limx→011+e−1x=

As x→0−⇒e1x→0limx→0−11+e−1x=limx→0−e1xe1x+1=00+1=0As x→0+⇒e−1x→0limx→0+11+e−1x==11+0=1Since both limits are not the same, so that the limit does not exist.