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anonymous
 2 years ago
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anonymous
 2 years ago
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anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0I think the answer is supposed to be zero but how I get there?

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0I 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
 2 years ago
Best ResponseYou've already chosen the best response.1You should see a pattern in the derivatives.

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

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

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

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0I kinda understand what kainui said

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

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0So does the limit even exist?

Kainui
 2 years 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

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0Ah, that's what I thought, thanks!
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