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jennisicle
 one year ago
Best ResponseYou've already chosen the best response.1I think the answer is supposed to be zero but how I get there?

jennisicle
 one year ago
Best ResponseYou've already chosen the best response.1I 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.

myininaya
 one year ago
Best ResponseYou've already chosen the best response.1You should see a pattern in the derivatives.

myininaya
 one year ago
Best ResponseYou've already chosen the best response.1you can do l'hospitals again

myininaya
 one year ago
Best ResponseYou've already chosen the best response.1well nevermind kainui did it for you

myininaya
 one year ago
Best ResponseYou've already chosen the best response.1You could have left it. Maybe it is hard for some people to reach that.

Kainui
 one year ago
Best ResponseYou've already chosen the best response.1The 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.

myininaya
 one year ago
Best ResponseYou've already chosen the best response.1If it is hard for you to see the constant kainui got. You can look at a an example with a lesser power

myininaya
 one year ago
Best ResponseYou've already chosen the best response.1Like what would f^(4)(x) look like if we had f(x)=x^5

jennisicle
 one year ago
Best ResponseYou've already chosen the best response.1I kinda understand what kainui said

Kainui
 one year ago
Best ResponseYou've already chosen the best response.1Sorry, I realized I wasn't being cryptic enough, so I went more cryptical. =P

jennisicle
 one year ago
Best ResponseYou've already chosen the best response.1So does the limit even exist?

Kainui
 one year ago
Best ResponseYou've already chosen the best response.1\[\frac{ d^n }{ dx^n }(x^n)=n!\] just for fun  Yeah, it even exists and it's even zero. @jennisicle

jennisicle
 one year ago
Best ResponseYou've already chosen the best response.1Ah, that's what I thought, thanks!
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