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- anonymous

.

- jamiebookeater

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- anonymous

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

- anonymous

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.

- myininaya

You should see a pattern in the derivatives.

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## More answers

- myininaya

you can do l'hospitals again

- myininaya

well nevermind kainui did it for you

- myininaya

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

- 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.

- myininaya

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

- myininaya

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

- anonymous

I kinda understand what kainui said

- myininaya

coolness

- Kainui

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

- anonymous

So does the limit even exist?

- Kainui

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

- anonymous

Ah, that's what I thought, thanks!

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