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Let $x_n = n(\sqrt{n^2+1} - n)\sin\dfrac{n\pi}8$ , $n\in\Bbb{N}$

Find $\limsup x_n$.

Hint: lim sup $x_n = \sup C(x_n)$.

How to make it into a fraction to find the cluster point of $x_n$?

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Please don't mix $x$ and $X$. Variables that stand for real numbers are usually lower case, so I edited your question to use $x$ everywhere. – Nate Eldredge Mar 17 '13 at 14:32

Expand: $n · \left( \sqrt{n²+1} - n \right) = n · \tfrac{1}{\sqrt{n²+1} + n}$ for all $n ∈ ℕ$.

Since $\tfrac{\sqrt{n²+1} + n}{n} = \sqrt{1+\tfrac{1}{n²}} + 1 \overset{n → ∞}{\longrightarrow} 2$, you have $n · \left( \sqrt{n²+1} - n \right) \overset{n → ∞}{\longrightarrow} \tfrac{1}{2}$.

Now $x_n = n · \left( \sqrt{n²+1} - n \right) · \sin \left(\tfrac{n · π}{8}\right) ≤ n · \left( \sqrt{n²+1} - n \right)$ for all $n ∈ ℕ$, so: \begin{align*} \limsup_{n→∞} x_n &= \limsup_{n→∞} \left( n · \left( \sqrt{n²+1} - n \right) · \sin \left(\tfrac{n · π}{8}\right) \right)\\ &≤ \limsup_{n→∞} \left( n · \left( \sqrt{n²+1} - n \right) \right)= \tfrac{1}{2} \end{align*}

But $\sin \left(\tfrac{(16n+4)· π}{8}\right) = \sin \left(n·2π + \tfrac{π}{2}\right) = \sin \left(\tfrac{π}{2}\right)= 1$ for all $n ∈ ℕ$.

This means $x_{16n+4} = (16n+4) · \left( \sqrt{(16n+4)²+1} - (16n+4) \right) · \sin \left(\tfrac{(16n+4)· π}{8}\right) \overset{n → ∞}{\longrightarrow} \tfrac{1}{2}$.

Therefore $\tfrac{1}{2}$ is a limit point of the sequence $(x_n)_{n ∈ ℕ}$ as well as an upper bound of the limit points of that sequence. So it’s the upper limit of that sequence.

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