$\lim\limits_{x \rightarrow 2}\{[2-x]+[x-2]-x\}$ is equal to

0 3 -3 does not exist

-3

RH Limit = $\lim\limits_{h \rightarrow 0}\{[2-(2+h)]+[(2+h)-2]-(2+h)\}=\lim\limits_{h \rightarrow 0}\{[-h]+[h]-2-h\}$ $=\lim\limits_{h \rightarrow 0}\{0-1-2+h\}=-3$ LH Limit = $\lim\limits_{h \rightarrow 0}\{[2-(2-h)]+[(2-h)-2]-(2-h)\}=\lim\limits_{h \rightarrow 0}\{0-1-2+h\} = -3$ Hence (3) is the correct answer.