[Math] Why can we replace an infinitesimal in a limit with an equivalent infinitesimal
infinitesimalslimitslimits-without-lhopital
I read the following in a website.
I want to know why we can replace one infinitesimal with an equivalent one. The idea seems intuitive but is there a formal proof?
Best Answer
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:
If you have a sufficiently smooth function $F$ from a real variable $t$ to real $n\times n$ matrices, then you can differentiate each element of the matrix with respect to $t$, and therefore give meaning to $F'(t)$.
Now if we know that $F(t)$ is always a rotation matrix and that $F(t_0)=I$, then it turns out that $F'(t_0)$ will always be skew-symmetric. And conversely, every skew-symmetric matrix will be the derivative of some $F$ that satisfies these conditions.
In this way we can consider the skew-symmetic $F'(t_0)$ to encode which way and how fast the new coordinates given by $F(t)$ rotate at time $t=t_0$.
Without the assumption that $F(t_0)=I$ we can still say that $F(t)^{-1}F'(t)$ and $F'(t)F(t)^{-1}$ are always skew-symmetric; these encode the instantaneous rotation at any time in two different ways.
Best Answer
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:
$$\lim_{x \to 0} \frac{\ln(1+4x)}{\sin(3x)}=\lim_{x \to 0} \frac{\ln(1+4x)}{\sin(3x)} \lim_{x \to 0} \frac{4x}{\ln(1+4x)} \lim_{x \to 0} \frac{\sin(3x)}{3x}=\lim_{x \to 0} \frac{4x}{3x}=4/3.$$