Find lim sup $x_n$ 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 $x_n$?
 A: 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.
