If f(x)=xe(1/x)-e(-1/x)e(1/x)+e(-1/x);x≠00;x=0, then which of the following is true

f0−=limx→0−xe1x−e-1xe1x+e-1x          =limx→0-xe2x-1e2x+1 =limx→0x0-10+1=0f0+=limx→0+x1−e−2x1+e-2x                      =limx→0+x1−01+0=0f(0)=0   (given) So, f is continuous everywhere f'(0-)=limk→0-he1h-e-1he1h+e-1h-0h =0-10+1=-1f'0+=limh→0+he1h−e-1he1h+e-1h−0h =1 ∴               f'0−≠f'0+