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

1. anonymous

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

2. 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

3. anonymous

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

4. anonymous

And then I have trouble differentiating this xD

5. IrishBoy123

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

6. anonymous

Yep! Exactly! Is that right?

7. IrishBoy123

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

8. anonymous

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

9. IrishBoy123

and cancel out x's

10. anonymous

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

11. anonymous

But then it's still $\frac{ \infty }{ }$

12. anonymous

I mean infinity/ infinity

13. IrishBoy123

so do L'Hopital again

14. IrishBoy123

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

15. anonymous

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

16. anonymous

Ahaha true! Kk :D

17. anonymous

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

18. IrishBoy123

no. keep differentitaing the top and you'll get it down to a constant

19. IrishBoy123

yes $$e^{x^2}$$ is a beast

20. anonymous

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

21. perl

Correct, you will eventually get $$1 / \infty$$

22. anonymous

And that's = 0 :D

23. anonymous

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