One other way is to see the polynomial expansion of e^x. It has terms of higher degree than 2 with +ve coefficients. So limit will be 0 for x

 

I put up this prob bcos of this instructive solution I saw somewhere:

 

Let f(x) = x³e^-x. Then f '(x) = (3x² - x³) e^-x < 0 for x > 3. Hence f(x) < f(3) for x > 3, so x² e^-x < f(3)/x for x > 3. Hence x² e^-x tends to zero.