Prove that $f( \limsup \limits_{x \to +\infty} x_n)$ = $\limsup \limits_{x \to +\infty} f(x_n)$ [closed]

The problem is divided into two questions:

$$x_n$$ is a real valued sequence.

• Prove that $$f( \limsup \limits_{n \to +\infty} x_n)$$ = $$\limsup \limits_{n \to +\infty} f(x_n)$$ given that $$f$$ is continuous and increasing.
• What can we say when $$f$$ is decreasing.

If it were a regular limit I can just pass the $$\lim$$ inside the function and we're done. But this is $$\limsup$$ .

For any sequence $$(y_{n})$$ let $$y_{n}^{+} = \sup_{k\geq n}y_{k}$$. Let $$L = \limsup x_{n}$$ be a real number. Since $$L = \lim x_{n}^{+}$$ and $$f$$ is continuous, we have $$f(L) = \lim f(x_{n}^{+}).$$ Now we show that $$f(x_{n}^{+}) = f(x_{n})^{+}$$ for each $$n$$. Since $$x_{n}^{+}\geq x_{k}$$ for each $$k\geq n$$ and $$f$$ is increasing, we have $$f(x_{n}^{+})\geq f(x_{n})^{+}$$. For any $$\varepsilon > 0$$ there exists $$k\geq n$$ such that $$x_{n}^{+}-\varepsilon < x_{k}$$. Again using the fact that $$f$$ is increasing we get $$f(x_{n}^{+}-\varepsilon)\leq f(x_{k})\leq f(x_{n})^{+}$$. Letting $$\varepsilon\rightarrow 0$$ and using the continuity of $$f$$ we get $$f(x_{n}^{+})\leq f(x_{n})^{+}$$.
If $$f$$ is decreasing then $$(-f)$$ is increasing and thus $$(-f)(\limsup x_{n}) = \limsup\:(-f)(x_{n})$$. This reduces to $$f(\limsup x_{n}) = \liminf f(x_{n})$$ after simplification.