\liminf_{n\to\infty} u_n=\limsup_{n\to\infty} u_n=L if and only if \lim_{n\to\infty} u_n=L.

For u_n=\sin n we have \liminf_{n\to\infty} u_n=-1 and \limsup_{n\to\infty} u_n=1

\liminf_{n\to\infty} u_n\leq \limsup_{n\to\infty} u_n .

For u_n=2+(-1)^n\left(1+\frac{1}{n}\right ) we have \liminf_{n\to\infty} u_n=1 and \limsup_{n\to\infty} u_n=3

\limsup_{n\to\infty} u_n always exists.

\liminf_{n\to\infty} u_n always exists.

For u_n=(-1)^n we have \liminf_{n\to\infty} u_n=-1 and \limsup_{n\to\infty} u_n=1