Solving an Infinite Limit with L'Hôpital's Rule

[sacwchiri]

Im trying to solve this problem using l'hopital but amm not sure how to do it

lim
X->infinite x^3 * e^(-x^2)

soo this infinite * e^-infinite... but from there I am not sure if you can use it to solve this...

 

[ice109]

Im trying to solve this problem using l'hopital but amm not sure how to do it

lim
X->infinite x^3 * e^(-x^2)

soo this infinite * e^-infinite... but from there I am not sure if you can use it to solve this...

you have to put expression as the ratio of two functions first

like this

\frac{x^3}{e^{x^2}}

then you can apply l'hospitale

 

[sacwchiri]

yeah i thought about that but amm not sure how much is e^-infinite

 

[cepheid]

yeah i thought about that but amm not sure how much is e^-infinite

This question makes no sense.

As ice109 said, apply L'Hospital's rule. Do you know what L'Hospital's rule is? If so, apply it to this question.

 

[HallsofIvy]

yeah i thought about that but amm not sure how much is e^-infinite

That sounds like you are saying "how infinite is it"! What you mean is "what number is e^-infinite". (Strictly speaking that is also meaningless- you cannot evaluate a function of real numbers "at infinity", you can only take limits at infinity.) You know, I hope, that e^x increases without bound (i.e. "goes to infinity") as x goes to infinity. You should also know that "e^(-x²) MEANS 1/e^(x²). If A goes to infinity, what does 1/A go to?