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In general: How do I figure out how fast a function blows up at a certain point or infinity?

How fast does $\ln x$ blow up at $0$? Does it blow up as fast as $1/x$, $1/x^2$, or maybe faster than any $1/x^n$?

How do I answer such a question about other types of functions? How do I answer such a question for infinity?

I know that $\ln x$ blows up very slowly at infinity, slower than any $x^n, n>0$. How do I justify this information?

Edit: After some thinking I realized that lnx is e^x flipped about y=x. This means lnx is asymptotic to the -y axis as e^-x is asymptotic to +x axis. This is an intuitive justification of lnx blowing up slower than any x^-n, n>0

Gappy Hilmore
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  • The easy way is just using l'Hospital's rule. Consider the ratio $\frac{\log x}{x^n}$ as $x\to+\infty$ or the product $x^n\log x$ as $x\downarrow 0$. You can evaluate those limits easily by the aforementioned rule of calculus. – Giuseppe Negro May 11 '15 at 00:07
  • Try looking at limits of ratios: $\lim_{x \to \infty} \frac{\ln x}{x^a}.$ – matt biesecker May 11 '15 at 00:08
  • I know how to answer this question for two particular functions. lnx vs 1/x was only an example/motivation. I want to know if there is a classification of how functions blow up. – Gappy Hilmore May 11 '15 at 00:25
  • Suppose $\lim_{x \to 0} |f(x)| = \lim_{x \to 0} |g(x)| = +\infty$. We say that $f$ "blows up" faster than $g$ if $\lim_{x \to 0}\bigg|\frac{f(x)}{g(x)} \bigg|= +\infty$. – shalop May 11 '15 at 00:29
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    $\log x = -\log (1/x)$ so if you know how fast $\log$ grows towards $x=\infty$ you automatically knows the behaviour at $x=0$. – Winther May 11 '15 at 00:49
  • Consider http://math.stackexchange.com/q/55468/160028 for explaining why exponential grow faster than polynomial – Elimination Jul 22 '15 at 09:20

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