Help with l'hopitals rule please

Ive been doing this problem for over an hour and I keep getting infinity as an answer but the answer is 0
Please help
(x^2)e^(-x^2)/2

This how the problem reads x squared times e to the negative 1/2 x squared

The e is being raised to the 1/2 times x^2

Thankmu so much 10 points

I think you mean (x^2)e^(-x^2/2).
lim x^2 e^(-x^2/2)
x-> ?

What is x approaching?

I'll assum x -> ∞.

lim x^2 e^(-x^2/2) =
x-> ∞
lim x^2/e^(x^2/2) =...now use L'Hopital's rule because numerator and denominator approach ∞
x-> ∞
lim 2x/(e^(x^2/2) * x) =
x-> ∞
lim 2/e^(x^2/2) = 0
x-> ∞

I'm assuming that you are looking for the limit as x goes to infinity? So the expression can be re-written as (x^2) / (e^((x^2)/2). Take the derivative of top and bottom separately to get (2x)/(x*e^((x/2)/2) which reduces to 2 / e^((x/2)/2) which indeed goes to 0 as x -> infinity.

What's up, Jr.

The problem is equivalent to finding the limit of the following, right?

y = lim x→∞ x² / (e^[x²/2])

If so, then y = lim x→∞ x² / (e^[x²/2]) = lim x→∞ (2x)/[(2x)(e^[x²/2])] = lim x→∞ 1/(e^[x²/2]) = 0.

Hope that helps!