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$g(n)=n^a$, $f(n)=n^b$, where $a$ and $b$ are real numbers such that: $0 \lt a \lt b$.

I want to use L'Hopital's technique to prove that $g(n)$ belongs to $f(n)$.

My attempt: $$ \lim_{n\to\infty}\frac{n^a}{n^{a+1}}.$$

I'm using $a+1$ to state that $b \gt a$.

The problem is that I don't know if using $a+1$ is correct.

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    What do you mean by "belongs to f(n)"? Using a+1 is incorrect. You don't need l'Hôpital, just properties of exponents.2011-01-19
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    By belongs I mean Cf(n) is an upper bound of g(n) for some constant C>0.2011-01-19
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    In other words, you want to show that $g(x) = O(f(x))$.2011-01-19
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    @PEV Yes! thats what I want to prove2011-01-19
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    No, you cannot assume that $b\geq a+1$ just from assuming that $b\gt a$.2011-01-19
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    Many values of $C$ work. Do you have a specific $C$ that you suspect will do the trick?2011-01-19

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$\frac{g(n)}{f(n)}=\frac{n^a}{n^b}=n^{(a-b)}$ where $a-b<0$ So what is the limit as $n\rightarrow \infty $?

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    Why do we need to even mention limits at all?2011-01-19
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    is it infinit? ∞2011-01-19
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    @Wilbert: No, it's not.2011-01-19
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    n tends to cero2011-01-19