Find limx→1f(x), if f(x)={x2−1,x≤1−x2−1,x>1
limx→1−f(x)=limx→1−x2−1
Let x =1 -h, where h→0
=limh→0(−1−h)2−1=1−1=0
Now, limx→1+f(x)
limx→1+f(x)=limx→1+−x2−1=limh→0
Let x =1-h, where h→0
(1+h)2−1=−2
Since, limx→1−f(x)≠limx→1+f(x)
∴limx→1f(x) does not exist.