limx→0(e1/x−1)(e1/x)+1
0
1
-1
Does not exist
f(x)=(e1/x−1)(e1/x)+1, then limx→0+ f(x) = limh→0(e1/h−1)(e1/h+1) =limh→0(e1/h−1)(e1/h)+1 = limh→0e1/h×(1−(1e1/h)e1h×(1+1e1/h) = 1 Similarly limx→0−f(x) = -1. Hence limit does not Exist
If f(x)=xsin(1x),x≠0, then limx→0f(x)=