Question

Limit Problem!

What is the limit of, as x approach infinity, of (x^3)*e^(-x^2)?

If someone could help me answer this problem, I'd greatly appreciate it! Thanks!

Answer 1

you know Lhopital's Rule?

Answer 2

mhmm, that's what i've been trying to use

Answer 3

you'll prolly have to apply it more than once.

Answer 4

I've been trying it. But I can't arrive at an answer.

Answer 5

how many times did you apply it?

Answer 6

I just looked. I'm pretty sure twice.

Answer 7

Looking over at my computation, I think the answer is infinity. Can you verify?

Answer 8

you can write the expression as \[\large \frac{x^3}{e^{x^2}} \]

Answer 9

Oh yeah, you're right! I didn't think of it that way. Lemme try apply the rule once more after the derivative.

Answer 10

the denominator will still have that e^(x^2) the numerator will end up a constant....
shouldn't the limit be 0 then?

Answer 11

I'll will try and compute first using the quotient rule, correct?

Answer 12

no... quotient rule is not Lhopital's Rule.

Answer 13

just take derivative of the top
then derivative of the bottom

Answer 14

But don't I need to differentiate first?

Answer 15

Since it's a quotient, shouldn't I use the quotient rule though?

Answer 16

\[\large lim \frac{f}{g} =lim\frac{f'}{g'}\]
this is Lhopitals Rule

Answer 17

where f=x^3
and g=e^(x^2)

Answer 18

uh-huh. then i got that to equal:

\[3x ^{2}/e ^{x ^{2}}\]

Answer 19

your denominator, g' = 2xe^(x^2)

Answer 20

it goes to 0

Answer 21

you can tell as you take more and more derivatives the top goes to 0 where the denominator never will

Answer 22

then it looks like you'll have to apply it once more since you still have an indeterminate form after applying Lhopital's rule the first time.

Answer 23

But then the 2x in the denominator can cancel out with the numerator. Wouldn't it be:

3x/2e^x^2

Answer 24

yes...

Answer 25

but you still have an indeterminate form here still...

Answer 26

so .... once more....:)

Answer 27

like a drill sargeant.

Answer 28

Okay, I did it once more. And I got 3/4xe^x^2.

Answer 29

so as x approaches infinity.... what does the valu of that expression approach?

Answer 30

http://www.wolframalpha.com/input/?i=3x%5E2%2Fe%5E%7Bx%5E2%7D

Answer 31

0! Gotcha. But is my last differentiation correct?

Answer 32

yes it is...:)

Answer 33

Ah, great! Thank you so much! That was not as bad as I first thought it would be. I just completely blanked out with the negative exponent. I forgot you could place it in the denominator. But I understand now. Thank you!

Answer 34

yw...:)