$\liminf, \limsup$ and continuous functions

Assuming $f$ is continuous, NO.
Let $f(x)=-x$, $(x_n)=(0,1,0,1,0,1,\ldots)$.
Then $(f(x_n))=(0,-1,0,-1,\ldots)$.
Thus $\lim\inf x_n=0$ but $\lim\inf f(x_n)=-1\neq 0 = f(0)$.

If $f$ is also monotone increasing, in addition to continuous, then you get what you want, since in this case $$f\left(\liminf_{n\rightarrow \infty} x_n\right) = \liminf_{n\rightarrow \infty} f(x_n) \quad \textrm{and} \quad f\left(\limsup_{n\rightarrow \infty} x_n\right) = \limsup_{n\rightarrow \infty} f(x_n).$$