#lim_(x->0+) (xe^(1/x))/(1+e^(1/x))#
#= lim_(x->0+) x/(1/e^(1/x)+1)#
#= lim_(x->0+) x/(e^(-1/x)+1)#
#=0/(0+1)= 0#
Now consider #lim_(x->0^-) e^(1/x)#
Limit chain rule states that:
If #lim_(u->b) f(u) = L# and #lim_(x->a) g(x) = b#
Then #lim_(x->a) f(g(x)) =L#
Here, #g(x) = 1/x and f(u) = e^u#
#lim_(x->0^-) 1/x =-oo#
#lim_(u->-oo) e^u =0#
Hence, #lim_(x->0^-) e^(1/x) =0# [Limit chain rule]
Now considering the original limit as #x->0^-#
#lim_(x->0^-) (xe^(1/x))/(1+e^(1/x))= (0*0)/(1+0) =0#
Thus, #lim_(x->0) (xe^(1/x))/(1+e^(1/x)) =0#
