This article is not a calculus tutorial, but rather a collection of my own views during my studies and discussions with fellow enthusiasts on related issues.
When picking up any "Calculus" textbook, one is met with professional and rigorous mathematical language, which many find daunting. Indeed, because the calculus founded by Newton and Leibniz was not rigorous, it triggered the Second Mathematical Crisis. Through the efforts of the French mathematician Cauchy and the German mathematician Weierstrass, calculus achieved unprecedented rigor, overcoming the Second Mathematical Crisis. With the subsequent Third Mathematical Crisis, mathematics became even more rigorous.
However, for beginners, rigorous calculus is quite difficult to comprehend. Therefore, we might as well follow the order in which calculus was created—that is, the sequence from "unrigorous to rigorous." This not only allows us to learn more efficiently but also increases our interest in learning mathematics.
Generally speaking, the sequence of a calculus course is: Sets, Mappings, Functions, Limits, Derivatives, Differentials, Integrals, ... and so on. The first three parts—Sets, Mappings, and Functions—are only there to provide rigorous definitions; the actual problem-solving begins with limits (though the preceding foundations should not be ignored).
The meaning of a limit is: in a function f(x), as x approaches a pre-given x_0 (or \infty) to an infinitely close degree (here "infinitely close" does not mean "equal to," which was the main step in overcoming the mathematical crisis), we find the value of f(x). If f(x) approaches a definite value, the limit is said to exist; otherwise, the limit does not exist.
Note: When x_0 = \infty or x_0 = 0, there is a distinction between positive and negative. That is, x_0 = +\infty, x_0 = -\infty or x_0 = +0, x_0 = -0. The condition for the existence of a limit, besides approaching a stable value, is that the limits from the positive and negative sides must be equal. For example, \lim_{x \to +\infty} f(x) = \lim_{x \to -\infty} f(x) is required for the limit to exist. If \lim_{x \to +\infty} f(x) \neq \lim_{x \to -\infty} f(x), then the limit does not exist.
The operational process of calculus is: first operate and simplify, then substitute the value.
Below are some practice problems regarding limits that I have solved. These are just the problems I felt were worth writing about and are not exhaustive. The solution process might not be strictly rigorous, but it is effective—this is my style ^_^:
(1) \lim_{n \to \infty} (\sqrt{n^4+n+1}-n^2)(n+3)
Solution:
First, one should be familiar with a formula:
\sqrt{a^2+b}=a+\frac{b}{2a+\frac{b}{2a+\frac{b}{2a+\dots}}}.
I derived this based on an equation iteration method. It is very
convenient for solving limits. As follows: \lim_{n \to \infty}
(\sqrt{n^4+n+1}-n^2)(n+3) = \lim_{n
\to \infty}
\left(n^2+\frac{n+1}{2n^2+\frac{n+1}{2n^2+\dots}}-n^2\right)(n+3)
= \lim_{n \to \infty}
\frac{(n+1)(n+3)}{2n^2+\frac{n+1}{2n^2+\dots}}
Since \lim_{n \to \infty} \frac{n+1}{2n^2} \to 0, we can ignore the subsequent terms and only consider: \lim_{n \to \infty} \frac{(n+1)(n+3)}{2n^2} = \lim_{n \to \infty} \frac{1}{2} (1+1/n)(1+3/n) = 1/2
(2) \lim_{x \to \infty} \frac{\sqrt[3]{x} \sin x}{x+1}
Solution: \frac{\sqrt[3]{x} \sin x}{x+1} < \frac{\sqrt[3]{x} \sin x}{x} = x^{-2/3}\sin x As x \to \infty, we have x^{-2/3} \to 0 and |\sin x| \leq 1, so \lim_{x \to \infty} \frac{\sqrt[3]{x} \sin x}{x+1} \to 0
(3) \lim_{n \to 0} \frac{e^n-1}{n}=1
Solution:
Since e^n=1+n+\frac{n^2}{2!}+\frac{n^3}{3!}+\frac{n^4}{4!}+\dots,
we have \lim_{n \to 0} \frac{e^n-1}{n}
= \lim_{n \to 0}
\frac{n+\frac{n^2}{2!}+\frac{n^3}{3!}+\frac{n^4}{4!}+\dots}{n}
= \lim_{n \to 0}
1+\frac{n}{2!}+\frac{n^2}{3!}+\frac{n^3}{4!}+\dots = 1
(4) \lim_{x \to 0} \frac{\sqrt{1+x \sin x}-1}{e^{x^2}-1}
Solution:
This is a problem I consider relatively difficult; it requires the
conclusion from (3). \lim_{x \to 0}
\frac{\sqrt{1+x \sin x}-1}{e^{x^2}-1} = \lim_{x \to 0} \frac{\left(\frac{x \sin
x}{2+\frac{x \sin x}{2+\dots}}\right)}{e^{x^2}-1} Since \lim_{x \to 0} x \sin x \to 0, we ignore the
subsequent terms and only consider: \lim_{x
\to 0} \frac{x \sin x}{(e^{x^2}-1)\cdot 2} = \lim_{x \to 0} \frac{\sin x / x}{(e^{x^2}-1)/x^2
\cdot 2} Given that \lim_{x \to 0}
\frac{\sin x}{x} \to 1 and \lim_{x \to
0} \frac{e^{x^2}-1}{x^2} \to 1, we have: \lim_{x \to 0} \frac{\sqrt{1+x \sin
x}-1}{e^{x^2}-1} \to 1/2
More content will be added successively to explore further topics with everyone.
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