Question 435783
{{{lim(x->infinity, x^2e^(-x)) = lim(x->infinity, x^2/e^x)}}}. This is indeterminate so we use L'Hopital's rule:


{{{lim(x->infinity, x^2/e^x) = lim(x->infinity, 2x/e^x) = lim(x->infinity, 2/e^x)}}}. This limit is equal to zero, so the limit of the original expression is also equal to zero.