Limits: Solving x^3e^(-x^2) with L'Hospital's Rule

[grothem]

Lim x$$\rightarrow$$$$\infty$$ x^3 $$\times$$ e^(-x^2)

I moved x^3 down to make it a fraction and then I used L'Hospital's rule, but I can't come up with the right answer. Not sure what I'm doing wrong

 

[grothem]

that looks kind of confusing...

its the limit of x^3 * e^(-x^2) as x approaches infinity

 

[sutupidmath]

well, i guess this is going to be zero, since $$e^{x^2}$$ grows much faster than x^3

 

[Mystic998]

What do you mean you "moved it down"?

Regardless, I suggest rewriting $e^{-x^{2}}$ using the facts you know about negative exponents. Probably easier that way.

 

[sutupidmath]

and if you want to use l'hospital's rule, you need to do it three times from what i can see!

 

[grothem]

ah ok...I used l'hospital's rule 3 times and ended up with

lim (3/(4x^3*e^(x^2)) = 0

 

[sutupidmath]

ah ok...I used l'hospital's rule 3 times and ended up with

lim (3/(4x^3*e^(x^2)) = 0

Well the bottom will defenitely not be what u got!
[e^(x^2)]'=(x^2)'e^(x^2)]=[2xe^(x^2)] not take again the derivative and apply the product rule

(uv)'=u'v+v'u