Understanding Orders of Infinitesimals

Question

I'm having trouble understanding the following piece of text:

Problems in Mathematical Analysis, B. Demidovich

For one thing, I don't understand the first sentence. $\lim\limits_{x \to \infty}{\frac{ax^n+bx^{n-k}}{x^n}} = a$, it would seem that the lower order is actually irrelevant, contrary to the statement (if I understand it correctly).

Also, the example jumps from problem to answer, without the solution. That really doesn't help me. How is $\ln (1+2x) \sim 2x$ and $\sqrt[3]{x^3+2x^4} \sim \sqrt[3]{x^3}$?

Answer 1

This text is an old-style explanation of the rules of asymptotic calculus.

For you first point, the lower order is not irrelevant as it is about the limit at infinity, whereas the explanation are about expressions which tend to $0$.

For your second point, $\;\ln(1+2x)\sim 2x$ (supposing in a neighbourhood of $0$) comes from the high-school limit: $$\lim_{u\to 0} \frac{\ln(1+u)}{u}=1,\quad\text{which we translate as }\enspace\ln(1+u)\sim_{0}u.$$

Similarly, $\lim\limits_{x\to 0}\dfrac{x^3+2x^4}{x^3}=1$, so we say that $x^3+2x^4\sim_0x^3$.

These equivalences relations between functions are compatible with multiplication, division, exponentiation, composition by log (under mild conditions), but not with addition and subtraction.

A general rule, as stated in your quotation, asserts that a polynomial is equivalent to its highest degree term near infinity, and to its lowest degree term near $0$.

Note that, on the contrary, $\lim\limits_{x\to+\infty}\dfrac{x^3+2x^4}{x^3}=+\infty$, and we say that $x^3=o(x^3+2x^4)$ near $\infty$ in the Landau notation.

Answer 2

If we divide top and bottom by the highest order factor of x we get $$\lim_{x\to\infty}\frac{a+bx^{-k}}{1}=0 $$ since $x^{-k}\to 0$ as $x\to\infty.$ In this case, $x^{n}$ is the only relevant order of $x$ in the limit. This is different than the idea that $$\Delta x \gg \left(\Delta x\right)^2\qquad as\quad x\to 0.$$

As far as the approximations you have listed, these are true by virtue of the functions' respective Maclaurin series expansions. There most likely are some geometric arguments you could use to arrive at the same result at this stage of the game that sidesteps the use of differentiation and I'd be interested to see if someone else could provide that.

Answer 3

I think of infinitesimals like polynomials of $x = \frac{1}{y}$. A higher order polynomial goes to zero faster than a lower order polynomial when $x \to 0$, just like it goes to infinity faster than a lower one when $x \to \infty$.

So $\sqrt{x^3 + 2x^4} \sim \sqrt{x^3}$ for infinitesimals (as $x \to 0$) in the sense that

$$\lim_{x\to0}\frac{\sqrt{x^3 + 2x^4}}{\sqrt{x^3}}= \lim_{y\to\infty}\frac{\sqrt{\frac{1}{y^3} + \frac{2}{y^4}}}{\sqrt{\frac{1}{y^3}}} = \lim_{y\to\infty}{\sqrt{1 + \frac{2}{y}}} = \lim_{x\to0} \sqrt{1+2x} = 1$$

just like $\sqrt{x^3 + 2x^4} \sim \sqrt{2 x^4}$ for polynomials as $x \to \infty$:

$$\lim_{x\to\infty}\frac{\sqrt{x^3 + 2x^4}}{\sqrt{2x^4}} = 1$$