If f(x)=|x+2|tan −1⁡(x+2), x≠−22,          x=-2then f(x) is

limx→−2−f(x)=limx→−2−x+2tan−1x+2=limx→−2−−x+2tan−1x+2=−1    ∵limx→0−tan−1xx=1limx→−2+f(x)=limx→−2+x+2tan−1x+2=limx→−2+x+2tan−1x+2=1∴limx→−2f(x) does not exist.