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How does $\lim\limits_{n \to \infty} \frac{f(n)}{g(n)} = 0$ imply that $f(n) \in \mathcal{O}(g(n))$? I'm having trouble understanding that.

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By definition, $f(x)\in\mathcal{O}(g(x))$ if and only if there exists some $N>0$ such that $$|f(x)| \le N|g(x)|$$ for all $x$ greater than some $x_0$. Given the above limit, for any $\epsilon >0$ we have $$\left|\frac{f(x)}{g(x)}\right| < \epsilon$$ for $x$ greater than some $x_0$. If we take $N=\epsilon$ then we have the required $N$ for $$|f(x)| \le N|g(x)|$$ so this shows that $f(x)\in\mathcal{O}(g(x))$. In fact, we have $f(x)\in\mathcal{o}(g(x))$ since the above holds for every $N>0$.

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Thanks! You are using the $\delta$-$\epsilon$ definition. I think other limits will give other bounds. –  staame Sep 29 '12 at 23:03
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Hint: If the limit is $0$, then if $n$ is large enough, we have $$\left|\frac{f(n)}{g(n}\right|\lt 1,$$ meaning that $|f(n)|\lt |g(n)|$.

If, as is common in Computer Science, you restrict "big $O$" statements to non-negative functions, one can remove the absolute value signs.

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