Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

We can talk about how "quickly" an infinite series approaches $0$ by talking about an asymptotic bound on its terms - a series that is $O(1/x)$ converges more slowly than one that is $O(1/x^2)$, etc.

I am confused about what we mean when we say that a function rapidly approaches a limit at a (finite) point. For example, consider the functions $x$ and $x^2$ as $x$ approaches $0$. I think we say that $x^2$ approaches $0$ "more quickly" than $x$ does as $x\to 0$ because $x^2$ is $o(x)$ as $x\to 0$, so $x^2$ reaches any arbitrary "closeness to $0$" before $x$ does as $x\to 0$.

On the other hand, close to $0$, it also seems that $x$ is approaching $0$ "more quickly" than $x^2$ in the sense that the magnitude of derivative of $x$ is larger than that of $x^2$.

So my question is which one of these does "more quickly" generally mean in informal math-language, and if anyone has any insight into the intuition here.

share|improve this question
    
$x^2$ approaches 0 "more quickly" than $x$ because $\lim_{x\rightarrow 0^+}{\frac{x}{x^2}}=+\infty$. That is, the ratio of $x$ to $x^2$ gets arbitrarily large. More generally, you can say that $f(x)$ approaches $x_0$ "more quickly" than $g(x)$ does if $\left|\lim_{x\rightarrow x_0}{\frac{f(x)}{g(x)}}\right|<1$. –  Hayden Mar 20 at 18:16
    
Note that L'Hopital means that the [$x^2=o(x)$ as $x\to 0$] definition and the derivative definition are equivalent. –  Alyosha Mar 20 at 18:18

1 Answer 1

up vote 2 down vote accepted

Notice that you are not using the size of the derivative in discussing limits at $\infty.$ So, you don't emphasize the derivative in limits at $0.$ You just talk about which function gets close to $0$ earlier, which at $0^+$ means for larger $x.$ Also, at $\infty^+,$ the derivative information (in absolute value) agrees with the size information; note that we are thinking of a point moving to the right along the $x$ axis.

I suppose it is fair to say that the derivative information is backwards of this, at $0^+, $ and this is happening because approaching $0^+$ we are moving to the left. The slowest approach to $0$ you are likely to see is Holder continuity such as $\sqrt x,$ where the derivative is actually infinite at the origin. `Slow' meaning: you need to have $x < 0.0001$ to get $\sqrt x < 0.01$

share|improve this answer
    
You could slow the convergence even more by considering $\sqrt[n]{x}$ for larger $n$. –  Baby Dragon Mar 20 at 18:30
    
yes; still Holder –  Will Jagy Mar 20 at 18:31

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.