AP Calculus: Lecture Notes

These are the notes of my AP Calculus lectures. The course runs in this order:

$$\text{Preliminaries}\to \text{Limits}\to \text{Differentiation}\to \text{Integration}\to \text{Series of functions (BC)}$$

For the preliminaries we follow the preliminary chapter of Adams & Essex, Calculus: A Complete Course (sections P.1–P.7), together with exponential, logarithmic and inverse functions from its Chapter 3. From limits on we follow Barron's AP Calculus. The examples are the ones we worked through on the board. After each part there is a quiz of 15 multiple-choice questions, in the style of the AP exam.

Part	Topic	Status
P.1	Real numbers, inequalities, absolute value	covered
P.2–P.3	Coordinates, lines, circles, parabolas, ellipses	covered
P.4–P.6	Functions, combining functions, polynomials	covered
P.7	Trigonometric functions	covered
3.1–3.5	Exponential, logarithmic and inverse functions	covered
Unit 1	Limits and continuity	covered
Units 2–10	Derivatives, integrals, differential equations, series	coming later

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P.1 Real Numbers and the Real Line

Number Systems

• Natural numbers: $\mathrm{\mathbb{N}}=\{(0),1,2,3,4,\dots \}$ (some books start at $0$).
• Integers: $\mathrm{\mathbb{Z}}=\{\dots ,-2,-1,0,1,2,\dots \}$.
• Rational numbers: $\mathrm{\mathbb{Q}}=\{{\displaystyle \frac{m}{n}}:m,n\in \mathrm{\mathbb{Z}},n\ne 0\}$.

The rational numbers are dense, but they have gaps. Split $\mathrm{\mathbb{Q}}$ into two sets: $A$, the rationals that are negative or have $x^2<2$, and $B$, the positive rationals with $x^2>2$. Then $A\cup B=\mathrm{\mathbb{Q}}$ and $A\cap B=\varnothing$, but there is no rational number at the place where $A$ ends and $B$ begins: that place is $\sqrt{2}$, which is not rational.

Decimal Expansions

Every real number has a decimal expansion. Every irrational number has an infinite decimal expansion, but not every infinite decimal is irrational: a decimal that repeats is always rational.

Example. Write $x=0.\overline{132}=0.132132\dots$ as a fraction. Multiply by $1000$, since the block $132$ has three digits: $$1000x=132.\overline{132}=132+0.\overline{132}=132+x\Rightarrow 999x=132\Rightarrow x=\frac{132}{999}=\frac{44}{333}.$$

Order in the Real Numbers

$x<y$ means $y-x>0$, and $x\le y$ means $y-x\ge 0$. The rules for working with inequalities:

1. If $x<y$, then $x+c<y+c$ for every real $c$. (Adding or subtracting a number keeps the direction.)
2. If $x<y$ and $c>0$, then $xc<yc$. If $x<y$ and $c<0$, then $xc>yc$: multiplying by a negative number reverses the inequality. For example $1<10$, but $-1>-10$.
3. If $a>0$, then ${\displaystyle \frac{1}{a}}>0$.
4. If $0<a<b$, then ${\displaystyle \frac{1}{b}}<{\displaystyle \frac{1}{a}}$.

Intervals

Notation	Set	Name
$(a,b)$	$a<x<b$	open interval
$[a,b]$	$a\le x\le b$	closed interval
$[a,b)$, $(a,b]$	$a\le x<b$, $a<x\le b$	half-open (half-closed) interval
$(a,\mathrm{\infty })$, $(-\mathrm{\infty },a]$	$x>a$, $x\le a$	infinite intervals
$(-\mathrm{\infty },\mathrm{\infty })$	all of $\mathrm{ℝ}$	the real line

$\mathrm{\infty }$ is not a number, so it never gets a square bracket.

Solving Inequalities

Example (a). $2x+1>x+3$. Subtract $x$, then $1$: $x>2$. Solution: $(2,\mathrm{\infty })$.

Example (b). $-{\displaystyle \frac{x}{3}}\ge 2x-1$. Multiply by $-3$ (reverse!): $x\le -6x+3$, so $7x\le 3$ and $x\le {\displaystyle \frac{3}{7}}$. Solution: $(-\mathrm{\infty },\frac{3}{7}]$.

Example (c). ${\displaystyle \frac{2}{x-1}}\ge 5$. We cannot just multiply by $x-1$, because its sign is unknown. Two cases:

• $x-1>0$: then $2\ge 5(x-1)=5x-5$, so $7\ge 5x$ and $x\le \frac{7}{5}$. Together with $x>1$: $(1,\frac{7}{5}]$.
• $x-1<0$: then the inequality reverses, $2\le 5x-5$, so $x\ge \frac{7}{5}$. This contradicts $x<1$: no solution.

Solution: $(1,\frac{7}{5}]$.

Another solution. Bring everything to one side: $$\frac{2}{x-1}-5\ge 0\Rightarrow \frac{2-5(x-1)}{x-1}\ge 0\Rightarrow \frac{7-5x}{x-1}\ge 0.$$ A quotient is positive when numerator and denominator have the same sign. $7-5x\ge 0$ and $x-1>0$ gives $(1,\frac{7}{5}]$; $7-5x\le 0$ and $x-1<0$ has no solution.

Example (d). $3x-1<5x+3\le 2x+15$. Solve both parts: $$3x-1<5x+3⟺-2<x,5x+3\le 2x+15⟺x\le 4.$$ Both must hold: $(-2,\mathrm{\infty })\cap (-\mathrm{\infty },4]=(-2,4]$.

Quadratic Inequalities

A quadratic is a polynomial of degree two, $a{x}^2+bx+c$ with $a\ne 0$.

Example. $x^2-5x+6<0$. Factor: $(x-3)(x-2)<0$. A product is negative when the two factors have opposite signs:

• $x-3>0$ and $x-2<0$: $x>3$ and $x<2$, no solution.
• $x-3<0$ and $x-2>0$: $2<x<3$.

Solution: $(2,3)$.

A general method. To find the roots of $a{x}^2+bx+c$ we use the discriminant $$\mathrm{\Delta }=b^2-4ac,x_{1,2}=\frac{-b\pm \sqrt{\mathrm{\Delta }}}{2a}.$$

$\mathrm{\Delta }$	Roots
$\mathrm{\Delta }>0$	two different real roots
$\mathrm{\Delta }=0$	one double root: $a{x}^2+bx+c=a(x-x_1{)}^2$
$\mathrm{\Delta }<0$	no real roots

Examples.

• $x^2-5x+6$: $a=1$, $b=-5$, $c=6$, $\mathrm{\Delta }=25-24=1$, so $x_1=\frac{5+1}{2}=3$ and $x_2=\frac{5-1}{2}=2$.
• $2x^2+5x-3$: $\mathrm{\Delta }=5^2-4(2)(-3)=25+24=49$, so $x_{1,2}=\frac{-5\pm 7}{4}$: $x_1=\frac{1}{2}$, $x_2=-3$. Indeed $2x^2+5x-3=(2x-1)(x+3)$.
• $x^2-4x+4$: $\mathrm{\Delta }=16-16=0$, one double root $x=-\frac{b}{2a}=2$; indeed $x^2-4x+4=(x-2{)}^2$.

Proof of the Quadratic Formula

Motivation: $(a\pm b{)}^2=a^2\pm 2ab+b^2$. This lets us complete the square: $$x^2-6x+2=0⟺(x-3{)}^2-9+2=0⟺(x-3{)}^2=7⟺x-3=\pm \sqrt{7}⟺x=3\pm \sqrt{7}.$$

Sign Charts

If $a{x}^2+bx+c$ has two roots $x_1<x_2$, it has the sign of $a$ outside the roots and the opposite sign between them: for $a>0$ the signs are $+-+$, for $a<0$ they are $-+-$.

The same idea works for any polynomial whose roots we know: mark the roots on a line and find the sign on each piece. The sign changes at a root of odd multiplicity and does not change at a root of even multiplicity.

Example. $f(x)=(x-2{)}^2(x-1{)}^3(x+1)$. The roots are $-1$, $1$ (triple) and $2$ (double). For large $x$ everything is positive, so going from right to left the signs are $+$ (after 2), $+$ (no change at the double root 2), $-$ (change at 1), $+$ (change at $-1$).

• $f(x)>0$: $(-\mathrm{\infty },-1)\cup (1,2)\cup (2,\mathrm{\infty })$.
• $f(x)\le 0$: $[-1,1]\cup \{2\}$.

Example. ${\displaystyle \frac{(x-1{)}^3{x}^2}{(x-2{)}^2}}<0$. Roots of the numerator: $0$ (double) and $1$ (triple); $x=2$ is not allowed. The signs, from the left: $-$, $-$ (no change at $0$), $+$, $+$ (no change at $2$).

• $<0$: $(-\mathrm{\infty },0)\cup (0,1)$.
• $\ge 0$: $\{0\}\cup [1,2)\cup (2,\mathrm{\infty })$.

The Absolute Value

$|x|$ is the distance from $x$ to $0$ on the real line; for example $|-3|=(-1)(-3)=3$. More generally, $|x-a|$ is the distance between $x$ and $a$.

Equations and inequalities with $|\cdot |$. For $k>0$:

	means	solution
$|x|=k$	$x=k$ or $x=-k$	two points
$|x|<k$	$-k<x<k$	$(-k,k)$
$|x|>k$	$x>k$ or $x<-k$	$(-\mathrm{\infty },-k)\cup (k,\mathrm{\infty })=\mathrm{ℝ}\setminus [-k,k]$

The same with $\le$ and $\ge$, endpoints included. Replacing $x$ by $x-a$: $$|x-a|<D⟺-D<x-a<D,|x-a|>D⟺x-a>D\text{ or }x-a<-D.$$

Examples.

• $|x|=7$: $x=7$ or $x=-7$.
• $|x|<1$: $(-1,1)$. $|x|>2$: $(-\mathrm{\infty },-2)\cup (2,\mathrm{\infty })$.
• $|x-3|=4$: $x-3=4$ or $x-3=-4$, so $x=7$ or $x=-1$: the two points at distance $4$ from $3$.
• $|x-1|\le 2$: $-2\le x-1\le 2$, so $-1\le x\le 3$: $[-1,3]$.
• $|x-3|>5$: $x>8$ or $x<-2$: $(-\mathrm{\infty },-2)\cup (8,\mathrm{\infty })$.
• $|x-a|=0$ only for $x=a$.

Properties.

• $|a|=|-a|$; for example $|x-y|=|y-x|$ and $|x^2-3x+y+2|=|-x^2+3x-y-2|$.
• $|ab|=|a||b|$; for example $|(x-2)(x-3)|=|x-2||x-3|$.
• $\left|{\displaystyle \frac{a}{b}}\right|={\displaystyle \frac{|a|}{|b|}}$ for $b\ne 0$.

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Quiz 1: Real Numbers, Inequalities and Absolute Value

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P.2 Cartesian Coordinates in the Plane

The two axes divide the plane into four quadrants, numbered I to IV counterclockwise starting from the upper right. The point $(a,b)$ is $a$ units along the $x$-axis and $b$ units along the $y$-axis.

The Distance Formula

The distance between $P_1(x_1,y_1)$ and $P_2(x_2,y_2)$ is the hypotenuse of a right triangle with legs $\mathrm{\Delta }x=x_2-x_1$ and $\mathrm{\Delta }y=y_2-y_1$, so by Pythagoras $$d(P_1,P_2)=\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^2}.$$

Example. $P_1(3,4)$, $P_2(7,8)$: $d=\sqrt{(7-3{)}^2+(8-4{)}^2}=\sqrt{32}=4\sqrt{2}$.

The Equation of a Circle

A point $(x,y)$ is on the circle with centre $(0,0)$ and radius $1$ exactly when its distance to the centre is $1$: $$\sqrt{(x-0{)}^2+(y-0{)}^2}=1⟺x^2+y^2=1\text{(the unit circle)}.$$ In general, the circle with centre $(a,b)$ and radius $r$ is $$(x-a{)}^2+(y-b{)}^2=r^2.$$

Examples.

• $(x+2{)}^2+(y-1{)}^2=4$ is the circle with centre $(-2,1)$ and radius $2$.
• $x^2-2x+y^2+4y=20$. Complete the squares: $(x-1{)}^2-1+(y+2{)}^2-4=20$, so $(x-1{)}^2+(y+2{)}^2=25=5^2$: the circle with centre $(1,-2)$ and radius $5$.

The Equation of a Parabola

$y=x^2$ is the basic parabola, with its vertex at the origin. Moving the vertex to $(a,b)$ gives $$y-b=(x-a{)}^2,\text{that is,}y=(x-a{)}^2+b.$$

Example. $y=x^2+2x+4=(x+1{)}^2-1+4=(x+1{)}^2+3$, so $y-3=(x+1{)}^2$: a parabola with vertex $(-1,3)$.

Straight Lines

The slope of a line is $$m=\frac{\mathrm{\Delta }y}{\mathrm{\Delta }x}=\frac{\text{rise}}{\text{run}}=\tan \theta ,$$ where $\theta$ is the angle the line makes with the positive $x$-axis. A line rising $2$ units for every $1$ unit to the right has $m=2$; a line falling $1$ unit for every $1$ unit has $m=-1$.

Point-slope form. If the line has slope $m$ and passes through $P_1(x_1,y_1)$, then for any other point $(x,y)$ on it $$m=\frac{y-y_1}{x-x_1}\Rightarrow y-y_1=m(x-x_1)\Rightarrow y=m(x-x_1)+y_1.$$

Example. Through $(1,7)$ with slope $2$: $y-7=2(x-1)$, so $y=2x+5$.

Two-point form. Through $P_1(x_1,y_1)$ and $P_2(x_2,y_2)$: first the slope $m={\displaystyle \frac{y_2-y_1}{x_2-x_1}}$, then the point-slope form with either point. Both points give the same equation.

Example. Through $(1,7)$ and $(3,1)$: $m=\frac{1-7}{3-1}=-3$. With $(1,7)$: $y-7=-3(x-1)$, so $y=-3x+10$. With $(3,1)$: $y-1=-3(x-3)$, again $y=-3x+10$.

Perpendicular lines. Two lines with slopes $m_1$ and $m_2$ are perpendicular when $$m_1\cdot m_2=-1.$$

Example. The line through $(1,4)$ perpendicular to $y=3x+1$. Here $m_1=3$, so $m_2=-\frac{1}{3}$ and $y-4=-\frac{1}{3}(x-1)$, that is, $y=-{\displaystyle \frac{x}{3}}+{\displaystyle \frac{13}{3}}$.

Intercepts. The $y$-intercept is where the line meets the $y$-axis (put $x=0$); the $x$-intercept is where it meets the $x$-axis (put $y=0$). For $y=3x+1$: $x=0$ gives $y=1$, and $y=0$ gives $x=-\frac{1}{3}$. So the line passes through $(0,1)$ and $(-\frac{1}{3},0)$.

Intercept form. The line with $x$-intercept $a$ and $y$-intercept $b$ (through $(a,0)$ and $(0,b)$) is $$\frac{x}{a}+\frac{y}{b}=1.$$

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P.3 Graphs of Quadratic Equations

Circles and Disks

The circle $(x-a{)}^2+(y-b{)}^2=r^2$ has centre $(a,b)$ and radius $r$. An equality describes a curve, an inequality describes a region:

• $x^2+y^2=0$ is only the point $(0,0)$.
• $x^2+y^2=4$ is the circle of radius $2$.
• $\sqrt{x^2+y^2}\le 2$, that is, $x^2+y^2\le 4$, is the closed disk of radius $2$: the circle together with its inside.
• $x^2+y^2<4$ is the open disk (the inside only).

On the circle $x^2+y^2=4$ lie, for example, $(2,0)$, $(0,2)$, $(-2,0)$ and $(1,\sqrt{3})$.

Shifting a Graph

Replace	by	The graph moves
$x$	$x-a$ ($a>0$)	$a$ units to the right
$x$	$x+a$ ($a>0$)	$a$ units to the left
$y$	$y-b$ ($b>0$)	$b$ units up
$y$	$y+b$ ($b>0$)	$b$ units down

Examples.

• $x^2+y^2=1\to (x-1{)}^2+y^2=1$: one unit to the right. $\to (x-1{)}^2+(y+2{)}^2=1$: then two units down.
• $y=x\to y=x-3$: three units to the right.
• $y=x^2\to y-2=(x+1{)}^2$: one unit to the left and two units up; the vertex moves to $(-1,2)$.

Scaling a Graph

Multiplying the function by a number stretches or squeezes the graph vertically: $y=2x^2$ is narrower than $y=x^2$, $y=\frac{1}{2}{x}^2$ is wider, and $y=-x^2$ is $y=x^2$ reflected in the $x$-axis.

Ellipses

We can write the equation of a circle as $$x^2+y^2=9⟺\frac{x^2}{3^2}+\frac{y^2}{3^2}=1.$$ Changing one of the two denominators squeezes the circle into an ellipse: $$\frac{x^2}{3^2}+\frac{y^2}{2^2}=1$$ still meets the $x$-axis at $\pm 3$, but meets the $y$-axis at $\pm 2$. In general, $$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$$ is the ellipse through $(\pm a,0)$ and $(0,\pm b)$.

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Quiz 2: Coordinates, Lines and Graphs

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P.4 Functions and Their Graphs

Definition of a Function

Think of a function as a machine: $x$ goes in (from the domain), $y=f(x)$ comes out (into the range). In calculus we study functions on numbers, mostly on the real numbers $\mathrm{ℝ}$.

Examples of functions.

1. $f(x)=1$: a constant function. Its graph is the horizontal line $y=1$.
2. $f(x)=x^2$.
3. $x^2+y^2=4$ is not a function of $x$: for $x=1$ we get both $y=\sqrt{3}$ and $y=-\sqrt{3}$.
4. Motion along a line: the position at time $t$ is $x(t)=3t^2$. Its velocity is $v(t)=6t$ and its acceleration is $a(t)=6$ (we will see why when we study derivatives).
5. The area of a rectangle with sides $d$ and $2d$: $A(d)=2d^2$.
6. The volume of a cone of height $10$ and base radius $r$: $V(r)=\frac{1}{3}\pi r^2\cdot 10$.

The Vertical Line Test

The Domain Convention

Two functions are equal if (1) their rules are the same and (2) their domains are the same. For example, $$f:[0,\mathrm{\infty })\to [0,\mathrm{\infty }),f(x)=x^2\text{and}g:\mathrm{ℝ}\to [0,\mathrm{\infty }),g(x)=x^2$$ are different functions: the graph of $f$ is only the right half of the parabola.

Finding the maximal domain of a function is tested on the AP exam. Watch out for square roots (the inside must be $\ge 0$) and denominators (they must be $\ne 0$).

• $f(x)=\sqrt{x}$: $D(f)=[0,\mathrm{\infty })$.
• $f(x)={\displaystyle \frac{x^2}{x-1}}$: $D(f)=\mathrm{ℝ}\setminus \{1\}$.

Graphs of Basic Functions

You should know these graphs by heart; most of the functions in the course are built from them.

• $y=x^2$ looks like every even power $y=x^{2n}$, and $y=x^3$ like every odd power $y=x^{2n+1}$ (flatter near $0$, steeper far away).
• $y=\frac{1}{x^2}$ looks like $y=\frac{1}{x^{2n}}$, and $y=\frac{1}{x}$ like $y=\frac{1}{x^{2n+1}}$.
• The graph of $y=-\sqrt{1-x^2}$ is the lower half of the unit circle, and $y=\sqrt{1-x^2}$ the upper half.

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P.5 Combining Functions

Sums, Differences, Products and Quotients

Example. $f(x)=\sqrt{x}$, $g(x)=x^2-1$: $$(f\cdot g)(x)=\sqrt{x}(x^2-1),(f+g)(x)=\sqrt{x}+x^2-1,\left(\frac{f}{g}\right)(x)=\frac{\sqrt{x}}{x^2-1}.$$ $D(f)=[0,\mathrm{\infty })$ and $g(x)=0$ at $x=\pm 1$, so $D(f/g)=[0,\mathrm{\infty })\setminus \{1\}=[0,1)\cup (1,\mathrm{\infty })$.

Recall: the domain (tanım kümesi) is the set of inputs we may use; the range (değer kümesi) is the set of values the function actually takes.

• $f(x)=|x|$: domain $\mathrm{ℝ}$, range $[0,\mathrm{\infty })$.
• $f(x)=\sqrt{x}$: domain $[0,\mathrm{\infty })$, range $[0,\mathrm{\infty })$.
• $f(x)=\sqrt{x-3}$: domain $[3,\mathrm{\infty })$, range $[0,\mathrm{\infty })$.

Composite Functions

Example. $f(x)=\sqrt{x}$, $g(x)=x+1$:

Function	Formula	Domain
$f\circ g$	$f(x+1)=\sqrt{x+1}$	$[-1,\mathrm{\infty })$
$g\circ f$	$g(\sqrt{x})=\sqrt{x}+1$	$[0,\mathrm{\infty })$
$f\circ f$	$\sqrt{\sqrt{x}}=x^{1/4}$	$[0,\mathrm{\infty })$
$g\circ g$	$(x+1)+1=x+2$	$\mathrm{ℝ}$

$g(x)=x+1$ is defined for all $x$, but $g(x)$ is in the domain of $f$ only if $x+1\ge 0$, that is, $x\ge -1$. Note that $f\circ g\ne g\circ f$ in general.

Piecewise Defined Functions

Sometimes a function is given by different formulas on different parts of its domain. The absolute value is one example: $$|x|=\{\begin{array}{ll}x,& x\ge 0,\\ -x,& x<0.\end{array}$$

Example. $f(x)=\{\begin{array}{ll}x-1,& x\le 1,\\ x^2-1,& x>1.\end{array}$ Then $f(3)=3^2-1=8$ and $f(0)=0-1=-1$.

Example. $f(x)=\{\begin{array}{ll}3,& x<-1,\\ -x,& -1\le x<2,\\ \sqrt{x-2},& x\ge 2.\end{array}$

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P.6 Polynomials and Rational Functions

Polynomials

Examples. $p(x)=1+2x+x^2$ has degree $2$; $p(x)=3$ and $p(x)=\pi$ have degree $0$; $p(x)=x+3$ has degree $1$.

Division of Polynomials

Polynomials can be divided like numbers, with a quotient and a remainder of lower degree than the divisor.

Example. Divide $2x^3-3x^2+3x+4$ by $x^2+1$. At each step, divide the leading term by $x^2$ and subtract:

1. ${\displaystyle \frac{2x^3}{x^2}}=2x$: $$(2x^3-3x^2+3x+4)-2x(x^2+1)=-3x^2+x+4.$$
2. ${\displaystyle \frac{-3x^2}{x^2}}=-3$: $(-3x^2+x+4)-(-3)(x^2+1)=x+7$.
3. $x+7$ has lower degree than $x^2+1$, so we stop.

So the quotient is $2x-3$ and the remainder is $x+7$: $$2x^3-3x^2+3x+4=(x^2+1)(2x-3)+(x+7),\frac{2x^3-3x^2+3x+4}{x^2+1}=2x-3+\frac{x+7}{x^2+1}.$$ Writing a rational function this way will be useful later, for limits at infinity and for integration.

Roots, Zeros and Factors

Examples. $p(x)=x^2-2x+1$ has the root $1$, since $p(1)=0$. $p(x)=x^3-4x$ has the root $0$ (and also $2$ and $-2$).

If $(x-r{)}^k$ is a factor of $P$ but $(x-r{)}^{k+1}$ is not, $r$ is a root of multiplicity $k$.

Example. $f(x)=x^3(x-1{)}^2(x+4{)}^4(x-3)$ has the roots $0,1,-4,3$ with multiplicities $3,2,4,1$.

How to Find Roots

1. Degree 1: $ax+b=0$ gives $x=-{\displaystyle \frac{b}{a}}$.
2. Degree 2: $a{x}^2+bx+c$, with $\mathrm{\Delta }=b^2-4ac$:
   • $\mathrm{\Delta }>0$: two real roots $x_{1,2}={\displaystyle \frac{-b\pm \sqrt{\mathrm{\Delta }}}{2a}}$;
   • $\mathrm{\Delta }=0$: one real root $x_1=x_2=-{\displaystyle \frac{b}{2a}}$ of multiplicity $2$;
   • $\mathrm{\Delta }<0$: no real roots.
3. Degree 3 and higher: there are formulas for degrees 3 and 4, but we do not use them. (The discriminant of the cubic $a{x}^3+b{x}^2+cx+d$ is $\mathrm{\Delta }=18abcd-4b^{3}d+b^2{c}^2-4a{c}^3-27a^2{d}^2$; it tells the nature of the roots.) Instead, try small integers $0,\pm 1,\pm 2,\pm 3,\dots$; once a root $r$ is found, divide by $x-r$ and continue with a polynomial of lower degree.

Examples.

• $x^2-2x+1$: $\mathrm{\Delta }=4-4=0$, $x_1=x_2=1$, and $x^2-2x+1=(x-1{)}^2$.
• If the roots are rational, we may find them by factoring: $x^2-7x+6=(x-6)(x-1)$, roots $6$ and $1$; $2x^2-3x-5=(2x-5)(x+1)$, roots $\frac{5}{2}$ and $-1$.
• $3x^2-16x+28$: $\mathrm{\Delta }=256-336=-80<0$, no real roots.
• $3x^2-16x+5$: $\mathrm{\Delta }=256-60=196$, so $x_{1,2}={\displaystyle \frac{16\pm 14}{6}}$: $x_1=5$, $x_2=\frac{1}{3}$. Indeed $3x^2-16x+5=(3x-1)(x-5)$.
• $p(x)=x^3+6x^2+11x+6$. Try $x=-1$: $p(-1)=-1+6-11+6=0$, so $x+1$ is a factor. Dividing, $p(x)=(x+1)(x^2+5x+6)=(x+1)(x+2)(x+3)$, and the roots are $-1,-2,-3$.

Factorization Rules

• $a^2-b^2=(a-b)(a+b)$. For example $x^2-2=(x-\sqrt{2})(x+\sqrt{2})$. This works not only for polynomials: $x-4=(\sqrt{x}-2)(\sqrt{x}+2)$, so ${\displaystyle \frac{x-4}{\sqrt{x}+2}}=\sqrt{x}-2$.
• $x^3-a^3=(x-a)(x^2+xa+a^2)$.
• In general, $$a^n-b^n=(a-b)(a^{n-1}+a^{n-2}b+a^{n-3}{b}^2+\cdots +a{b}^{n-2}+b^{n-1}).$$
• $a^3+b^3=(a+b)(a^2-ab+b^2)$; for odd $n$, also $x^5+b^5=(x+b)(x^4-x^{3}b+x^2{b}^2-x{b}^3+b^4)$.

The Binomial Expansion

$$(x+a{)}^2=x^2+2xa+a^2,(x+a{)}^3=x^3+3x^{2}a+3x{a}^2+a^3.$$ For example $(x+1{)}^2=x^2+2x+1$ and $(x+1{)}^3=x^3+3x^2+3x+1$. In general $$(a+b{)}^n=(\genfrac{}{}{0ex}{}{n}{n})a^n+(\genfrac{}{}{0ex}{}{n}{n-1})a^{n-1}b+(\genfrac{}{}{0ex}{}{n}{n-2})a^{n-2}{b}^2+\cdots +(\genfrac{}{}{0ex}{}{n}{1})a{b}^{n-1}+(\genfrac{}{}{0ex}{}{n}{0})b^n,$$ where ${\displaystyle (\genfrac{}{}{0ex}{}{n}{k})}$ is the number of combinations of $n$ things taken $k$ at a time: $$(\genfrac{}{}{0ex}{}{n}{k})=\frac{n!}{k!(n-k)!},(\genfrac{}{}{0ex}{}{n}{n})=(\genfrac{}{}{0ex}{}{n}{0})=1,(\genfrac{}{}{0ex}{}{n}{n-1})=(\genfrac{}{}{0ex}{}{n}{1})=n,(\genfrac{}{}{0ex}{}{n}{a})=(\genfrac{}{}{0ex}{}{n}{b})\text{ if }a+b=n.$$ The coefficients can also be read off Pascal's triangle, where each number is the sum of the two above it: $$\begin{array}{c}1\\ 11\\ 121\\ 1331\\ 14641\\ 15101051\\ 1615201561\end{array}$$

Example. $$(x+2{)}^5=x^5+5x^4\cdot 2+10x^3\cdot 2^2+10x^2\cdot 2^3+5x\cdot 2^4+2^5=x^5+10x^4+40x^3+80x^2+80x+32.$$

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Quiz 3: Functions and Polynomials

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P.7 Trigonometric Functions

Trigonometric Ratios

In a right triangle with an acute angle $\alpha$: $$\cos \alpha =\frac{\text{adjacent}}{\text{hypotenuse}},\sin \alpha =\frac{\text{opposite}}{\text{hypotenuse}},\tan \alpha =\frac{\text{opposite}}{\text{adjacent}},\cot \alpha =\frac{\text{adjacent}}{\text{opposite}}.$$

Special Triangles

From these two triangles:

	$0^{\circ }$	${30}^{\circ }$	${45}^{\circ }$	${60}^{\circ }$	${90}^{\circ }$
$\sin$	$0$	$\frac{1}{2}$	$\frac{1}{\sqrt{2}}$	$\frac{\sqrt{3}}{2}$	$1$
$\cos$	$1$	$\frac{\sqrt{3}}{2}$	$\frac{1}{\sqrt{2}}$	$\frac{1}{2}$	$0$
$\tan$	$0$	$\frac{1}{\sqrt{3}}$	$1$	$\sqrt{3}$	undefined
$\cot$	undefined	$\sqrt{3}$	$1$	$\frac{1}{\sqrt{3}}$	$0$

If $a+b={90}^{\circ }$, then $\cos a=\sin b$: the cosine of an angle is the sine of its complement.

Radians

${360}^{\circ }=2\pi$ radians, so $$\text{radians}=\text{degrees}\cdot \frac{\pi }{180},\text{degrees}=\text{radians}\cdot \frac{180}{\pi }.$$ For example ${30}^{\circ }=\frac{\pi }{6}$, ${45}^{\circ }=\frac{\pi }{4}$, ${60}^{\circ }=\frac{\pi }{3}$, ${90}^{\circ }=\frac{\pi }{2}$, ${180}^{\circ }=\pi$. In calculus, angles are always in radians.

In physics, a velocity $v$ at angle $\theta$ has the components $v_x=v\cos \theta$ and $v_y=v\sin \theta$.

The Unit Circle

Since the point is on the unit circle, ${\sin }^{2}x+{\cos }^{2}x=1$.

Examples.

• $\cos {120}^{\circ }$: the point at ${120}^{\circ }$ makes ${60}^{\circ }$ with the negative $x$-axis, so it is $(-\frac{1}{2},\frac{\sqrt{3}}{2})$. Thus $\cos {120}^{\circ }=-\frac{1}{2}$ and $\sin {120}^{\circ }=\frac{\sqrt{3}}{2}$.
• $\sin (-{135}^{\circ })$: ${135}^{\circ }={90}^{\circ }+{45}^{\circ }$, so $-{135}^{\circ }=-\frac{3\pi }{4}$ ends in the third quadrant at $(-\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}})$. Thus $\sin (-{135}^{\circ })=-\frac{1}{\sqrt{2}}$ and $\cos (-{135}^{\circ })=-\frac{1}{\sqrt{2}}$.

Important Trigonometric Identities

• If $a+b=\frac{\pi }{2}$, then $\cos a=\sin b$.
• $\cos (-x)=\cos x$: cosine is an even function. $\sin (-x)=-\sin x$: sine is an odd function.
• ${\sin }^{2}x+{\cos }^{2}x=1$.
• Addition formulas: $$\sin (a+b)=\sin a\cos b+\cos a\sin b,\cos (a+b)=\cos a\cos b-\sin a\sin b.$$
• Double-angle formulas: $$\sin 2x=2\sin x\cos x,\cos 2x={\cos }^{2}x-{\sin }^{2}x.$$
• Half-angle formulas (from ${\cos }^{2}x-{\sin }^{2}x=\cos 2x$ and ${\cos }^{2}x+{\sin }^{2}x=1$, adding or subtracting): $${\cos }^{2}x=\frac{1+\cos 2x}{2},{\sin }^{2}x=\frac{1-\cos 2x}{2}.$$
• $-1\le \sin x\le 1$ and $-1\le \cos x\le 1$: the range of both is $[-1,1]$, and the domain of both is $(-\mathrm{\infty },\mathrm{\infty })$.

Graphs of the Trigonometric Functions

$\sin x$ and $\cos x$ are periodic with period $2\pi$. $\tan x$ has period $\pi$ and is undefined where $\cos x=0$, at $x=\frac{\pi }{2}+k\pi$; there its graph has vertical asymptotes.

The Sine and Cosine Laws

In a triangle $ABC$ with sides $a$, $b$, $c$ opposite the angles $A$, $B$, $C$: $$\text{Sine law:}\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}.$$ $$\text{Cosine law:}{a}^2=b^2+c^2-2bc\cos A,b^2=a^2+c^2-2ac\cos B,c^2=a^2+b^2-2ab\cos C.$$

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Exponential and Logarithmic Functions

Exponential Functions

For example $3^{7/4}=\sqrt[4]{3^7}$. Irrational exponents are approximated by rational ones: $3^{1.7},3^{1.73},3^{1.732},\dots$ approach $3^{\sqrt{3}}$; in the same way $3^{\pi }$ makes sense. This is why we need $a>0$: $(-2{)}^{7/2}=\sqrt{(-2{)}^7}$ is not a real number.

The exponential function with base $a$ is $f(x)=a^x$ ($a>0$), defined for all $x\in (-\mathrm{\infty },\mathrm{\infty })$.

The graph of $y=a^x$ always passes through $(0,1)$. For $a>1$ it is increasing, for $0<a<1$ it is decreasing, and for $a=1$ it is the constant line $y=1$. The larger $a$ is, the steeper the graph for $x>0$.

Logarithmic Functions

For example ${\log }_{3}9=2$ because $3^2=9$, and ${\log }_{2}1=0$ because $2^0=1$.

$\ln e=1$, $\ln e^2=2$, $\ln \frac{1}{e}=-1$. (Here $e\approx 2.7$ and $\pi \approx 3.1$; $\ln \pi$ is a little more than $1$.)

$a^x$ and ${\log }_{a}x$ are inverse functions of each other: $$a^{{\log }_{a}x}=x(x>0),{\log }_a{a}^x=x,e^{\ln x}=x,\ln e^x=x.$$

Some of these follow from the others:

• (3) follows from (5): ${\log }_a\frac{1}{x}={\log }_a{x}^{-1}=-{\log }_{a}x$.
• (4) follows from (2) and (3): ${\log }_a\frac{x}{y}={\log }_a(x\cdot y^{-1})={\log }_{a}x+{\log }_a{y}^{-1}={\log }_{a}x-{\log }_{a}y$.

By change of base, any logarithm can be computed with $\ln$ or $\log ={\log }_{10}$: $${\log }_{a}x=\frac{\ln x}{\ln a}=\frac{\log x}{\log a}=\frac{{\log }_{2}x}{{\log }_{2}a}.$$

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Inverse Functions

One-to-One Functions

The function $f(x)=x^3$ takes only one value for each $x$, as every function does. But it also has a second property: different values of $x$ always give different values $f(x)$. Geometrically, every horizontal line meets its graph at most once.

Examples.

• $f(x)=x^2$ is not one-to-one: $f(-1)=f(1)=1$ but $-1\ne 1$. From the definition: $f(a)=f(b)$ means $a^2=b^2$, that is, $|a|=|b|$, so $a=b$ or $a=-b$.
• $f(x)=x^3$ is one-to-one: $f(a)=f(b)⟺a^3=b^3⟺a=b$.
• $f(x)=\sin x$ is not one-to-one: horizontal lines between $-1$ and $1$ meet its graph infinitely often.

The Inverse Function

Careful: $f^{-1}$ is not ${\displaystyle \frac{1}{f}}$.

How to find the inverse. Write $y=f(x)$ and solve for $x$.

Example. $f(x)=2x+1$. From $y=2x+1$ we get $x={\displaystyle \frac{y-1}{2}}$, so $$f^{-1}(x)=\frac{x-1}{2}.$$ $f$ sends $1\mapsto 3$ and $0\mapsto 1$; $f^{-1}$ sends them back: $f^{-1}(3)=1$ and $f^{-1}(1)=0$.

For any inverse function, $$(f\circ f^{-1})(x)=x\text{and}(f^{-1}\circ f)(x)=x.$$ In the example: $f(f^{-1}(x))=2\cdot \frac{x-1}{2}+1=x$ and $f^{-1}(f(x))=\frac{(2x+1)-1}{2}=x$.

Restricting the Domain to Get an Inverse

$f(x)=x^2$ on $\mathrm{ℝ}$ has no inverse. But the function $F(x)=x^2$ with domain $[0,\mathrm{\infty })$ (the right half of the parabola) is one-to-one, so it has an inverse: from $y=x^2$, $x\ge 0$, we get $x=\sqrt{y}$, and $F^{-1}(x)=\sqrt{x}$. Note that $f\ne F$, because their domains differ.

The graph of $f^{-1}$ is the graph of $f$ reflected in the line $y=x$. In the same way $\ln x$ and $e^x$ are inverses of each other.

Inverse Trigonometric Functions

The inverse sine. $\sin x$ is not one-to-one. We keep only the part of its graph on $[-\frac{\pi }{2},\frac{\pi }{2}]$: there it takes every value between $-1$ and $1$ exactly once. This restricted function is written $\mathrm{Sin}x$.

Examples.

• $\arcsin \frac{1}{2}={\sin }^{-1}\frac{1}{2}=\frac{\pi }{6}$, because $\sin \frac{\pi }{6}=\frac{1}{2}$ and $\frac{\pi }{6}$ is in $[-\frac{\pi }{2},\frac{\pi }{2}]$.
• $\arcsin (-\frac{\sqrt{3}}{2})=-\frac{\pi }{3}$.
• ${\sin }^{-1}(-\frac{1}{\sqrt{2}})=-\frac{\pi }{4}$ and ${\sin }^{-1}(-1)=-\frac{\pi }{2}$.
• ${\sin }^{-1}2$ is not defined: $2$ is not in the range of sine.

The inverse cosine. Similarly we restrict $\cos x$ to $[0,\pi ]$.

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Quiz 4: Trigonometric, Exponential, Logarithmic and Inverse Functions

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Limits

Definition and First Examples

$x$ goes to $c$ but is never equal to $c$: the value $f(c)$ itself plays no role, and $f$ need not even be defined at $c$.

One-sided limits. $x$ can approach $c$ from the left (through values less than $c$) or from the right (through values greater than $c$): $$\underset{x\to c^{-}}{lim}f(x)\text{(left-hand limit)},\underset{x\to c^{+}}{lim}f(x)\text{(right-hand limit)}.$$

Example. $\underset{x\to 3}{lim}(x+1)=4$.

Example. $f(x)={\displaystyle \frac{x^2-3x+2}{x-1}}$ is not defined at $x=1$. But for $x\ne 1$, $$f(x)=\frac{(x-1)(x-2)}{x-1}=x-2,\text{so}\underset{x\to 1}{lim}f(x)=\underset{x\to 1}{lim}(x-2)=-1.$$ A table confirms it: $f(0.9)=-1.1$, $f(0.99)=-1.01$, $f(0.999)=-1.001$. Far from $1$ the values are different: $f(3)=1$, $f(1.5)=-0.5$.

Example. In the graph below, $\underset{x\to 4^{-}}{lim}f(x)=3\ne 5=\underset{x\to 4^{+}}{lim}f(x)$, so $\underset{x\to 4}{lim}f(x)$ does not exist.

The Greatest Integer Function

For example $[1.7]=1$, $[2]=2$, $[-3.7]=-4$. Its graph is a staircase (a step function). At every integer the left and right limits differ; for example $$\underset{x\to 1^{-}}{lim}[x]=0\ne 1=\underset{x\to 1^{+}}{lim}[x],\text{so}\underset{x\to 1}{lim}[x]\text{ does not exist (DNE)}.$$

Constant functions. For $f(x)=k$ the limit is always $k$: $\underset{x\to c}{lim}k=k$. For example $\underset{x\to 8}{lim}3=3$, $\underset{x\to e}{lim}{\pi }^2={\pi }^2$, $\underset{x\to \pi /4}{lim}\sqrt{e}=\sqrt{e}$.

A Worked Example

$$f(x)=\{\begin{array}{ll}x+1,& -2<x<0,\\ 2,& x=0,\\ -x,& 0<x<2,\\ 0,& x=2,\\ x-4,& 2<x\le 4.\end{array}$$ Does $f$ have a limit at (a) $x=-2$, (b) $x=0$, (c) $x=2$, (d) $x=4$?

The domain is $(-2,4]$, and its boundary points are $-2$ and $4$. At a boundary point we check only the one-sided limit from inside the domain.

• (a) $\underset{x\to -2^{+}}{lim}f(x)=-1$, so the limit exists and equals $-1$.
• (b) $\underset{x\to 0^{-}}{lim}f(x)=1\ne 0=\underset{x\to 0^{+}}{lim}f(x)$: the limit does not exist.
• (c) $\underset{x\to 2^{-}}{lim}f(x)=-2=\underset{x\to 2^{+}}{lim}f(x)$, so $\underset{x\to 2}{lim}f(x)=-2$ (although $f(2)=0$).
• (d) $x=4$ is a boundary point: $\underset{x\to 4^{-}}{lim}f(x)=0$, so $\underset{x\to 4}{lim}f(x)=0$.

More Examples

Example. Prove that $\underset{x\to 0}{lim}|x|=0$. The domain of $|x|$ is $(-\mathrm{\infty },\mathrm{\infty })$ and $$|x|=\{\begin{array}{ll}-x,& x\le 0,\\ x,& x>0,\end{array}\underset{x\to 0^{-}}{lim}(-x)=0=\underset{x\to 0^{+}}{lim}x.$$ Since the left and right limits are equal, $\underset{x\to 0}{lim}|x|=0$.

Example. What is the domain of $f(x)=\sqrt{\sin x}$? We need $\sin x\ge 0$: $$D(f)=\cdots \cup [-2\pi ,-\pi ]\cup [0,\pi ]\cup [2\pi ,3\pi ]\cup \cdots =\bigcup _{k\in \mathrm{\mathbb{Z}}}[2k\pi ,\pi +2k\pi ].$$

Example. For $f(x)=x^2\sin (2x)$, $\underset{x\to 2}{lim}f(x)=4\sin 4\approx -3.03$ (just substitute; $\sin 4$ in radians).

Example. Let $f(x)=\{\begin{array}{ll}{x}^2+a^2,& x\le 1,\\ 5,& x>1.\end{array}$ If $f$ has a limit at $x=1$, what values can $a$ take?

The two one-sided limits must be equal: $$\underset{x\to 1^{-}}{lim}(x^2+a^2)=1+a^2,\underset{x\to 1^{+}}{lim}5=5\Rightarrow 1+a^2=5\Rightarrow a^2=4\Rightarrow a=\pm 2.$$ For example, for $a=2$, $f(x)=x^2+4$ for $x\le 1$ and $5$ for $x>1$.

The Formal Definition of a Limit

$\epsilon$ and $\delta$ are small distances. However small an $\epsilon$ we are given, we must find a $\delta$ such that every $x$ within $\delta$ of $a$ (except $a$ itself) has $f(x)$ within $\epsilon$ of $L$. In the picture: the horizontal band $(L-\epsilon ,L+\epsilon )$ is given; we must find a vertical band $(a-\delta ,a+\delta )$ such that the piece of the graph inside the vertical band stays inside the horizontal band.

Here $|x^2-4|<1$ means $\sqrt{3}<x<\sqrt{5}$; the distance from $2$ to the nearer end is $min(2-\sqrt{3},\sqrt{5}-2)=\sqrt{5}-2\approx 0.236$. Any larger $\delta$ fails. Try it yourself with the epsilon-delta tool.

Infinite Limits

Since $\pm \mathrm{\infty }$ is not a number, the limit does not exist in these cases; writing $=\mathrm{\infty }$ says how it fails to exist.

Example. $f(x)={\displaystyle \frac{1}{x}}$. Some investigation: $$\frac{1}{0.1}=10,\frac{1}{0.01}=100,\frac{1}{0.0001}=10000;\frac{1}{-0.1}=-10,\frac{1}{-0.01}=-100,\frac{1}{-0.0001}=-10000.$$ So $\underset{x\to 0^{+}}{lim}{\displaystyle \frac{1}{x}}=+\mathrm{\infty }$ and $\underset{x\to 0^{-}}{lim}{\displaystyle \frac{1}{x}}=-\mathrm{\infty }$; the one-sided limits are different, and $\underset{x\to 0}{lim}{\displaystyle \frac{1}{x}}$ does not exist (we cannot even write $\pm \mathrm{\infty }$).

Examples.

• $\underset{x\to 1}{lim}{\displaystyle \frac{4}{x-1}}$: check the one-sided limits. $\underset{x\to 1^{+}}{lim}{\displaystyle \frac{4}{x-1}}={\displaystyle \frac{4}{0^{+}}}=+\mathrm{\infty }$ and $\underset{x\to 1^{-}}{lim}{\displaystyle \frac{4}{x-1}}={\displaystyle \frac{4}{0^{-}}}=-\mathrm{\infty }$. Hence the limit does not exist.
• $\underset{x\to 4}{lim}{\displaystyle \frac{2}{(x-4{)}^2}}$: from both sides $(x-4{)}^2\to 0^{+}$, so both one-sided limits are $+\mathrm{\infty }$. The limit still does not exist, but we can write $\underset{x\to 4}{lim}{\displaystyle \frac{2}{(x-4{)}^2}}=+\mathrm{\infty }$.
• $\underset{x\to -1}{lim}{\displaystyle \frac{-3}{(x+1{)}^3}}$: $x+1$ is negative to the left of $-1$ and positive to the right, and so is its cube. So $\underset{x\to -1^{-}}{lim}={\displaystyle \frac{-3}{0^{-}}}=+\mathrm{\infty }$ and $\underset{x\to -1^{+}}{lim}={\displaystyle \frac{-3}{0^{+}}}=-\mathrm{\infty }$.
• $g(x)={\displaystyle \frac{1}{(x-1{)}^2}}$ near $x=1$: $\underset{x\to 1}{lim}g(x)=+\mathrm{\infty }$.
• $\underset{x\to \pi /2}{lim}{\displaystyle \frac{1}{\sin (x-\pi /2)}}$: as $x\to {\frac{\pi }{2}}^{+}$, $x-\frac{\pi }{2}\to 0^{+}$ and $\sin \to 0^{+}$, so the limit is $+\mathrm{\infty }$; as $x\to {\frac{\pi }{2}}^{-}$ it is $-\mathrm{\infty }$.
• $\underset{x\to 0}{lim}{\displaystyle \frac{\cos x-4}{1-\cos x}}$: the numerator goes to $-3$, and $1-\cos x\to 0^{+}$ from both sides (since $\cos x\le 1$). So both one-sided limits are ${\displaystyle \frac{-3}{0^{+}}}=-\mathrm{\infty }$.
• $\underset{x\to \pi /2}{lim}{\displaystyle \frac{\sin x-3}{\cos x}}$: the numerator goes to $-2$. From the left $\cos x\to 0^{+}$, giving $-\mathrm{\infty }$; from the right $\cos x\to 0^{-}$, giving $+\mathrm{\infty }$.
• $\underset{x\to 0}{lim}{\displaystyle \frac{\cos (e^x-1)}{\sin (e^x-1)}}$: the numerator goes to $\cos 0=1$. For $x\to 0^{-}$, $e^x<1$, so $e^x-1\to 0^{-}$ and $\sin (e^x-1)\to 0^{-}$: the limit is $-\mathrm{\infty }$. For $x\to 0^{+}$ it is $+\mathrm{\infty }$.

Tip. If you can see that the limit is infinite but are not sure of the sign, use your calculator (in radians): compute $f$ at a value very close to $a$, once from the left and once from the right.

Limits at Infinity

Example. $f(x)={\displaystyle \frac{1}{x}}$: ${\displaystyle \frac{1}{10}}=0.1$, ${\displaystyle \frac{1}{100}}=0.01$, ${\displaystyle \frac{1}{1000}}=0.001$, so $\underset{x\to \mathrm{\infty }}{lim}{\displaystyle \frac{1}{x}}=0$, and likewise $\underset{x\to -\mathrm{\infty }}{lim}{\displaystyle \frac{1}{x}}=0$. In general, $$\frac{\text{number}}{\pm \mathrm{\infty }}=0.$$

Example. $h(x)=1+{\displaystyle \frac{3}{x-2}}$: $\underset{x\to \mathrm{\infty }}{lim}h(x)=1+0=1$. The line $y=1$ is a horizontal asymptote.

Infinite limits at infinity. If $f(x)$ becomes infinite as $x$ becomes infinite, we write $\underset{x\to \pm \mathrm{\infty }}{lim}f(x)=+\mathrm{\infty }$ or $-\mathrm{\infty }$. For example $$\underset{x\to \mathrm{\infty }}{lim}(x^2+1)=\mathrm{\infty },\underset{x\to -\mathrm{\infty }}{lim}(-x{)}^3=\mathrm{\infty },\underset{x\to \mathrm{\infty }}{lim}(-x^2)=-\mathrm{\infty }.$$

Limits of Polynomials and Rational Functions

A polynomial is $p(x)=a_0+a_{1}x+a_2{x}^2+\cdots +a_n{x}^n$, of degree $n$ if $a_n\ne 0$, and a rational function is ${\displaystyle \frac{p(x)}{q(x)}}$ with polynomials $p$ and $q$.

Limit Laws

The theorem above follows from a few rules that let us split a complicated limit into simple ones.

Since $\underset{x\to c}{lim}k=k$ and $\underset{x\to c}{lim}x=c$, laws 1–3 and 5 give $\underset{x\to c}{lim}p(x)=p(c)$ for every polynomial, and law 4 gives the same for rational functions: $$\underset{x\to c}{lim}\frac{p(x)}{q(x)}=\frac{p(c)}{q(c)}\text{whenever }q(c)\ne 0.$$

Example. $$\underset{x\to -1}{lim}\frac{3x^2-2x-1}{x^2+1}=\frac{3+2-1}{1+1}=\frac{4}{2}=2.$$ The denominator is not $0$ at $x=-1$, so we simply substitute.

Evaluating Limits Algebraically

The quotient law needs $M\ne 0$. When we substitute and get

• a number: that is the limit;
• ${\displaystyle \frac{\text{nonzero number}}{0}}$: the limit is infinite (or does not exist); check the two one-sided limits with the sign rules of the remark above;
• ${\displaystyle \frac{0}{0}}$: this is an indeterminate form. It tells us nothing yet: the limit may be any number, infinite, or may not exist. We have to rewrite the expression first.

The rewriting always uses the same idea: for $x\ne c$ the expression equals a simpler one, and since the limit never uses the value at $x=c$, the two have the same limit. The common tricks are factoring, multiplying by the conjugate, and combining fractions.

Factoring

Example. $$\underset{x\to 3}{lim}\frac{x^2-9}{x-3}=\underset{x\to 3}{lim}\frac{(x-3)(x+3)}{x-3}=\underset{x\to 3}{lim}(x+3)=6.$$

Example. Here we use the sum of cubes, $a^3+b^3=(a+b)(a^2-ab+b^2)$ (and the difference of cubes $a^3-b^3=(a-b)(a^2+ab+b^2)$): $$\underset{x\to -2}{lim}\frac{x^3+8}{x^2-4}=\underset{x\to -2}{lim}\frac{(x+2)(x^2-2x+4)}{(x-2)(x+2)}=\underset{x\to -2}{lim}\frac{x^2-2x+4}{x-2}=\frac{4+4+4}{-4}=-3.$$

Example. Cancelling can also leave a nonzero number over zero: $$\underset{x\to 0}{lim}\frac{x}{x^3}=\underset{x\to 0}{lim}\frac{1}{x^2}=+\mathrm{\infty }\text{(the limit does not exist)},$$ because $\underset{x\to 0^{+}}{lim}\frac{1}{x^2}=\underset{x\to 0^{-}}{lim}\frac{1}{x^2}=+\mathrm{\infty }$. With one more power the signs differ: $$\underset{x\to 0}{lim}\frac{x}{x^4}=\underset{x\to 0}{lim}\frac{1}{x^3}:\underset{x\to 0^{+}}{lim}\frac{1}{x^3}=\frac{1}{0^{+}}=+\mathrm{\infty },\underset{x\to 0^{-}}{lim}\frac{1}{x^3}=\frac{1}{0^{-}}=-\mathrm{\infty }.$$

Example. $\underset{x\to 1}{lim}{\displaystyle \frac{x^2-1}{x^2-1}}=1$. The function is not defined at $x=\pm 1$, but everywhere else it equals $1$, so its limit at $1$ is $1$.

Example. (This kind of limit will define the derivative.) $$\underset{\mathrm{\Delta }x\to 0}{lim}\frac{(3+\mathrm{\Delta }x{)}^2-3^2}{\mathrm{\Delta }x}=\underset{\mathrm{\Delta }x\to 0}{lim}\frac{9+6\mathrm{\Delta }x+\mathrm{\Delta }{x}^2-9}{\mathrm{\Delta }x}=\underset{\mathrm{\Delta }x\to 0}{lim}\frac{\mathrm{\Delta }x(6+\mathrm{\Delta }x)}{\mathrm{\Delta }x}=\underset{\mathrm{\Delta }x\to 0}{lim}(6+\mathrm{\Delta }x)=6.$$

Combining Fractions

Example. $$\underset{h\to 0}{lim}\frac{1}{h}(\frac{1}{2+h}-\frac{1}{2})=\underset{h\to 0}{lim}\frac{1}{h}\cdot \frac{2-(2+h)}{2(2+h)}=\underset{h\to 0}{lim}\frac{1}{h}\cdot \frac{-h}{2(2+h)}=\underset{h\to 0}{lim}\frac{-1}{2(2+h)}=-\frac{1}{4}.$$

Multiplying by the Conjugate

The conjugate of $\sqrt{A}-\sqrt{B}$ is $\sqrt{A}+\sqrt{B}$, and their product has no square roots: $(\sqrt{A}-\sqrt{B})(\sqrt{A}+\sqrt{B})=A-B$.

Example. $$\underset{x\to 0}{lim}\frac{\sqrt{1+2x}-\sqrt{1-3x}}{x}=\underset{x\to 0}{lim}\frac{(\sqrt{1+2x}-\sqrt{1-3x})(\sqrt{1+2x}+\sqrt{1-3x})}{x(\sqrt{1+2x}+\sqrt{1-3x})}$$ $$=\underset{x\to 0}{lim}\frac{(1+2x)-(1-3x)}{x(\sqrt{1+2x}+\sqrt{1-3x})}=\underset{x\to 0}{lim}\frac{5x}{x(\sqrt{1+2x}+\sqrt{1-3x})}=\frac{5}{1+1}=\frac{5}{2}.$$

Example. $$\underset{x\to 4}{lim}\frac{\sqrt{x+5}-3}{x-4}=\underset{x\to 4}{lim}\frac{(\sqrt{x+5}-3)(\sqrt{x+5}+3)}{(x-4)(\sqrt{x+5}+3)}=\underset{x\to 4}{lim}\frac{x+5-9}{(x-4)(\sqrt{x+5}+3)}=\underset{x\to 4}{lim}\frac{1}{\sqrt{x+5}+3}=\frac{1}{6}.$$

Example. In the same way (the square root is now in the denominator), $$\underset{x\to 4}{lim}\frac{x-4}{\sqrt{x+5}-3}=\underset{x\to 4}{lim}\frac{(x-4)(\sqrt{x+5}+3)}{x+5-9}=\underset{x\to 4}{lim}(\sqrt{x+5}+3)=6.$$

Limits at Infinity of Rational Functions

Let $$\frac{P(x)}{Q(x)},P(x)=a_0+a_{1}x+\cdots +a_n{x}^n(a_n\ne 0),Q(x)=b_0+b_{1}x+\cdots +b_m{x}^m(b_m\ne 0),$$ so $\mathrm{deg}P=n$ and $\mathrm{deg}Q=m$. For example, $P(x)=3+7x+9x^2$ has $\mathrm{deg}P=2$ and $a_2=9$, and $Q(x)=3x^3-2x+1$ has $\mathrm{deg}Q=3$ and $b_3=3$.

Method. Factor out the highest power of $x$ from the numerator and from the denominator. Every term ${\displaystyle \frac{k}{x^j}}$ ($j\ge 1$) then goes to $0$, because $$\frac{1}{\pm \mathrm{\infty }}=0,\frac{k}{\pm \mathrm{\infty }}=0.$$

Case I: $\mathrm{deg}Q>\mathrm{deg}P$. Then $\underset{x\to \pm \mathrm{\infty }}{lim}{\displaystyle \frac{P(x)}{Q(x)}}=0$. $$\underset{x\to \mathrm{\infty }}{lim}\frac{2x^2+1}{3x^3-2x^2+x+1}=\underset{x\to \mathrm{\infty }}{lim}\frac{x^2(2+\frac{1}{x^2})}{x^3(3-\frac{2}{x}+\frac{1}{x^2}+\frac{1}{x^3})}=\underset{x\to \mathrm{\infty }}{lim}\frac{1}{x}\cdot \frac{2+\frac{1}{x^2}}{3-\frac{2}{x}+\frac{1}{x^2}+\frac{1}{x^3}}=0\cdot \frac{2}{3}=0.$$ Another example: $\underset{x\to -\mathrm{\infty }}{lim}{\displaystyle \frac{-17x}{3x^2}}=\underset{x\to -\mathrm{\infty }}{lim}{\displaystyle \frac{-17}{3x}}=0$.

Case II: $\mathrm{deg}P>\mathrm{deg}Q$. Then $\underset{x\to \pm \mathrm{\infty }}{lim}{\displaystyle \frac{P(x)}{Q(x)}}=\pm \mathrm{\infty }$ (the limit does not exist; $\pm \mathrm{\infty }$ is not a number). The sign is the sign of ${\displaystyle \frac{a_n{x}^n}{b_m{x}^m}}$ for large $|x|$. $$\underset{x\to \mathrm{\infty }}{lim}\frac{x^3+1}{x^2+2}=+\mathrm{\infty },\underset{x\to \mathrm{\infty }}{lim}\frac{x^3+1}{2-x^2}=-\mathrm{\infty },\underset{x\to -\mathrm{\infty }}{lim}\frac{x^3-5}{1+x^2}=-\mathrm{\infty }.$$ In the second one the leading terms give $\frac{x^3}{-x^2}=-x\to -\mathrm{\infty }$; in the third, $\frac{x^3}{x^2}=x\to -\mathrm{\infty }$.

Case III: $\mathrm{deg}P=\mathrm{deg}Q$. Then the limit is the ratio of the leading coefficients: $\underset{x\to \pm \mathrm{\infty }}{lim}{\displaystyle \frac{P(x)}{Q(x)}}={\displaystyle \frac{a_n}{b_n}}$. $$\underset{x\to \mathrm{\infty }}{lim}\frac{10x^2-19}{2x^2+5x-3}=\underset{x\to \mathrm{\infty }}{lim}\frac{x^2(10-\frac{19}{x^2})}{x^2(2+\frac{5}{x}-\frac{3}{x^2})}=\frac{10}{2}=5.$$

So in Case III the graph has the horizontal asymptote $y=\frac{a_n}{b_n}$, in Case I the asymptote $y=0$, and in Case II none.

Square roots at $-\mathrm{\infty }$ (a common AP trap). $\sqrt{x^2}=|x|$, and for $x<0$ this is $-x$, not $x$. $$\underset{x\to \mathrm{\infty }}{lim}\frac{\sqrt{4x^2+1}}{x+3}=\underset{x\to \mathrm{\infty }}{lim}\frac{|x|\sqrt{4+\frac{1}{x^2}}}{x(1+\frac{3}{x})}=\frac{2}{1}=2,\underset{x\to -\mathrm{\infty }}{lim}\frac{\sqrt{4x^2+1}}{x+3}=\underset{x\to -\mathrm{\infty }}{lim}\frac{-x\sqrt{4+\frac{1}{x^2}}}{x(1+\frac{3}{x})}=-2.$$ This graph has two different horizontal asymptotes, $y=2$ and $y=-2$.

The Squeeze Theorem

$f$ is trapped between two functions that go to the same place, so it has to go there too.

Example. $\underset{x\to \mathrm{\infty }}{lim}{\displaystyle \frac{\sin x}{x}}=0$. Since $-1\le \sin x\le 1$, for $x>0$ $$-\frac{1}{x}\le \frac{\sin x}{x}\le \frac{1}{x},\text{and}\underset{x\to \mathrm{\infty }}{lim}(-\frac{1}{x})=\underset{x\to \mathrm{\infty }}{lim}\frac{1}{x}=0.$$ The same argument works for $x\to -\mathrm{\infty }$.

Example. $\underset{x\to 0}{lim}{x}^2\cos {\displaystyle \frac{1}{x}}=0$. We cannot use the product law, because $\underset{x\to 0}{lim}\cos \frac{1}{x}$ does not exist (it oscillates). But $-x^2\le x^2\cos \frac{1}{x}\le x^2$, and both sides go to $0$.

Some Basic Limits

These limits come up again and again (and will give the derivatives of $\sin x$, $\cos x$ and $e^x$).

1. The limit of ${\displaystyle \frac{\sin x}{x}}$ at $0$.

Examples. The trick is to make the argument of $\sin$ and the denominator equal.

• $\underset{x\to 0}{lim}{\displaystyle \frac{\sin 4x}{4x}}$: put $u=4x$; $x\to 0⟺u\to 0$, so $\underset{x\to 0}{lim}{\displaystyle \frac{\sin 4x}{4x}}=\underset{u\to 0}{lim}{\displaystyle \frac{\sin u}{u}}=1$.
• $\underset{x\to 0}{lim}{\displaystyle \frac{\sin 3x}{x}}=\underset{x\to 0}{lim}3\cdot {\displaystyle \frac{\sin 3x}{3x}}=3\cdot 1=3$.
• $\underset{x\to 0}{lim}{\displaystyle \frac{\sin 5x}{\sin 2x}}=\underset{x\to 0}{lim}{\displaystyle \frac{\frac{\sin 5x}{5x}\cdot 5x}{\frac{\sin 2x}{2x}\cdot 2x}}={\displaystyle \frac{5}{2}}$.
• $\underset{x\to 0}{lim}{\displaystyle \frac{\tan x}{x}}=\underset{x\to 0}{lim}{\displaystyle \frac{\sin x}{x}}\cdot {\displaystyle \frac{1}{\cos x}}=1\cdot 1=1$.
• $\underset{x\to 0}{lim}{\displaystyle \frac{1-\cos x}{x}}=0$. Multiply by $1+\cos x$: $$\frac{1-\cos x}{x}=\frac{1-{\cos }^{2}x}{x(1+\cos x)}=\frac{\sin x}{x}\cdot \frac{\sin x}{1+\cos x}\to 1\cdot \frac{0}{2}=0.$$

2. The limit of ${\displaystyle \frac{\sin x}{x}}$ at $\pm \mathrm{\infty }$ is $0$, by the squeeze theorem (the first example above).

3. The number $e$.

This is a $1^{\mathrm{\infty }}$ form: the base goes to $1$, the exponent to $\mathrm{\infty }$. Such a limit is not automatically $1$. Some consequences:

• $\underset{n\to \mathrm{\infty }}{lim}{\left({\displaystyle \frac{1+n}{n}}\right)}^n=\underset{n\to \mathrm{\infty }}{lim}{(1+{\displaystyle \frac{1}{n}})}^n=e$.
• $\underset{n\to \mathrm{\infty }}{lim}{\left({\displaystyle \frac{n}{n+1}}\right)}^n=\underset{n\to \mathrm{\infty }}{lim}{\displaystyle \frac{1}{{(1+\frac{1}{n})}^n}}={\displaystyle \frac{1}{e}}$.
• $\underset{n\to \mathrm{\infty }}{lim}{(1+{\displaystyle \frac{1}{n}})}^{2n}=\underset{n\to \mathrm{\infty }}{lim}{\left({(1+{\displaystyle \frac{1}{n}})}^n\right)}^2=e^2$.
• More generally, $\underset{x\to \mathrm{\infty }}{lim}{(1+{\displaystyle \frac{a}{x}})}^x=e^a$, and $\underset{h\to 0}{lim}(1+h{)}^{1/h}=e$ (put $h=\frac{1}{x}$).

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Quiz 5: Limits

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Continuity

If a function is continuous on an interval, we can draw its graph there without lifting the pencil from the paper: the graph has no holes, breaks or jumps. Conceptually, if $f$ is continuous at $x=c$, then the closer $x$ is to $c$, the closer $f(x)$ is to $f(c)$. The following definition makes this precise.

Continuity at a Point

The three conditions are all hidden in one equation: $$\underset{(2)\text{exists}}{\underbrace{\underset{x\to c}{lim}f(x)}}\underset{(3)}{=}\underset{(1)\text{exists}}{\underbrace{f(c)}}.$$ On the AP exam, a justification of continuity has to check all three, for example: "$f(2)=5$, $\underset{x\to 2^{-}}{lim}f(x)=\underset{x\to 2^{+}}{lim}f(x)=5$, so $\underset{x\to 2}{lim}f(x)=f(2)$ and $f$ is continuous at $x=2$."

Continuous Functions

• Polynomials are continuous everywhere, at every real number: $p(x)=a_0+a_{1}x+\cdots +a_n{x}^n$ is defined for every $x\in \mathrm{ℝ}$ and $\underset{x\to c}{lim}p(x)=p(c)$.
• Rational functions ${\displaystyle \frac{P(x)}{Q(x)}}$ are continuous everywhere they are defined, that is, at every $x$ with $Q(x)\ne 0$.
• $|x|$ is continuous at every point.
• Trigonometric, inverse trigonometric, exponential and logarithmic functions are continuous at every point where they are defined.
• $f(x)=\sqrt[n]{x}=x^{1/n}$ ($n\ge 2$ an integer) is continuous at every point where it is defined.
• Sums, differences, products and constant multiples of continuous functions are continuous, and so are quotients where the denominator is not $0$ (by the limit laws).
• If $g$ is continuous at $c$ and $f$ is continuous at $g(c)$, then $f\circ g$ is continuous at $c$. In particular we can move a limit inside a continuous function: $$\underset{x\to c}{lim}f(g(x))=f(\underset{x\to c}{lim}g(x));\text{for example}\underset{x\to 0}{lim}\cos \left(\frac{\sin x}{x}\right)=\cos 1.$$
• The greatest integer function $[x]$ is continuous at every point except the integers. For example $\underset{x\to 1^{+}}{lim}[x]=1\ne 0=\underset{x\to 1^{-}}{lim}[x]$, so $\underset{x\to 1}{lim}[x]$ does not exist: the left-hand and right-hand limits are different.

Kinds of Discontinuities

1. Jump discontinuity. Both one-sided limits exist at $c$, but they are different: $$\underset{x\to c^{+}}{lim}f(x)\ne \underset{x\to c^{-}}{lim}f(x).$$

Example. $f(x)=\{\begin{array}{ll}{x}^2+1,& x<1,\\ x^3+2x+1,& x\ge 1.\end{array}$ Here $\underset{x\to 1^{-}}{lim}f(x)=2$ and $\underset{x\to 1^{+}}{lim}f(x)=4=f(1)$, so $f$ has a jump at $x=1$.

Example. $[x]$ has a jump discontinuity at every integer.

2. Removable discontinuity. $\underset{x\to c}{lim}f(x)$ exists, but $f(c)$ is not defined or $f(c)\ne \underset{x\to c}{lim}f(x)$. The graph has a hole, and we can “remove” the discontinuity by defining (or redefining) $f(c)$ as the limit.

Example. $f(x)={\displaystyle \frac{x^2-1}{x-1}}$: $f(1)$ is not defined, but $$\underset{x\to 1}{lim}\frac{x^2-1}{x-1}=\underset{x\to 1}{lim}\frac{(x-1)(x+1)}{x-1}=2.$$ The function $$F(x)=\{\begin{array}{ll}{\displaystyle \frac{x^2-1}{x-1}},& x\ne 1,\\ 2,& x=1\end{array}$$ is continuous at $x=1$, since $\underset{x\to 1}{lim}F(x)=\underset{x\to 1}{lim}f(x)=2=F(1)$. (In fact $F(x)=x+1$ for every $x$.)

3. Infinite discontinuity. At least one of the one-sided limits is infinite: $$\underset{x\to c^{+}}{lim}f(x)=\pm \mathrm{\infty }\text{or}\underset{x\to c^{-}}{lim}f(x)=\pm \mathrm{\infty }.$$ The line $x=c$ is then a vertical asymptote.

Example. $f(x)={\displaystyle \frac{3x^2+2x+1}{x^2-5}}$ has infinite discontinuities where $x^2=5$. At $x=\sqrt{5}$ the numerator is $16+2\sqrt{5}>0$, and $x^2-5\to 0^{+}$ from the right and $0^{-}$ from the left, so $$\underset{x\to {\sqrt{5}}^{+}}{lim}f(x)=+\mathrm{\infty },\underset{x\to {\sqrt{5}}^{-}}{lim}f(x)=-\mathrm{\infty }.$$ The same happens at $x=-\sqrt{5}$.

Examples

Example. Discuss the continuity of the function of the worked example in the limits section, $$f(x)=\{\begin{array}{ll}x+1,& -2<x<0,\\ 2,& x=0,\\ -x,& 0<x<2,\\ 0,& x=2,\\ x-4,& 2<x\le 4.\end{array}$$

• On each open piece $f$ is a polynomial, so it is continuous there.
• $x=0$: $\underset{x\to 0^{-}}{lim}f(x)=1\ne 0=\underset{x\to 0^{+}}{lim}f(x)$. Jump discontinuity.
• $x=2$: $\underset{x\to 2}{lim}f(x)=-2$, but $f(2)=0$. Removable discontinuity (redefine $f(2)=-2$).
• $x=4$: $\underset{x\to 4^{-}}{lim}f(x)=0=f(4)$, so $f$ is continuous from the left at the endpoint $4$.
• $x=-2$ is not in the domain, so we do not ask about continuity there.

Example. $f(x)={\displaystyle \frac{x-1}{x^2+x}}$ is not continuous at $x=0$, because $f(0)$ is not defined (the same for $x=-1$: $f(-1)$ is not defined). Both are infinite discontinuities: at $x=0$ the numerator goes to $-1$ and the denominator $x(x+1)$ to $0$.

Example. Discuss the continuity of the function in the figure.

• $x=1$: $\underset{x\to 1}{lim}f(x)=2$ exists, but $f(1)=0\ne 2$. Not continuous: a removable discontinuity.
• $x=3$: $\underset{x\to 3^{-}}{lim}f(x)=1$ and $\underset{x\to 3^{+}}{lim}f(x)=2$, so $\underset{x\to 3}{lim}f(x)$ does not exist. Not continuous: a jump.
• At the corner $x=2$ the graph changes direction but does not break, so $f$ is continuous there. It is continuous at every other point of $[0,5]$ too.

Example. Is $f(x)=\frac{1}{2}{x}^4-\sqrt{3}{x}^2+7$ continuous at $x=-1$? Yes: $f$ is a polynomial, and polynomials are continuous at every real number.

Example. Is $g(x)={\displaystyle \frac{1}{x-3}}$ continuous (a) at $x=3$, (b) at $x=0$?

• (a) No. $g(3)$ is not defined, and $\underset{x\to 3^{+}}{lim}{\displaystyle \frac{1}{x-3}}=+\mathrm{\infty }$, $\underset{x\to 3^{-}}{lim}{\displaystyle \frac{1}{x-3}}=-\mathrm{\infty }$: an infinite discontinuity.
• (b) Yes. $g$ is a rational function and its denominator is not $0$ at $x=0$: $\underset{x\to 0}{lim}g(x)=-\frac{1}{3}=g(0)$.

Making a Piecewise Function Continuous

A typical AP question gives a piecewise function with an unknown constant and asks for the value that makes it continuous. On each piece the function is a polynomial, so the only point to check is where the formula changes; there the two one-sided limits and the value must agree.

Example. For which $k$ is $f(x)=\{\begin{array}{ll}kx+1,& x\le 2,\\ x^2-k,& x>2\end{array}$ continuous everywhere? $$\underset{x\to 2^{-}}{lim}f(x)=f(2)=2k+1,\underset{x\to 2^{+}}{lim}f(x)=4-k.$$ Continuity at $2$ needs $2k+1=4-k$, so $3k=3$ and $k=1$. (Compare the earlier example where $x^2+a^2$ and $5$ met at $x=1$: there $a=\pm 2$.)

The Intermediate Value Theorem

The theorem says that such a $c$ exists, not where it is, and there may be several. Continuity on the whole closed interval is essential: $f(x)=\frac{1}{x}$ has $f(-1)=-1$ and $f(1)=1$, but it never takes the value $0$, because it is not continuous at $x=0$.

Example. Show that $x^3+x-1=0$ has a solution between $0$ and $1$.

Let $f(x)=x^3+x-1$. It is a polynomial, so it is continuous on $[0,1]$. Since $f(0)=-1<0$ and $f(1)=1>0$, the value $N=0$ lies between $f(0)$ and $f(1)$, and by the Intermediate Value Theorem there is a $c$ in $(0,1)$ with $f(c)=0$.

Example (AP style). $f$ is continuous on $[1,5]$ with $f(1)=4$, $f(3)=-2$ and $f(5)=1$. Then $f(x)=0$ has at least two solutions in $(1,5)$: one in $(1,3)$, where $f$ goes from $4$ to $-2$, and one in $(3,5)$, where it goes from $-2$ to $1$.

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Quiz 6: Limit Techniques and Continuity

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Flashcards

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