## jennisicle one year ago .

1. jennisicle

I think the answer is supposed to be zero but how I get there?

2. jennisicle

I took the deribative of the top and the bottom but I just ended up at (100n^99)/(e^n) and don't know where to go from here.

3. myininaya

You should see a pattern in the derivatives.

4. myininaya

you can do l'hospitals again

5. myininaya

well nevermind kainui did it for you

6. myininaya

You could have left it. Maybe it is hard for some people to reach that.

7. Kainui

The point is, L'H rule says the limit of a ratio is the same as the limit of the ratio of their derivatives as well. So we can just see that eventually n^100 will eventually become a constant while the bottom will always be a function of n.

8. myininaya

If it is hard for you to see the constant kainui got. You can look at a an example with a lesser power

9. myininaya

Like what would f^(4)(x) look like if we had f(x)=x^5

10. jennisicle

I kinda understand what kainui said

11. myininaya

coolness

12. Kainui

Sorry, I realized I wasn't being cryptic enough, so I went more cryptical. =P

13. jennisicle

So does the limit even exist?

14. Kainui

$\frac{ d^n }{ dx^n }(x^n)=n!$ just for fun -- Yeah, it even exists and it's even zero. @jennisicle

15. jennisicle

Ah, that's what I thought, thanks!