limit problem with e (without using l'hopitals rule)

[kangsang24]

Without using l'hopitals rule how could I solve => lim as x approaches infinity of x^2(e^-x). I know the answer is 0 but how do i use algebraic manipulation to achieve this. I started off by rewriting it as x^2/e^x.. then what??

 

[Deleted member 4993]

Without using l'hopitals rule how could I solve => lim as x approaches infinity of x^2(e^-x). I know the answer is 0 but how do i use algebraic manipulation to achieve this. I started off by rewriting it as x^2/e^x.. then what??

Do you know the expansion of e^x - as a infinite series polynomial?

Alternatively, start with evaluating:

\(\displaystyle \dfrac{(x+h)^2e^{(-x-h)} - x^2e^{-x}}{h}\)

 

[pka]

Without using l'hopitals rule how could I solve => lim as x approaches infinity of x^2(e^-x). I know the answer is 0 but how do i use algebraic manipulation to achieve this. I started off by rewriting it as x^2/e^x.. then what??

Without knowing where you are in your studies it is impossible to say how you should proceed.

I would note that the series \(\displaystyle \displaystyle\sum\limits_{x = 1}^\infty {{x^2}{e^{ - x}}} \) converses therefore it must be the case that \(\displaystyle {\displaystyle\lim _{x \to \infty }}{x^2}{e^{ - x}} = 0\)

 

[kangsang24]

Do you know the expansion of e^x - as a infinite series polynomial?

Alternatively, start with evaluating:

\(\displaystyle \dfrac{(x+h)^2e^{(-x-h)} - x^2e^{-x}}{h}\)

We actually did not go over this concept that you mentioned nor did we cover L'hopitals rule (I cannot use concepts we did not learn in this semester). Do you think there is even a possible way of solving without L'hopital or the concept you mentioned above?