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For every finite case, I can find a $c$ where $2^n = n^c$, so why is this true?

$$\lim_{n \rightarrow \infty} \frac{2^n}{n^c} = 0$$

From the finite cases it seems like $2^n$ grows faster because we can find a $c$ to match it at any $n$.

Jack Avante
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    I'm sorry, it is false. – Bernard Apr 21 '20 at 11:28
  • It is false?.. It's in a textbook with example problems with solutions. Suppose they made a mistake then? – Jack Avante Apr 21 '20 at 11:31
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    Do not confuse $\lim\limits_{c\to\infty}\lim\limits_{n\to\infty}\frac{2^n}{n^c}$ with $\lim\limits_{n\to\infty}\lim\limits_{c\to\infty}\frac{2^n}{n^c}$ with $\lim\limits_{n\to\infty}\frac{2^n}{n^c}$. These are all different problems. You specifically asked about the last one, and in this last one $c$ is a single value, whatever it happened to be, that once we started the problem never changes from then on. – JMoravitz Apr 21 '20 at 11:36

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Because $c$ then will depend on $n$. The whole idea of what makes a polynomial into a polynomial is that its power is not dependent on the argument, i.e. in $x^5$, $5$ is independent of $x$.

If I keep varying the exponent with $x$, you end up with something like $x^{f(x)}$, which is a general form that can grow faster than exponentials. But that does not make it into a polynomial.

gt6989b
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