Free Abelian Groups

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Before we discuss homology groups, we review some basics of abelian group theory.

The group operation for an abelian group is denoted by $+$. The unit element is denoted by $0$.

Let $G_1$ and $G_2$ be abalian groups. A map $f: G_1\longrightarrow G_2$ is said to be a homomorphism if $$f(x+y)=f(x)+f(y),\ x,y\in G_1.$$ If $f$ is also a bijection (i.e one-to-one and onto), $f$ is called an isomorphism. If there is an isomorphism $f: G_1\longrightarrow G_2$, $G_1$ is said to be isomorphic to $G_2$ and we write $G_1\stackrel{f}{\cong} G_2$ or simply $G_1\cong G_2$.

Example. Define a map $f: \mathbb Z\longrightarrow\mathbb Z_2=\{0,1\}$ by $$f(2n)=0\ \mbox{and}\ f(2n+1)=1.$$ Then $f$ is a homomorphism.

A subset $H\subset G$ is a subgroup if it is a group with respect to the group operation of $G$.

Example. For any $k\in\mathbb N$, $k\mathbb Z=\{kn: n\in\mathbb Z\}$ is a subgroup of $\mathbb Z$.

Example. $\mathbb Z_2=\{0,1\}$ is not a subgroup of $\mathbb Z$.

Let $H$ be a subgroup of $G$. Define a relation on $G$ by $$\forall x,y\in G,\ x\sim y\ \mbox{if}\ x-y\in H.$$ Then $\sim$ is an equivalence relation on $G$. The equivalence class of $x\in G$ is denoted by $[x]$, i.e. \begin{eqnarray*}[x]&=&\{y\in G: y\sim x\}\\&=&\{y\in G: y-x\in H\}.\end{eqnarray*} Let $G/H$ be the quotient set $$G/H=\{[x]: x\in G\}.$$ Define an operation $+$on $G/H$ by $$[x]+[y]=[x+y],\ \forall [x],[y]\in G/H.$$ Then $G/H$ becomes an abelian group with this operation.

Example. $\mathbb Z/2\mathbb Z=\{[0],[1]\}$. Define $\varphi: \mathbb Z/2\mathbb Z\longrightarrow\mathbb Z_2$ by $$\varphi([0])=0\ \mbox{and}\ \varphi([1])=1.$$ Then $\mathbb Z/2\mathbb Z\cong\mathbb Z_2$. In general, for every $k\in\mathbb N$, $\mathbb Z/k\mathbb Z\cong\mathbb Z_k$.

Lemma 1. Let $f: G_1\longrightarrow G_2$ be a homomorphism. Then

(a) $\ker f=\{x\in G_1: f(x)=0\}=f^{-1}(0)$ is a subgroup of $G_1$.

(b) ${\mathrm im}f=\{f(x): x\in G_1\}$ is a subgroup of $G_2$.

Theorem 2 [Fundamental Theorem of Homomorphism]. Let $f: G_1\longrightarrow G_2$ be a homomorphism. Then $$G_1/\ker f\cong{\mathrm im}f.$$

Example. Let $f: \mathbb Z\longrightarrow\mathbb Z_2$ be defined by $$f(2n)=0,\ f(2n+1)=1.$$ Then $\ker f=2\mathbb Z$ and ${\mathrm im}f=\mathbb Z_2$. By Fundamental Theorem of Homomorphism, $$\mathbb Z/2\mathbb Z\cong\mathbb Z_2.$$

Take $r$ elements $x_1,x_2,\cdots,x_r$ of $G$. The elements of $G$ of the form $$n_1x_1+n_2x_2+\cdots+n_rx_r\ (n_i\in\mathbb Z,\ 1\leq i\leq r)$$ form a subgroup of $G$, which we denote $\langle x_1,\cdots,x_r\rangle$. $\langle x_1,\cdots,x_r\rangle$ is called a subgroup of $G$ generated by the generators $x_1,\cdots,x_r$. If $G$ itself is generated by finite lelements, $G$ is said to be finitely generated. If $n_1x_1+\cdots+n_rx_r=0$ is satisfied only when $n_1=\cdots=n_r=0$, $x_1,\cdots,x_r$ are said to be linearly independent.

Definition. If $G$ is fintely generated by $r$ linearly independent elements, $G$ is called a free abelian group of rank $r$.

Example. $\mathbb Z$ is a free abelian group of rank 1 generated by 1 (or $-1$).

Example. Let $\mathbb Z\oplus\mathbb Z=\{(m,n):m,n\in\mathbb Z\}$. The $\mathbb Z\oplus\mathbb Z$ is a free abelian group of rank 2 generated by $(1,0)$ and $0,1)$. More generally, $$\stackrel{r\ \mbox{copies}}{\overbrace{\mathbb Z\oplus\mathbb Z\oplus\cdots\oplus\mathbb Z}}$$ is a free abelian group of rank $r$.

Example. $\mathbb Z_2=\{0,1\}$ is fintely generated by 1 but is not free. $1+1=0$ so 1 is not linearly independent.

If $G=\langle x\rangle=\{0,\pm x,\pm 2x,\cdots\}$, $G$ is called a cyclic group. If $nx\ne 0$ $\forall n\in\mathbb Z\setminus\{0\}$, it is an infinite cyclic group. If $nx=0$ for some $n\in\mathbb Z\setminus\{0\}$, it is a finite cyclic group. Let $G=\langle x\rangle$ and let $f:\mathbb Z\longrightarrow G$ be a homomorphism defined by $f(k)=kx$, $k\in\mathbb Z$. $f$ is an epimorphism (i.e. onto homomorphism), so by Fundamental Theorem of Homomorphism, $$G\cong\mathbb Z/\ker f.$$ If $G$ is a finite group, then there exists the smallest positive integer $N$ such that $Nx=0$. Thus $$\ker f=\{0,\pm N,\pm 2N,\cdots\}=N\mathbb Z.$$ Hence $$G\cong\mathbb Z/N\mathbb Z\cong\mathbb Z_N.$$ If $G$ is an infinite cyclic group, $\ker f=\{0\}$. Hence, $$G\cong\mathbb Z/\{0\}\cong\mathbb Z.$$

Lemma 3. Let $G$ be a free abelian group of rank $r$, and let $H$ be a subgroup of $G$. Then one may always choose $p$ generators $x_1,\cdots,x_p$ out of $r$ generators of $G$ so that $k_1x_1,\cdots,k_px_p$ generate $H$. Hence, $$H\cong k_1\mathbb Z\oplus\cdots\oplus k_p\mathbb Z$$ and $H$ is of rank $p$.

Theorem 4 [Fundamental Theorem of Finitely Generated Abelian Groups] Let $G$ be a finitely generated abelian group with $m$ generators. Then $$G\cong\stackrel{r}{\overbrace{\mathbb Z\oplus\cdots\oplus\mathbb Z}}\oplus \mathbb Z_{k_1}\oplus\cdots\oplus\mathbb Z_{k_p}$$ where $m=r+p$. The number $r$ is called the rank of $G$.

Proof. Let $G=\langle x_1, \cdots,x_m\rangle$ and let $f: \mathbb Z\oplus\cdots\oplus\mathbb Z\longrightarrow G$ be the surjective homomorphism $$f(n_1,\cdots,n_m)=n_1x_1+\cdots +n_mx_m.$$ Then by Fundamental Theorem of Homomorphism $$\mathbb Z\oplus\cdots\oplus\mathbb Z/\ker f\cong G.$$ $\stackrel{m}{\overbrace{\mathbb Z\oplus\cdots\oplus\mathbb Z}}$ is a free abelian group of rank $m$ and $\ker f$ is a subgroup of $\mathbb Z\oplus\cdots\oplus\mathbb Z$, so by Lemma 3 $$\ker f\cong k_1\mathbb Z\oplus\cdots\oplus k_p\mathbb Z.$$ Define $\varphi:\stackrel{p}{\overbrace{\mathbb Z\oplus\cdots\oplus\mathbb Z}}/k_1\mathbb Z\oplus \cdots\oplus k_p\mathbb Z\longrightarrow\mathbb  Z/k_1\mathbb Z\oplus\cdots\oplus\mathbb Z/k_p\mathbb Z$ by $$\varphi((n_1,\cdots,n_p)+k_1\mathbb Z\oplus\cdots\oplus k_p\mathbb Z)=(n_1+k_1\mathbb Z,\cdots,n_p+k_p\mathbb Z).$$ Then $$\stackrel{p}{\overbrace{\mathbb Z\oplus\cdots\oplus\mathbb Z}}/k_1\mathbb Z\oplus\cdots\oplus k_p\mathbb Z\stackrel{\varphi}{\cong}\mathbb Z/k_1\mathbb Z\oplus\cdots\oplus\mathbb Z/k_p\mathbb Z.$$ Hence, \begin{eqnarray*}G&\cong&\stackrel{m}{\overbrace{\mathbb Z\oplus\cdots\oplus\mathbb Z}}/\ker f\\&\cong&\stackrel{m}{\overbrace{\mathbb Z\oplus\cdots\oplus\mathbb Z}}/k_1\mathbb Z\oplus\cdots\oplus k_p\mathbb Z\\&\cong&\stackrel{m-p}{\overbrace{\mathbb Z\oplus\cdots\oplus\mathbb Z}}\oplus\mathbb Z/k_1\mathbb Z\oplus\cdots\oplus Z/k_p\mathbb Z\\&\cong&\stackrel{m-p}{\overbrace{\mathbb Z\oplus\cdots\oplus\mathbb Z}}\oplus\mathbb Z_{k_1}\oplus\cdots\oplus\mathbb Z_{k_p}.\end{eqnarray*}

Derivatives

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In this lecture, I am going to introduce you a new idea, which was discovered by Sir Issac Newton and Gottfried Leibiz, to find the slope of a tangent line. This is in fact a quite ingenious idea as you will see. Let a function $y=f(x)$ be given. We want to find the slope of a line tangent to the graph of $y=f(x)$ at a point $x=a$. First consider another point on the $x$-axis that is away from $x=a$. If the distance from $x=a$ to this point is $h$, then the point can be written as $x=a+h$. Let $P(a,f(a))$ and $Q(a+h,f(a+h))$. Then the slope of line segment $\overline{PQ}$ is given by $$\frac{f(a+h)-f(a)}{h}.$$

Now we continuously change $h$ so that it gets smaller and smaller close to $0$, consequently the point $a+h$ gets closer to $a$. We want to see how the rate $\frac{f(a+h)-f(a)}{h}$ changes as $h\to 0$. To illustrate the situation better, I will use a specific example, say $f(x)=x^2$ with $a=2$. First we take $h=1$. The following picture shows you the graph of $f(x)=x^2$ (in black), where $1.5\leq x\leq 3$ and the line through $P(2,4)$ and $Q(2+h,(2+h)^2)$ (in blue), and the line tangent to the graph $f(x)=x^2$ at $x=2$ (in red).

Next we take $h=0.5$. Then the picture becomes

For $h=0.1$, the picture becomes

As one can clearly see, the line through $P(2,4)$ and $Q(2+h,(2+h)^2)$ gets closer to the tangent line as $h$ gets smaller close to $0$. We can still do better. For $h=0.001$, the picture becomes

The line through $P(2,4)$ and $Q(2+h,(2+h)^2)$ and the tangent line now appear to be overlapping. From this observation, we can see that the rate $\frac{f(a+h)-f(a)}{h}$ gets closer and closer to the slope of tangent line as $h$ gets smaller and smaller close to $0$. In fact, the slope would be exactly the limit of $\frac{f(a+h)-f(a)}{h}$ as $h$ approaches $0$. Denote the limit by $f'(a)$. Then $$f'(a)=\lim_{h\to 0}\frac{f(a+h)-f(a)}{h}.$$ $f'(a)$ is called the derivative of $f(x)$ at $x=a$. One may wonder why we need another name for the slope of a tangent line. The reason is that as we will see later the slope of a tangent line can mean something else in different contexts. Let $x=a+h$. Then $x\to a$ as $h\to 0$. So $f'(a)$ can be also written as $$f'(a)=\lim_{x\to a}\frac{f(x)-f(a)}{x-a}.$$ The equation of tangent line to $y=f(x)$ at $x=a$ is then given by $$y-f(a)=f'(a)(x-a).$$

Example. Find the equation of tangent line to the graph of $f(x)=x^2$ at $x=2$.

Solution. First we need to find $f'(2)$, i.e. the slop of the tangent line. \begin{eqnarray*}f'(2)&=&\lim_{h\to 0}\frac{f(2+h)-f(2)}{h}\\&=&\lim_{h\to 0}\frac{(2+h)^2-4}{h}\\&=&\lim_{h\to 0}\frac{4+4h+h^2-4}{h}\\&=&\lim_{h\to 0}(4+h)\\&=&4.\end{eqnarray*}

Of course, we can also use the alternative definition of $f'(a)$ to calculate the slope:\begin{eqnarray*}f'(2)&=&\lim_{x\to 2}\frac{f(x)-f(2)}{x-2}\\&=&\lim_{x\to 2}\frac{x^2-4}{x-2}\\&=&\lim_{x\to 2}\frac{(x+2)(x-2)}{x-2}\\&=&\lim_{x\to 2}(x+2)\\&=&4.\end{eqnarray*}

The equation of tangent line is then $y-4=4(x-2)$ or $y=4x-4$.

Remark. One may wonder which definition of $f'(a)$ to use. I would say that is the matter of a personal taste. For a polynomial function, one notable difference between the two definitions is that if you use the first definition, you will end up expanding a polynomial, while you will have to factorize a polynomial with the second definition. Since the expansion of a polynomial is easier than the factorization, you may want to use the first definition if you are not confident with factorizing polynomials.

Example. Find the equation of tangent line to the graph of $f(x)=x^5$ at $x=1$.

Solution. As we discussed in the previous lecture, this is an extremely difficult problem to solve by using only algebra if not impossible. But surprise! With the new method, this is more or less a piece of cake. First we calculate the slope $f'(1)$. \begin{eqnarray*}f'(1)&=&\lim_{h\to 0}\frac{(1+h)^5-1}{h}\\&=&\lim_{h\to 0}\frac{(1+h)^5-1}{h}\\&=&\lim_{h\to 0}\frac{1+5h+10h^2+10h^3+5h^4+h^5-1}{h}\\&=&\lim_{h\to 0}(5+10h+10h^2+5h^3+h^4)\\&=&5.\end{eqnarray*} Or by the second definition, \begin{eqnarray*}f'(1)&=&\lim_{x\to 1}\frac{f(x)-f(1)}{x-1}\\&=&\lim_{x\to 1}\frac{x^5-1}{x-1}\\&=&\lim_{x\to 1}\frac{(x-1)(x^4+x^3+x^2+x+1)}{x-1}\\&=&\lim_{x\to 1}(x^4+x^3+x^2+x+1)\\&=&5.\end{eqnarray*}Therefore the equation of the tangent line is given by $y-1=5(x-1)$ or $y=5x-4$. The following picture shows the graph of $y=x^5$ (in blue) and the graph of tangent line $y=5x-4$.

Finding a Line Tangent to a Curve

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Let us consider a simple geometry problem. Given a curve $y=f(x)$, we want to find a line tangent to the graph of $y=f(x)$ at $x=a$ (meaning the line meets the graph of $y=f(x)$ exactly at a point $(a,f(a))$ on a small interval containing $x=a$.

One may wonder at this point why finding a tangent line is a big deal. Well, it is in fact a pretty big deal besides mathematicians’ purely intellectual curiosities. There is a reason why Sir Issac Newton had to invent calculus of which crucial notion is the slope of a tangent line. It is still too early to talk about why it is important or useful. We will get there when we are ready.

We attempt to tackle the problem with an example first. Here is an example we want to consider

Example. Find the equation of a line tangent to the graph of $y=x^2$ at $x=2$.

Solution. To find the equation of a line, we need two ingredients: slope and $y$-intercept or slope and a point. We already know a point. We know that the line must pass through $(2,4)$. So all we need to find is its slope $m$. From algebra, we know that the equation of a line passing through $(2,4)$ with slope $m$ is given by $y-4=m(x-2)$ or $y=mx-2m+4$. Since $y=x^2$ and $y=mx-2m+4$ meet exactly at one point, the quadratic equation $x^2=mx-2m+4$ or $x^2-mx+2m-4=0$ must have exactly one solution. We have learned from the theory of quadratic equations that in that case the discriminant $D=b^2-4ac$ must be equal to $0$. That is, in our case $$D=m^2-4(2m-4)=m^2-8m+16=(m-4)^2=0.$$ Hence we determine that $m=4$ and the equation of the tangent line is $y=4x-4$.

So we see that finding the slope of a tangent line is not that difficult and that it does not require any new mathematics, or does it? Remember that we have not yet tackled our problem in general context. Before we get more ambitious, consider another example with a more complicated function, say $y=x^5$. Let us say that we want to find the line tangent to the graph of $y=x^5$ at $x=1$. Then the equation of the tangent line would be $y=mx-m+1$. In order for $y=x^5$ and the line $y=mx-m+1$ to meet exactly at one point, the quintic equation $x^5-mx+m-1=0$ must have exactly one solution. Our problem here is that we have no algebraic means, such as quadratic formula or discriminant, to use to determine the value of $m$. We are stuck here and there is no hope of tackling our simple geometry problem using only algebra. That is the reason why we have to cleverly devise a new way to tackle the problem. This is where we enter the realm of Calculus. The new idea to tackle the problem is not really new and it was already used by the ancient Greeks. And the world had to wait until it was rediscovered independently by Sir Issac Newton and by Gottfried Leibniz. I do not know if any of them actually knew about the ancient Greek idea.

Limits involving Infinity and Asymptotes

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So far we have mainly studied finite limits. Here we would like to discuss infinite limits. You may wonder why  we need to study infinite limits. They in fact do have important applications. One immediate application is that it provides  us some information on the shape of a function, i.e. it helps us sketch the graph of a function, as we will see later.

We first begin with the notion of vertical asymptotes.

Definition. The line $x=a$ is called a  vertical asymptote of the graph of $y=f(x)$ if $$\lim_{x\to a+}f(x)=\pm\infty,\ \mbox{or}\ \lim_{x\to a-}f(x)=\pm\infty.$$

Example. Find the vertical asymptotes of the graph of $y=\displaystyle\frac{x^2-3x+2}{x^3-4x}$.

Solution. The candidates for vertical asymptotes are the values of $x$ that make the denominator $0$. In our example, they are the root of the equation $x^3-4x=0$. Since $x^3-4x=x(x^2-4)=x(x+2)(x-2)$, we find three roots $x=-2,0,2$. However, some of them may not necessarily be vertical asymptotes. To check this, we calculate the limits: \begin{eqnarray*}\lim_{x\to 0+}\frac{x^2-3x+2}{x^3-4x}&=&\frac{2}{0-}\ \mbox{(Can you see why?)}\\&=&-\infty,\\\lim_{x\to 0-}\frac{x^2-3x+2}{x^3-4x}&=&\frac{2}{0+}\\&=&\infty,\\\lim_{x\to -2-}\frac{x^2-3x+2}{x^3-4x}&=&\frac{12}{0-}\\&=&-\infty,\\\lim_{x\to -2+}\frac{x^2-3x+2}{x^3-4x}&=&\frac{12}{0+}\\&=&\infty,\\\lim_{x\to 2}\frac{x^2-3x+2}{x^3-4x}&=&\lim_{x\to 2}\frac{(x-1)(x-2)}{x(x+2)(x-2)}\\&=&\lim_{x\to 2}\frac{x-1}{x(x+2)}\\&=&\frac{1}{8}.\end{eqnarray*}

So, we see that $x=0,-2$ are vertical asymptotes while $x=2$ is not.

Definition. A line $y=b$ is called a horizontal asymptote of the graph of $y=f(x)$ if $$\lim_{x\to\infty}f(x)=b,\ \mbox{or}\ \lim_{x\to -\infty}f(x)=b.$$

Example. Find the horizontal asymptotes of the graph of $y=\displaystyle\frac{5x^2+8x-3}{3x^2+2}$.

Solution. You can notice at once that the limit $\displaystyle\lim_{x\to\infty}\frac{5x^2+8x-3}{3x^2+2}$ is an $\frac{\infty}{\infty}$ type indeterminate form. So how do we calculate this kind of indeterminate form? First divide the numerator and the denominator by the highest power of $x$ appeared in the denominator:\begin{eqnarray*}\lim_{x\to\infty}\frac{5x^2+8x-3}{3x^2+2}&=&\lim_{x\to\infty}\frac{\frac{5x^2+8x-3}{x^2}}{\frac{3x^2+2}{x^2}}\\&=&\lim_{x\to\infty}\frac{5+\frac{8}{x}-\frac{3}{x^2}}{3+\frac{2}{x^2}}\\&=&\frac{5}{3}.\end{eqnarray*} The final answer is obtained by the limits $\displaystyle\lim_{x\to\infty}\frac{1}{x^n}=0,$ where $n$ is a positive integer.

Similarly, $\displaystyle\lim_{x\to\infty}\frac{5x^2+8x-3}{3x^2+2}=\frac{5}{3}$ using the limit $\displaystyle\lim_{x\to -\infty}\frac{1}{x^n}=0,$ where $n$ is a positive integer. The following picture contains the graphs of the function (in blue) and the horizontal asymptote (in red).

Figure 1. The graph of y=(5x^2+8x-3)/(3x^2+2)

Example. Find the horizontal asymptotes of $y=\displaystyle\frac{x^2-3x+2}{x^3-4x}$.

Solution. The limits $\displaystyle\lim_{x\to\pm\infty}\frac{x^2-3x+2}{x^3-4x}$ is $\frac{\infty}{\infty}$ type indeterminate form. So as we did in the previous example, we first divide the numerator and the denominator by the highest power of $x$ that is appeared in the denominator: \begin{eqnarray*}\lim_{x\to\infty}\frac{x^2-3x+2}{x^3-4x}&=&\lim_{x\to\infty}\frac{\frac{x^2-3x+2}{x^3}}{\frac{x^3-4x}{x^3}}\\&=&\lim_{x\to\infty}\frac{\frac{1}{x}-\frac{3}{x^2}+\frac{2}{x^3}}{1-\frac{4}{x^2}}\\&=&0.\end{eqnarray*}

Similarly you find that $\displaystyle\lim_{x\to -\infty}\frac{x^2-3x+2}{x^3-4x}=0$.

The following picture shows you the graph of the function (in blue), the horizontal and the vertical asymptotes (in red).

Figure 2. The graph of y=(x^2-3x+2)(x^3-4x)

Normally the graph of a function $y=f(x)$ never touches or crosses its horizontal asymptote while it gets closer and closer to its horizontal asypmtote as $x\to\infty$ or $x\to -\infty$. But there are exceptions as shown in the following example.

Example. Consider the function $f(x)=2+\displaystyle\frac{\sin x}{x}$. Using the Sandwich Theorem, one can show that $\displaystyle\lim_{x\to\pm\infty}\frac{\sin x}{x}=0$ and hence $\displaystyle\lim_{x\to\pm\infty}f(x)=2$. That is $y=2$ is a horizontal asymptote of the curve on both left and right. As you can see in the following picture, the graph crosses the horizontal asymptote $y=2$ infinitely many
times.

Figure 3. The graph of y=2+sin(x)/x.

There is another kind of asymptotes, called oblique (slanted) asymptotes. An oblique asymptote can be seen from a rational function $\frac{p(x)}{q(x)}$ where $\deg p(x)>\deg q(x)$. An oblique asymptote is in fact given by a dominating term of a rational function as you can see in the following example.

Example. Consider the rational function $f(x)=\displaystyle\frac{2x^2-3}{7x+4}$. By long division, we obtain \begin{eqnarray*}f(x)&=&\frac{2x^2-3}{7x+4}\\&=&\left(\frac{2}{7}x-\frac{8}{49}\right)+\frac{-115}{49(7x+4)}.\end{eqnarray*} As $x\to\pm\infty$, the remainder $\displaystyle\frac{-115}{49(7x+4)}\to 0$. Hence the graph of $f(x)$ gets closer to the graph of the linear function $y=\displaystyle \frac{2}{7}x-\frac{8}{49}$ as $x\to\pm\infty$. This linear function is an oblique (slanted) asymptote of the graph of $f(x)$. The following picture shows the graph of $f(x)$ (in blue) and both the vertical asymptote $x=-\frac{4}{7}$ and the oblique asymptote $y=\frac{2}{7}x-\frac{8}{49}$ (in red).

Figure 4. The graph of y=(2x^3-3)/(7x+4).

A closer look.

Figure 5. The graph of y=(2x^3-3)/(7x+4).

Continuity

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Intuitively speaking, we say a function is continuous at a point if its graph has no separation, i.e. there is no hole or breakage, at that point. Such notion of continuity can be defined explicitly as follows.

Definition: A function \(f(x)\) is said to be continuous at a point \(x=a\) if \[\lim_{x\to a}f(x)=f(a).\]

Note that the above definition assumes the existence of both \(\displaystyle\lim_{x\to a}f(x)\) and \(f(a)\).

There are 3 different types of discontinuities.

  • \(f(a)\) is not defined.

For example, consider the function\[f(x)=\frac{x^2-4}{x-2}.\] Clearly \(f(2)\) is not defined. However the limit \(\displaystyle\lim_{x\to 2}f(x)\) exists:\begin{eqnarray*}\lim_{x\to 2}\frac{x^2-4}{x-2}&=&\lim_{x\to 2}\frac{(x+2)(x-2)}{x-2}\\&=&\lim_{x\to 2}(x+2)=4.\end{eqnarray*} As a result the graph has a hole.

This kind of discontinuity is called a removable discontinuity, meaning that we can extend \(f(x)\) to a function which is continuous at \(x=a\) in the following sense: Define \(g(x)\) by\[g(x)=\left\{\begin{array}{ccc}f(x)\ \mbox{if}\ x\ne a,\\\lim_{x\to a}f(x)\ \mbox{if}\ x=a.\end{array}\right.\]Then \(g(x)\) is a continuous at \(x=a\). The function \(g(x)\) is called the continuous extension of \(f(x)\). What we just did is basically filling the hole and the filling is the limit \(\displaystyle\lim_{x\to a}f(x)\). For the above example, we define\[g(x)=\left\{\begin{array}{ccc}\frac{x^2-4}{x-2} &\mbox{if}& x\ne 2,\\4 &\mbox{if}& x=2.\end{array}\right.\] Then \(g(x)\) is continuous at \(x=2\) and in fact, it is identical to \(x+2\).

  • \(\displaystyle\lim_{x\to a}f(x)\) deos not exist.

Example. Let \(f(x)=\left\{\begin{array}{cc}2x-2,\ &1\leq x<2\\3,\ &2\leq x\leq 4.\end{array}\right.\) \(f(2)=3\) but \(\displaystyle\lim_{x\to 2}f(x)\) does not exist because \(\displaystyle\lim_{x\to 2-}f(x)=2\) while \(\displaystyle\lim_{x\to 2+}f(x)=3\).

  • \(f(a)\) is defined and \(\displaystyle\lim_{x\to a}f(x)\) exists, but \(\displaystyle\lim_{x\to a}f(x)\ne f(a)\).

Example. Let \(f(x)=\left\{\begin{array}{cc}\displaystyle\frac{x^2-4}{x-2},\ &x\ne 2\\3,\ &x=2.\end{array}\right.\) Then \(f(2)=3\) and \(\displaystyle\lim_{x\to 2}f(x)=4\).


From the properties of limits (Theorem 1, Lecture 4), we obtain the following properties of continuous functions.

Theorem 9. If functions \(f(x)\) and \(g(x)\) are continuous at \(x=a\), then

  1. \((f\pm g)(x)=f(x)\pm g(x)\) is continuous at \(x=a\).
  2. \(f\cdot g(x)=f(x)\cdot g(x)\) is continuous at \(x=a\).
  3. \(\displaystyle\frac{f}{g}(x)=\frac{f(x)}{g(x)}\) is continous at \(x=a\) provided \(g(a)\ne 0\).

There are some important classes of continous functions.

  • Every polynomial function \(p(x)=a_nx^n+a_{n-1}x^{n-1}+\cdots+a_0\) is continuous everywhere, because \(\displaystyle\lim_{x\to a}p(x)=p(a)\) for any \(-\infty<a<\infty\).
  • If \(p(x)\) and \(q(x)\) are polynomials, then the rational function \(\displaystyle\frac{p(x)}{q(x)}\) is continuous wherever it is defined \(\{x\in\mathbb{R}|q(x)\ne 0\}\).
  • \(y=\sin x\) and \(y=\cos x\) are continuous everywhere.
  • \(y=\tan x\) is continous where it is defined, i.e. everywhere except at the points \(x=\pm\frac{\pi}{2},\pm\frac{3\pi}{2},\pm\frac{5\pi}{2},\cdots\).
  • If \(n\) is a positive integer, then \(y=\root n\of{x}\) is continuous where it is defined. That is, if \(n\) is an odd integer, it is defined everywhere. If \(n\) is an even integer,it is defined on \([0,\infty)\), the set of all non-negative real numbers.

Recall that the composite function \(g\circ f(x)\) of two functions \(f(x)\) and \(g(x)\) (read \(f\) followed by \(g\)) is defined by \[g\circ f(x):=g(f(x)).\]

Theorem 10. Suppose that \(\displaystyle\lim_{x\to a}f(x)=L\) exists and \(g(x)\)is continuous function at \(x=L\). Then\[\lim_{x\to a}g\circ f(x)=g(\lim_{x\to a}f(x)).\]

It follows from Theorem 10 that the composite function of two continuous functions is again a continuous function.

Corollary 11. If \(f(x)\) is continuous at \(x=a\) and \(g(x)\) is continuous at \(f(a)\), the the composite function \(g\circ f(x)\) is continuous at \(x=a\).

Example. The function\(y=\sqrt{x^2-2x-5}\) is the composite function  \(g\circ f(x)\) of two functions \(f(x)=x^2-2x-5\) and \(g(x)=\sqrt{x}\). The function \(f(x)=x^2-2x-5\) is continuous everywhere while \(g(x)=\sqrt{x}\) is continuous on \([0,\infty)\), so by Corollary 11, the composite function \(g\circ f(x)=\sqrt{x^2-2x-5}\) is continuous on its repective domain which is \((-\infty,1-\sqrt{6}]\) or \([1+\sqrt{6},\infty)\). The following picture shows you the graph of \(y=\sqrt{x^2-2x-5}\) on the intervals  \((-\infty,1-\sqrt{6}]\) and \([1+\sqrt{6},\infty)\).

Continuous functions exhibit many nice properties. I would like to introduced a couple of them here. The first is the so-called Max-Min Theorem.

Theorem 12. [Max-Min Theorem] If \(f(x)\) is a continuous function on a closed interval \([a,b]\), \(f(x)\) attains its maximum value and minimum value on \([a,b]\).

Another important property is the so-called Intermediate Value Theorem (IVT). The IVT has an important application in the study of equations.

Theorem 13. [The Intermediate Value Theorem] If \(f(x)\) is continuous on a closed interval \([a,b]\) and \(f(a)\ne f(b)\), then , then \(f(x)\) takes on every value between \(f(a)\) and \(f(b)\). In other words, if \(f(a)<k<f(b)\) (assuming that \(f(a)<f(b)\)), then \(f(c)=k\) for some number \(a<c<b\).

It follows from Theorem 13 that

Corollary 14. If \(f(x)\) is continuous on a closed interval \([a,b]\) and \(f(a)\cdot f(b)<0\), then \(f(x)=0\) for some \(a<x<b\).

Using  Corollary 14 we can tell if a root of the equation \(f(x)=0\) can be found in some interval. For instance

Example. Show that the equation \(x^3-x-1=0\) has a root in the interval \([-1,2]\).

Solution. Let \(f(x)=x^3-x-1\). Then \(f(x)\) is continuous on \([-1,2]\). Since \(f(-1)=-1\) and \(f(2)=5\) have different signs, by Corollary 14 there is a root of \(x^3-x-1=0\) in the open interval \((-1,2)\).