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

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SUMMARY

The discussion focuses on comparing the asymptotic behavior of two functions, \( f(x) \) and \( g(x) \), using limits. It establishes that if \( \lim_{x\rightarrow \infty} f(x) / g(x) = \infty \), then \( f \) is asymptotically greater than \( g \), and if the limit is 0, \( f \) is asymptotically less. Additionally, the use of logarithms is highlighted, specifically \( \lim_{x\rightarrow \infty} \log(f(x) / g(x)) \), where \( \infty \) indicates \( f \) is greater and \( -\infty \) indicates \( f \) is smaller. The discussion concludes that the logarithm's monotonic nature preserves the relationship between the functions.

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OhMyMarkov
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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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