $0$ as $n 	o \infty$.For any $\lambda > 0$,there exists an $N$ such that for all $n > N$,$|\sin(\sqrt{n+1}) - \sin(\sqrt{n})| Thus,$A_\lambda$ contains all natural numbers greater than $N$,making $A_\lambda$ an infinite set.Consequently,the complement $A_\lambda^c$ contains only a finite number of elements (those $n \le N$ that do not satisfy the inequality).Therefore,$A_{1/2}^c, A_{1/3}^c, A_{2/5}^c$ are all finite sets.

For each positive real number $\lambda$,let $A_\lambda$ be the set of all natural numbers $n$ such that $|\sin(\sqrt{n+1}) - \sin(\sqrt{n})| < \lambda$. Let $A_\lambda^c$ be the complement of $A_\lambda$ in the set of all natural numbers. Then,

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