Function f is increasing and continuous. $(x_n)$ is a bounded sequence. $M:= \limsup x_n$. Prove that $ \limsup f(x_n) = f(M) $.
I have tried this method. For any $\epsilon>0$ given, we want to show $x_n>M+\epsilon$, for finitely many n; and $x_n<M+\epsilon$, for infinitely many n.
If $f$ is continuous, then $$ f(\limsup_{n \to \infty}x_{n}) = f(\lim_{N \to \infty}\sup_{n \geq N}x_{n}) = \lim_{N \to \infty}f(\sup_{n \geq N}x_{n}); $$ if in addition $f$ is increasing, then $f(\sup_{n \geq N}x_{n}) = \sup_{n \geq N}f(x_{n})$ for all $N \geq 1$; hence $$\lim_{N \to \infty}f(\sup_{n \geq N}x_{n}) = \lim_{N \to \infty}\sup_{n \geq N}f(x_{n}) = \limsup_{n \to \infty}f(x_{n}).$$
