# Asymptotic equivalence?

Let there be two functions $f(x)$ and $g(x)$. If we consider $\lim_{x \rightarrow x_{0}} \frac{f(x)}{g(x)} = k$, we say that

1. $k=1$, then $f(x)\sim g(x)$, $f(x)$ is equivalent to $g(x)$ as $x \rightarrow x_{0}$
2. $k=0$, then $f(x) = o(g(x))$, $f(x)$ is dominated by $g(x)$ asymptotically as as $x \rightarrow x_{0}$ .
3. $k=\infty$, then $f(x) = \omega (g(x))$, $f(x)$ dominates $g(x)$ asymptotically as $x \rightarrow x_{0}$

But what do we say when $1 < k < \infty$, e.g. $k=10$. Are such functions asymptotically equivalent? For example: are $x$ and $10x$ equivalent as $x \rightarrow \infty$?

Thank you!

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Big Theta here –  Ilya Mar 23 '12 at 14:08
Thank you! That is exactly what I was looking for. So, to sum up, $O$ is $o$ or $\sim$, $\Omega$ is $\omega$ or $\sim$, and $\Theta$ is when $f(x)$ is bounded above and below by $g(x)$ multiplied by some constant $k$, or in other words it is $O$ and $\Omega$ . Right? –  user825089 Mar 23 '12 at 14:25
When $f = O(g)$ the limit still may not exist; e.g. $f_n = (-1)^n$ and $g_n = f_n^2$ - then $\frac{f_n}{g_n}$ oscillates but $f = O(g)$ –  Ilya Mar 23 '12 at 14:46
I see, thank you! –  user825089 Mar 23 '12 at 14:55
Warning: $f(x) \in \Theta(g(x))$ does not imply $f(x)/g(x)$ has a limit neither, only that it's bounded. Eg: $f(x)=x(2+\sin x)$ $g(x)=x$ –  leonbloy Jun 4 at 18:15