# Proof of some limit identities

Limit is a bedrock of Calculus. Understanding limit and its identities requires proofs as well as understanding others does.

## Prerequisite

### Triangle inequality

#### Formula

$\forall{}a,\ b\in{}\R,\ \exists{}\ |a+b|\leq{}|a|+|b|$

#### Proof

1. $$a,\ b\geq{}0$$
$$|a+b|=a+b=|a|+|b|$$
2. $$a\geq{}0,\ b<0\ and\ a-|b|\geq{}0$$
$$|a+b|=a+b\leq{}a-b$$ since $$a<0$$
3. $$a\geq{}0,\ b<0\ and\ a-|b|\leq{}0$$
$$|a+b|=-a-b\leq{}a-b$$ since $$-a\leq{}a$$
4. $$a,\ b<0$$
$$|a+b|=-a-b=|a|+|b|$$

Since the formula means the same regardless of the order of $$a$$ and $$b$$, we proved it for other two alternative conditions consist of $$b\geq{}0$$ for condition 2 and 3.

### Formal definitions for limit

#### $$\lim\limits_{x\rarr{}c}f(x)=L$$ where $$c,\ L\in\R$$

1. Two-sided
if $$\forall{}\ \epsilon{}>0$$, $$\exists{}\ \delta{}>0$$ such that $$|f(x)-L|< \epsilon$$ whenever $$x\in{}A=(c-\delta{},c)\cup(c,c+\delta{})$$ where $$D_{f}\supseteq{}A$$, then we call $$\lim\limits_{x\rarr{}c}f(x)=L$$
2. Left-sided
if $$\forall{}\ \epsilon{}>0$$, $$\exists{}\ \delta{}>0$$ such that $$|f(x)-L|< \epsilon$$ whenever $$x\in{}A=(c-\delta{},c)$$ where $$D_{f}\supseteq{}A$$, then we call $$\lim\limits_{x\rarr{}c^{-}}f(x)=L$$
3. Right-sided
if $$\forall{}\ \epsilon{}>0$$, $$\exists{}\ \delta{}>0$$ such that $$|f(x)-L|< \epsilon$$ whenever $$x\in{}A=(c,c+\delta{})$$ where $$D_{f}\supseteq{}A$$, then we call $$\lim\limits_{x\rarr{}c^{+}}f(x)=L$$

#### $$\lim\limits_{x\rarr{}c}f(x)=\pm\infty$$ where $$c\in\R$$

1. Two-sided $$+\infty$$
if $$\forall{}\ \epsilon{}>0$$, $$\exists{}\ \delta{}>0$$ such that $$f(x)> \epsilon$$ whenever $$x\in{}A=(c-\delta{},c)\cup(c,c+\delta{})$$ where $$D_{f}\supseteq{}A$$, then we call $$\lim\limits_{x\rarr{}c}f(x)= +\infty{}$$
2. Left-sided $$+\infty$$
if $$\forall{}\ \epsilon{}>0$$, $$\exists{}\ \delta{}>0$$ such that $$f(x)> \epsilon$$ whenever $$x\in{}A=(c-\delta{},c)$$ where $$D_{f}\supseteq{}A$$, then we call $$\lim\limits_{x\rarr{}c^{-}}f(x)= +\infty{}$$
3. Right-sided $$+\infty$$
if $$\forall{}\ \epsilon{}>0$$, $$\exists{}\ \delta{}>0$$ such that $$f(x)> \epsilon$$ whenever $$x\in{}A=(c,c+\delta{})$$ where $$D_{f}\supseteq{}A$$, then we call $$\lim\limits_{x\rarr{}c^{+}}f(x)= +\infty{}$$
4. Two-sided $$-\infty$$
if $$\forall{}\ \epsilon{}<0$$, $$\exists{}\ \delta{}>0$$ such that $$f(x)< \epsilon$$ whenever $$x\in{}A=(c-\delta{},c)\cup(c,c+\delta{})$$ where $$D_{f}\supseteq{}A$$, then we call $$\lim\limits_{x\rarr{}c}f(x)= -\infty{}$$
5. Left-sided $$-\infty$$
if $$\forall{}\ \epsilon{}<0$$, $$\exists{}\ \delta{}>0$$ such that $$f(x)< \epsilon$$ whenever $$x\in{}A=(c-\delta{},c)$$ where $$D_{f}\supseteq{}A$$, then we call $$\lim\limits_{x\rarr{}c^{-}}f(x)= -\infty{}$$
6. Right-sided $$-\infty$$
if $$\forall{}\ \epsilon{}<0$$, $$\exists{}\ \delta{}>0$$ such that $$f(x)< \epsilon$$ whenever $$x\in{}A=(c,c+\delta{})$$ where $$D_{f}\supseteq{}A$$, then we call $$\lim\limits_{x\rarr{}c^{+}}f(x)= -\infty{}$$

## Proof of identities

### Constant Rule for Limits

#### Formula

if $$a,\ b$$ are constants, then: $\lim\limits_{x\rarr{}c}b=b$

#### Proof

\begin{aligned} \because{}&\forall{}\epsilon>0,\ |f(x)-L|=|b-b|=0< \epsilon\\ \therefore{}&\forall{}\delta{}>0,\ |f(x)-L|< \epsilon{} \end{aligned}

### Identity Rule for Limits

#### Formula

if $$a$$ is a constant, then: $\lim\limits_{x\rarr{}c}x=c$

#### Proof

\begin{aligned} \because{}&\forall{}\delta{} = \epsilon{}>0,\ 0 < |x-a| < \delta{}\rArr{}|f(x)-c|=|x-c|< \epsilon{}\\ \therefore{}&\lim\limits_{x\rarr{}c}x=c \end{aligned}

### Scalar Product Rule for Limits

#### Formula

Suppose $$\lim\limits_{x\rarr{}c}f(x)=L$$ where $$L$$ is finite. Let $$\lambda$$ be a constant, then: $\lim\limits_{x\rarr{}c}\lambda{}f(x)=\lambda{}\cdot{}\lim\limits_{x\rarr{}c}f(x)=\lambda{}L$

#### Proof

\begin{aligned} \because{}&\lim\limits_{x\rarr{}c}f(x)=L\\ \therefore{}&\forall{}\epsilon{}>0,\ \exists{}\delta{}>0\ such\ that\ |f(x)-L|< \frac{\epsilon}{|\lambda{}|}\ whenever\ 0 < |x-c|< \delta{}\\ \because&|\lambda{}f(x)-\lambda{}L|=|\lambda{}|\cdot{}|f(x)-L| < |\lambda{}|\cdot{}\frac{\epsilon}{|\lambda{}|}=\epsilon{}\\ \therefore&\lim\limits_{x\rarr{}c}\lambda{}f(x)=\lambda{}\cdot{}\lim\limits_{x\rarr{}c}f(x)=\lambda{}L \end{aligned}

### Sum Rule for Limits

#### Formula

Suppose that $$\lim\limits_{x\rarr{}c}f(x)=L$$ and $$\lim\limits_{x\rarr{}c}g(x)=M$$, then: $\lim\limits_{x\rarr{}c}(f+g)(x)=\lim\limits_{x\rarr{}c}f(x)+\lim\limits_{x\rarr{}c}g(x)=L+M$

#### Proof

\begin{aligned} \because{}& \lim\limits_{x\rarr{}c}f(x)=L\ and\ \lim\limits_{x\rarr{}c}g(x)=M\\ \therefore{}&\begin{cases}\begin{aligned} \forall{}\epsilon{}>0,\ \exists{}\ \delta_{f}\ such\ that\ |f(x)-L|< \frac{\epsilon{}}{2}\ whenever\ 0< |x-c|< \delta_{f}\\ \forall{}\epsilon{}>0,\ \exists{}\ \delta_{g}\ such\ that\ |g(x)-M|< \frac{\epsilon{}}{2}\ whenever\ 0< |x-c|< \delta_{g}\\ \end{aligned}\end{cases}\\ \because{}&\begin{aligned} \vert{} [f(x)+g(x)]-(L+M) \vert{}&=|f(x)-L+g(x)-M|\\ &\leq{}|f(x)-L|+|g(x)-M|\\ &< \frac{\epsilon{}}{2}+ \frac{\epsilon{}}{2}\\ &=\epsilon{} \end{aligned}\\ &Supposing\ 0 < |x-c|< \min{}(\delta_{f},\delta_{g})\\ \therefore{}&\lim\limits_{x\rarr{}c}(f+g)(x)=\lim\limits_{x\rarr{}c}f(x)+\lim\limits_{x\rarr{}c}g(x)=L+M \end{aligned}

### Difference Rule for Limits

#### Formula

Suppose that $$\lim\limits_{x\rarr{}c}f(x)=L$$ and $$\lim\limits_{x\rarr{}c}g(x)=M$$, then: $\lim\limits_{x\rarr{}c}(f-g)(x)=\lim\limits_{x\rarr{}c}f(x)-\lim\limits_{x\rarr{}c}g(x)=L-M$

#### Proof

Let $$h(x)=-g(x)$$: \begin{aligned} \because{}&\lim\limits_{x\rarr{}c}h(x)=\lim\limits_{x\rarr{}c}-g(x)=-\lim\limits_{x\rarr{}c}g(x)=-M\\ \therefore{}&\lim\limits_{x\rarr{}c}(f-g)(x)=\lim\limits_{x\rarr{}c}[f(x)+(-g)(x)]=\lim\limits_{x\rarr{}c}f(x)+\lim\limits_{x\rarr{}c}-g(x)=\lim\limits_{x\rarr{}c}f(x)+\lim\limits_{x\rarr{}c}h(x)=L-M \end{aligned}

### Product Rule for Limits

#### Formula

Suppose that $$\lim\limits_{x\rarr{}c}f(x)=L$$ and $$\lim\limits_{x\rarr{}c}g(x)=M$$, then: $\lim\limits_{x\rarr{}c}(fg)(x)=\lim\limits_{x\rarr{}c}f(x)\cdot{}\lim\limits_{x\rarr{}c}g(x)=LM$

#### Proof

\begin{aligned} \because{}& \lim\limits_{x\rarr{}c}f(x)=L\ and\ \lim\limits_{x\rarr{}c}g(x)=M\\ \therefore{}&\begin{cases}\begin{aligned} &\forall{}\epsilon{}>0,\ \exists{}\ \delta_{f}\ such\ that\ |f(x)-L|< \frac{\epsilon{}}{2(1+|M|)}\ whenever\ 0< |x-c|< \delta_{f}\\ &\forall{}\epsilon{}>0,\ \exists{}\ \delta_{g1}\ such\ that\ |g(x)-M|< \frac{\epsilon{}}{2(1+|L|)}\ whenever\ 0< |x-c|< \delta_{g1}\\ &\exists{}\ \delta_{g2}\ such\ that\ |g(x)-M|<1\ whenever\ 0< |x-c|< \delta_{g2} \end{aligned}\end{cases}\\ \because{}&\begin{aligned} \vert{} g(x) \vert{}&=|g(x)-M+M|\\ &\leq{}|g(x)-M|+|M|\\ &<1+|M|\\ \end{aligned}\\ \therefore{}&|g(x)|<1+|M|\\ \because{}&\begin{aligned} \vert{} f(x)g(x)-LM \vert{}&=|f(x)g(x)-Lg(x)+Lg(x)-LM|\\ &=|g(x)[f(x)-L]+L\cdot{}[g(x)-M]|\\ &\leq{}|g(x)[f(x)-L]|+|L\cdot{}[g(x)-M]|\\ &=|g(x)|\cdot{}|f(x)-L|+|L|\cdot{}|g(x)-M|\\ &< (1+|M|)\cdot{}\frac{\epsilon{}}{2(1+|M|)}+|L|\cdot{}\frac{\epsilon{}}{2(1+|L|)}\\ &= \epsilon{}\cdot{}(\frac{1}{2}+\frac{|L|}{2(1+|L|)})\\ &< \epsilon{} \end{aligned}\\ &Supposing\ 0 < |x-c|< \min{}(\delta_{f},\delta_{g1},\delta_{g2})\\ \therefore{}&\lim\limits_{x\rarr{}c}(fg)(x)=\lim\limits_{x\rarr{}c}f(x)\cdot{}\lim\limits_{x\rarr{}c}g(x)=LM \end{aligned}

### Quotient Rule for Limits

#### Formula

Suppose that $$\lim\limits_{x\rarr{}c}f(x)=L$$ and $$\lim\limits_{x\rarr{}c}g(x)=M$$ where $$M\not = 0$$, then: $\lim\limits_{x\rarr{}c}\frac{f(x)}{g(x)}=\frac{\lim\limits_{x\rarr{}c}f(x)}{\lim\limits_{x\rarr{}c}g(x)}=\frac{L}{M}$

#### Proof

\begin{aligned} \because{}& \lim\limits_{x\rarr{}c}g(x)=M\\ \therefore{}&\begin{cases}\begin{aligned} &\forall{}\epsilon{}>0,\ \exists{}\ \delta_{g1}\ such\ that\ |g(x)-M|< \frac{\epsilon{}|M|^{2}}{2}\ whenever\ 0< |x-c|< \delta_{g1}\\ &\forall{}\epsilon{}>0,\ \exists{}\ \delta_{g2}\ such\ that\ |g(x)-M|< \frac{|M|}{2}\ whenever\ 0< |x-c|< \delta_{g2}\\ \end{aligned}\end{cases}\\ \because{}&\begin{aligned} \vert{} M \vert{}&=|M-g(x)+g(x)|\\ &\leq{}|g(x)-M|+|g(x)|\\ &< \frac{|M|}{2}+|g(x)|\\ \end{aligned}\\ \therefore{}&|g(x)|> \frac{|M|}{2}\\ \therefore{}&\frac{1}{|g(x)|} < \frac{2}{|M|}\\ \because{}&\begin{aligned} \vert{} \frac{1}{g(x)}- \frac{1}{M} \vert{}&=|\frac{M-g(x)}{Mg(x)}|\\ &= \frac{1}{|g(x)|} \cdot{} \frac{|g(x)-M|}{|M|}\\ &< \frac{2}{|M|}\cdot{}\frac{\frac{\epsilon{}|M|^{2}}{2}}{|M|}\\ &= \epsilon{} \end{aligned}\\ &Supposing\ 0 < |x-c|< \min{}(\delta_{g1},\delta_{g2})\\ \therefore{}&\lim\limits_{x\rarr{}c}\frac{1}{g(x)}=\frac{1}{M}\\ \therefore{}&\lim\limits_{x\rarr{}c}\frac{f(x)}{g(x)}=\lim\limits_{x\rarr{}c}[f(x)\cdot{}\frac{1}{g(x)}]=\lim\limits_{x\rarr{}c}f(x)\cdot{}\lim\limits_{x\rarr{}c}\frac{1}{g(x)}=\frac{L}{M}\\ \end{aligned}

## One more thing

Limits are interesting right? While another problem is also intriguing. This problem is brought from my Pre-Calculus class today.

### Problem

Find the maximum value of $$t$$ which fulfills: \tag{1} \begin{cases}\begin{aligned} x+y+z+t&=4\\ x^{2}+y^{2}+z^{2}+t^{2}&=\frac{16}{3}\\ \end{aligned}\end{cases}

### Solution

Surely it is a algebraic problem. You may find different kinds of algebraic solutions to it, such as using Cauchy–Schwarz inequality. Nevertheless, it has a lot to do with geometry in $$\R^{3}$$.
Transform (1) into (2), we have: \tag{2} \begin{cases}\begin{aligned} x+y+z&=4-t\\ x^{2}+y^{2}+z^{2}&=\frac{16}{3}-t^{2}\\ \end{aligned}\end{cases} In $$\R^{3}$$, $$x+y+z=4-t$$ is a plane whose x-,y-,z- intercepts (we name them as $$A,B,C$$ respectively) are $$(4-t,0,0),(0,4-t,0),(0,0,4-t)$$ respectively. $$x^{2}+y^{2}+z^{2}=\frac{16}{3}-t^{2}$$ is a sphere whose center is origin and radius is $$\sqrt{\frac{16}{3}-t^{2}}$$. Since $$\sqrt{\frac{16}{3}-t^{2}}\in\R$$, $$t\in{}(-\sqrt{\frac{16}{3}},\sqrt{\frac{16}{3}})$$. We hereby name the plane as $$\alpha$$ and sphere as $$\beta$$.
The set of solution $$S$$ is all the intersections of $$\alpha$$ and $$\beta$$ in the space.
In order to have $$S \not = \varnothing$$, the radius of $$\beta$$ should be greater than or equal to the distance between $$O$$ and $$\alpha$$. In the other word:
$\tag{3} (4-t)\cdot{}\sin{\angle{}ACO}\leq{}\sqrt{\frac{16}{3}-t^{2}}$ We can evaluate $$\sin{\angle{}ACO}$$:
\begin{aligned} \sin{\angle{}ACO}&=\frac{ \frac{4-t}{cos{\frac{\pi}{4}}} }{ \sqrt{(4-t)^{2}+(\frac{4-t}{cos{\frac{\pi}{4}}})^{2}} }\\ &=\frac{\sqrt{3}}{3} \end{aligned} Thus, we can find the equivalent of (3): \begin{aligned} \frac{(4-t)^{2}}{3}&\leq{}\frac{16}{3}-t^{2}\\ -\frac{8}{3}t+\frac{4}{3}t^{2}&\leq{}0\\ -2t+t^{2}&\leq{}0\\ t&\in{}[0,2] \end{aligned} Therefore, we have the maximum of $$t$$, which is $$2$$. Here are two images which help you to comprehend:

## References

I don't want to have a formal one here, thus let me just refer to one page that inspired me a lot or dug me out.
Calculus/Proofs of Some Basic Limit Rules from wikibooks.