I read the following in a website. Image

I want to know why we can replace one infinitesimal with an equivalent one. The idea seems intuitive but is there a formal proof?

  • $\begingroup$ I doubt if you have a clear notion of what infinitesimals are (and their equivalence means). Your website is trying to present some sort of hand waving technique to evaluate limits and nothing more. Don't trust it and instead learn laws of limits which are used in evaluation of limits. $\endgroup$
    – Paramanand Singh
    Dec 30, 2017 at 6:47
  • 1
    $\begingroup$ @ParamanandSingh, here the term "infinitesimal" is used in a different sense from the one envisioned by Leibniz (and Robinson). Here it refers merely to a pair of functions whose ratio tends asymptotically to 1. $\endgroup$ Dec 30, 2017 at 20:19

2 Answers 2


Note that in general you cannot simply replace arbitrary quantities by infinitesimally equivalent ones. For example, $\lim_{x \to 0} \frac{\sin(x)-x}{x^3}$ is not zero, which is what you would get if you replaced $\sin(x)$ by $x$ in it. In most cases what you are doing when you replace infinitesimally equivalent quantities is multiplying the final result by $1$, writing $1$ as a limit of a ratio of infinitesimally equivalent quantities, and then dragging this limit inside your original one. So for example in the first problem in the image, the steps look like:

$$\lim_{x \to 0} \frac{\ln(1+4x)}{\sin(3x)}=\lim_{x \to 0} \frac{\ln(1+4x)}{\sin(3x)} \lim_{x \to 0} \frac{4x}{\ln(1+4x)} \lim_{x \to 0} \frac{\sin(3x)}{3x}=\lim_{x \to 0} \frac{4x}{3x}=4/3.$$

  • $\begingroup$ Great tip on using infinitesimals! $\endgroup$
    – Macrophage
    Dec 29, 2017 at 15:56
  • $\begingroup$ Could you clarify the last phrase in the third line, "you are multiplying is multiplying the final.."? $\endgroup$ Dec 29, 2017 at 16:11
  • $\begingroup$ And can we use infinitesimally equivalent quantities in any way to solve the first example that you have given? $\endgroup$ Dec 29, 2017 at 16:11
  • $\begingroup$ @VishalSubramanyam We did use infinitesimally equivalent quantities here. We noticed that $\sin(3x)$ and $\ln(1+4x)$ are infinitesimally equivalent to $3x$ and $4x$ respectively. Thus we can choose a "clever form of $1$" to multiply by, which ultimately replaces these "complicated" expressions by simpler, infinitesimally equivalent ones. In this case that clever form of $1$ is $\lim_{x \to 0} \frac{4x}{\ln(1+4x)} \lim_{x \to 0} \frac{\sin(3x)}{3x}$. $\endgroup$
    – Ian
    Dec 29, 2017 at 16:13
  • $\begingroup$ Could you also give more examples of the first type of limit where direct use of an equivalent infinitesimal would be a bad idea? $\endgroup$ Dec 29, 2017 at 16:13

enter image description here So basically it's because for equivalent infinitesimal expressions of a function, the limit of its ratio with the original function as x approach 0 can be proved to be 1. (sin x/x for example. Sorry, I'm typing on a phone so the format is crude.)

Source: Wikipedia - Indeterminate Form.

  • $\begingroup$ Please add the source of this screenshot. $\endgroup$
    – edm
    Dec 29, 2017 at 15:39
  • $\begingroup$ I have a doubt. Can we replace an infinitesimal with an equivalent one only if it is in a ratio with another infinitesimal? Can't we apply this in all limits? $\endgroup$ Dec 29, 2017 at 15:41
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    $\begingroup$ @VishalSubramanyam Sure, and you can mentally change any expression to be a ratio with 1. And the same proof still applies. $\endgroup$
    – Macrophage
    Dec 29, 2017 at 15:44
  • $\begingroup$ @VishalSubramanyam Like you have a and a' in the picture above equal to 1, the same proof still applies. So b=b' as x->0. No problem. $\endgroup$
    – Macrophage
    Dec 29, 2017 at 15:45
  • $\begingroup$ @Macrophage Thanks a lot $\endgroup$ Dec 29, 2017 at 15:46

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