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In my calculus class, whenever we try to find a limit of a sequence as it approaches infinity and it turns out to be like: $1^{\infty}$. We have to end up using L'Hopital's rule. I don't understand why it has to be L'Hopitaled, can't you just take the limit as the sequence approach $99999$ and the answer would be $1^{99999} = 1$ and you are done?

Why do we have to L'Hopital then?

Bobson Dugnutt
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Asker123
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    Are you calling $\frac{0}{0}=\frac{\infty}{\infty}=1$? Cause that's not true – Bobson Dugnutt Feb 21 '16 at 23:44
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    Consider $(\frac{99999}{100000})^{99999}$. – Andreas Blass Feb 21 '16 at 23:47
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    99999 $\ne \infty$. But... what limit do you have in mind. Obviously you don't have $\lim_{x \rightarrow \infty} 1^x$ in mind. If you mean something like $\lim f(x) = (x^2 + 7x +2)/(x^3 + 3)$ well you can't do $lim_{x\rightarrow \infty} f(x) = \lim_{x \rightarrow 9999} f(x) = 1.0008001700250007991496709251891e-4$ and then say "well, obviously that is 0 because ... c'mon..." – fleablood Feb 22 '16 at 00:02

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Consider the following: $(1+1/n)^n \to e,$ while $(1+1/n)^{n^2} \to \infty.$

zhw.
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