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:

nil

nil

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.