Question

How do you find the limit of `x(e^(-x))` as x approaches infinity using l'hospital's rule?

Answer 1

Hence `lim_(x->oo) x/e^x =oo/oo` we use the L'Hopital law to get

#lim_(x->oo) x/e^x =lim_(x->oo) (dx/dx)/(de^x/dx)= lim_(x->oo) 1/e^x=0#