- #1

- 23

- 0

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

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter grothem
- Start date

- #1

- 23

- 0

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

- #2

- 23

- 0

that looks kind of confusing...

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

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

- #3

- 1,631

- 4

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

- #4

- 206

- 0

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

- #5

- 1,631

- 4

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

- #6

- 23

- 0

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

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

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

- #7

- 1,631

- 4

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

Share: