Calculus Study Notes (I): Limits

微积分学习（一）：极限 苏剑林 · 2009-08-16 · Mathematics learning calculus limit

This is not a calculus textbook — it's a record of my own thoughts while studying the subject, meant to be discussed with like-minded people.

Pick up any calculus textbook and you'll be confronted with rigorous, professional mathematical language, which is enough to scare many people off. Indeed, the calculus founded by Newton and Leibniz was not rigorous, and this eventually triggered the Second Crisis in the Foundations of Mathematics. Thanks to the efforts of the French mathematician Cauchy and the German mathematician Weierstrass, calculus achieved an unprecedented level of rigor, overcoming that crisis. Together with the later Third Crisis, mathematics as a whole became even more rigorous.

But for beginners, this rigorous formulation of calculus is quite hard to digest. So it might make more sense to learn calculus in the order it was originally developed — that is, "informal first, rigorous later." This not only makes learning more efficient, but also makes the subject more interesting. more

Generally speaking, a calculus textbook proceeds in the order: sets, mappings, functions, limits, derivatives, differentials, integrals, ... going ever deeper. The parts on "sets, mappings, functions" mainly exist to give rigorous definitions; the actual problem-solving really begins with limits (though of course the earlier foundations shouldn't be neglected either).

The meaning of a limit is: as the function $f(x)$ has x approach some given value $x_0$ (or $\infty$) with ever-increasing closeness — note that here "approach" means getting infinitely close to, not equal to; this distinction is precisely one of the key steps in overcoming the crisis in the foundations of mathematics — we find the value of $f(x)$. If $f(x)$ tends toward a fixed value, we say the limit exists; otherwise, the limit does not exist.

Note: when $x_0=\infty$ or $x_0=0$, there is a distinction between approaching from the positive side and the negative side, i.e., $x_0=+\infty,x_0=-\infty$ or $x_0=+0,x_0=-0$. For the limit to exist, besides tending to a stable value, the one-sided limits (from the positive and negative directions) must also be equal to each other, e.g. $\lim_{x \to +\infty} f(x)=\lim_{x \to -\infty} f(x)$ — only then can we say the limit exists. If $\lim_{x \to +\infty} f(x) \neq \lim_{x \to -\infty} f(x)$, the limit does not exist.

The general procedure in calculus computations is: first manipulate and simplify the expression, and only then substitute in the value.

Below are a few exercises on finding limits, which I've worked through. These are just problems I thought were worth writing up — this is by no means a comprehensive list. My solutions may not be fully rigorous, but they work, and that's my style. ^_^

(1) $\lim_{n \to \infty} (\sqrt{n^4+n+1}-n^2)(n+3)$

Solution:

First we need a formula: $\sqrt{a^2+b}=a+\frac{b}{2a+\frac{b}{2a+\frac{b}{2a+...}}}$, which I derived using a kind of iterative method for equations. It's quite convenient for computing limits. As follows:

$$\lim_{n \to \infty} (\sqrt{n^4+n+1}-n^2)(n+3) \\ =\lim_{n \to \infty} (n^2+\frac{n+1}{2n^2+\frac{n+1}{2n^2+...}}-n^2)(n+3) \\ =\lim_{n \to \infty} \frac{(n+1)(n+3)}{2n^2+\frac{n+1}{2n^2+...}}$$

Since $\lim_{n \to \infty} \frac{n+1}{2n^2} \to 0$, we can ignore the remaining terms and only need to consider

$$\lim_{n \to \infty} \frac{(n+1)(n+3)}{2n^2} \\ =\lim_{n \to \infty} 1/2 (1+1/n)(1+3/n)=1/2$$

(2) $\lim_{x \to \infty} \frac{\sqrt[3]{x} sin x}{x+1}$

Solution:

$$\frac{\sqrt[3]{x} \sin x}{x+1} < \frac{\sqrt[3]{x} \sin x}{x} = x^{-2/3}\sin x$$

When $x \to \infty$, we have $x^{-2/3} \to 0,|sin x| \leq 1$, 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:

$e^n=1+n+\frac{n^2}{2!}+\frac{n^3}{3!}+\frac{n^4}{4!}+...$, so we have

$$\lim_{n \to 0} \frac{e^n-1}{n} \\ =\lim_{n \to 0} \frac{n+\frac{n^2}{2!}+\frac{n^3}{3!}+\frac{n^4}{4!}+...}{n} \\ =\lim_{n \to 0} 1+\frac{n}{2!}+\frac{n^2}{3!}+\frac{n^3}{4!}+...=1$$

(4) $\lim_{x \to 0} \frac{\sqrt{1+x sin x}-1}{e^{x^2}-1}$

Solution:

I think this one is fairly hard — it needs to make use of result (3):

$$\lim_{x \to 0} \frac{\sqrt{1+x \sin x}-1}{e^{x^2}-1} \\ = \lim_{x \to 0} \frac{(\frac{x \sin x}{2+\frac{x \sin x}{2+...}})}{e^{x^2}-1}$$

Since $\lim_{x \to 0} x sin x \to 0$, we can ignore the remaining terms and only need to consider:

$$\lim_{x \to 0} \frac{x \sin x}{(e^{x^2}-1)\cdot 2} \\ =\lim_{x \to 0} \frac{{\sin x}/{x}}{{e^{x^2}-1}/{x^2}\cdot 2}$$

We already know: $\lim_{x \to 0} \frac{sin x}{x} \to 1,\lim_{x \to 0} \frac{e^{x^2}-1}{x^2} \to 1$,

so

$$\lim_{x \to 0} \frac{\sqrt{1+x \sin x}-1}{e^{x^2}-1} \to 1/2$$

More to be added over time — looking forward to discussing further with everyone.