anonymous
  • anonymous
Evaluate the limit using L'Hospital's rule if necessary limx→∞ (x^4)e^(−x^2)
Mathematics
katieb
  • katieb
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anonymous
  • anonymous
\[\lim _{x \rightarrow \infty} x^4e ^{-x^2}\] That's what it looks like :)
anonymous
  • anonymous
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
anonymous
  • anonymous
I'm not sure if I'm right though...

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anonymous
  • anonymous
And then I have trouble differentiating this xD
IrishBoy123
  • IrishBoy123
you re-wrote it as \(\frac{x^4}{e^{x^2}}\)?
anonymous
  • anonymous
Yep! Exactly! Is that right?
IrishBoy123
  • IrishBoy123
do differentiate top and bottom...? what do you get?
anonymous
  • anonymous
Is it \[\frac{ 4x^3 }{ 2xe ^{x^2} }\] ?
IrishBoy123
  • IrishBoy123
and cancel out x's
anonymous
  • anonymous
Right! So... \[\frac{ 4x^2 }{ 2e ^{x^2} }\]
anonymous
  • anonymous
But then it's still \[\frac{ \infty }{ }\]
anonymous
  • anonymous
I mean infinity/ infinity
IrishBoy123
  • IrishBoy123
so do L'Hopital again
IrishBoy123
  • IrishBoy123
and you can factor the 4/2 as well :p not that it really matters
anonymous
  • anonymous
No matter how many times I repeat won't it just keep giving me the same thing?
anonymous
  • anonymous
Ahaha true! Kk :D
anonymous
  • anonymous
Ohhh actually eventually I get 1/infinity right?!? :O
IrishBoy123
  • IrishBoy123
no. keep differentitaing the top and you'll get it down to a constant
IrishBoy123
  • IrishBoy123
yes \(e^{x^2}\) is a beast
anonymous
  • anonymous
Haha awesomeness! And yeah it sure is a beast XD lol
perl
  • perl
Correct, you will eventually get \( 1 / \infty \)
anonymous
  • anonymous
And that's = 0 :D
anonymous
  • anonymous
Ahhh I see! Yeah that makes sense! Thanks so much for the tips! :)

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