May somebody give me a prove why $n^{r} < c^{n}$ when $r \in R_{>0}$, $ c \in R_{>1}$ and $ n \in N$ (for all $n$ at some point).
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what is, given real $\beta > 0,$ $$ \lim_{x \rightarrow + \infty} \frac{e^{\beta x}}{x} ?$$
It is not necessary to use L'Hospital, it is enough to see that the first derivative is $$ \frac{\left( \beta x - 1 \right) e^{\beta x}}{x^2} $$ and the second is $$ \frac{\left( \beta^2 x^2 - 2 \beta x +2 \right) e^{\beta x}}{x^3}. $$ That is, for $x > 0,$ the second derivative is strictly positive. The original function begins with a vertical asymptote at $x=0,$ descends to a minimum at $x = 1 / \beta,$ but then increases without bound as $x \rightarrow + \infty$
With $\beta = \frac{1}{3},$ here is a nice graph for $x > 0.$
Will Jagy
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well, it is positive infinity, right? – get rekt m8 Jan 30 '17 at 02:34
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@HunterDJohny do you know a way to prove that? – Will Jagy Jan 30 '17 at 02:38
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I know that the numerator grows faster than the denominater, but i cannot say that the numerator is always bigger, so i'm not sure. – get rekt m8 Jan 30 '17 at 03:11
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@HunterDJohny LHospitals rule. Also, if $\beta$ is small enough, the ratio will be smaller than $1$ for some values of $x,$ but then grow without bound as $x$ increases still more – Will Jagy Jan 30 '17 at 03:16
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Is there any way to proove it without using the derivation? – get rekt m8 Jan 30 '17 at 03:20