limx→0 e1/x−1e1/x+1,is

Let f(x)=e1/x−1e1/x+1.Then ,(LHL of f(x) at x=0)=limx→0− f(x)=limh→0− f(0−h)=limh→0 e−1/h−1e−1/h+1=limh→0 1e1/h−11e1/h+1=−1 e1/h→∞⇒1e1/h→0and, (RHL of f(x) at x=0 )=limx→0+ f(x)=limh→0 f(0+h)=limh→0 e1/h−1e1/h+1=limh→0 1−1e1/h1+e1/h           [Dividing Nr and Dr by e1/h]=1−01+0=1 Clearly, limx→0− f(x)≠limx→0+ f(x) Hence, limx→0 f(x) does not exist.