MHB How Does the Limit of Logarithms Compare Two Functions Asymptotically?

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The discussion focuses on comparing the asymptotic behavior of two functions, f(x) and g(x), using limits. It establishes that if the limit of f(x) / g(x) as x approaches infinity is infinity, then f is asymptotically greater than g, while a limit of 0 indicates the opposite. The conversation also explores using the limit of the logarithm of the ratio, stating that if this limit approaches infinity, f is greater, and if it approaches negative infinity, f is smaller. Additionally, it highlights that the logarithm is a monotonic increasing function, preserving the relationship between f and g. This approach is noted for its utility in analyzing the growth rates of functions.
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Hello everyone!

I'm considering the problem of comparing the asymptotic behavior of two functions $f(x)$ and $g(x)$.

We can say that if $\lim _{x\rightarrow \infty} f(x) / g(x) = \infty$ then f is asymptotically greater than g. Similarly, if this limit is 0, then f is asymptotically less than g.

Now, how about if use $\lim _{x\rightarrow \infty} \log (f(x) / g(x))$, if the limit is $\infty$, then f is greater, and if the limity is $-\infty$, f is smaller, right?

Can we conclude from $\lim _{x\rightarrow \infty} \log f(x) / \log g(x)$ something about the respective behavior of f and g? It sure does help to look at logs of fast-growing functions...
 
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Re: Limit of Ration of Two Functions

The logarithm is monotonic increasing and changes no relationship. This is why logarithms (and square roots) are popular transformations.
 
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