Question

Evaluate the limit using L'Hospital's rule if necessary
limx→∞ (x^4)e^(−x^2)

Answer 1

\[\lim _{x \rightarrow \infty} x^4e ^{-x^2}\] That's what it looks like :)

Answer 2

After simply plugging in 'infinity' for x, I get 'infinity/infinity' which is an indeterminate form and so I can use L'Hopital's Rule

Answer 3

I'm not sure if I'm right though...

Answer 4

And then I have trouble differentiating this xD

Answer 5

you re-wrote it as \(\frac{x^4}{e^{x^2}}\)?

Answer 6

Yep! Exactly! Is that right?

Answer 7

do differentiate top and bottom...? what do you get?

Answer 8

Is it \[\frac{ 4x^3 }{ 2xe ^{x^2} }\] ?

Answer 9

and cancel out x's

Answer 10

Right! So... \[\frac{ 4x^2 }{ 2e ^{x^2} }\]

Answer 11

But then it's still \[\frac{ \infty }{ }\]

Answer 12

I mean infinity/ infinity

Answer 13

so do L'Hopital again

Answer 14

and you can factor the 4/2 as well :p not that it really matters

Answer 15

No matter how many times I repeat won't it just keep giving me the same thing?

Answer 16

Ahaha true! Kk :D

Answer 17

Ohhh actually eventually I get 1/infinity right?!? :O

Answer 18

no.

keep differentitaing the top and you'll get it down to a constant

Answer 19

yes

\(e^{x^2}\) is a beast

Answer 20

Haha awesomeness! And yeah it sure is a beast XD lol

Answer 21

Correct, you will eventually get \( 1 / \infty \)

Answer 22

And that's = 0 :D

Answer 23

Ahhh I see! Yeah that makes sense! Thanks so much for the tips! :)