Here's the question you clicked on:
jennisicle
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I think the answer is supposed to be zero but how I get there?
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.
You should see a pattern in the derivatives.
you can do l'hospitals again
well nevermind kainui did it for you
You could have left it. Maybe it is hard for some people to reach that.
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.
If it is hard for you to see the constant kainui got. You can look at a an example with a lesser power
Like what would f^(4)(x) look like if we had f(x)=x^5
I kinda understand what kainui said
Sorry, I realized I wasn't being cryptic enough, so I went more cryptical. =P
So does the limit even exist?
\[\frac{ d^n }{ dx^n }(x^n)=n!\] just for fun -- Yeah, it even exists and it's even zero. @jennisicle
Ah, that's what I thought, thanks!