1a
1b
§1.1 Function Machines
Course Administration
sketch x  [f]  y = f(x), input, output
Math Placement
Syllabus (show text)
WebAssign, class keys on syllabus
FDOC_self_enrollment.ppt
Your first homework assignment:
(1)
(2)
(3)
self enroll in WebAssign
Intro to WebAssign
Math 171 week #1A
Due this Friday at 11:59pm for MW and TuTh
tutorials.
Due this Saturday at 11:59pm for WF tutorials.
First tutorial meeting this week is cancelled.
domain={possible inputs}
range={possible outputs}
x is the independent variable (represents an input
value)
y is the dependent variable (represents an output
value)
A function is a rule that assigns to each element in its
domain exactly one element in its range.
Example. 𝑓(𝑥) = 𝑥 2 , −1 ≤ 𝑥 ≤ 1. domain =
[−1,1]. range = [0,1]. ■
2a
2b
Interval Notation
𝑥 ∈ [−1,1] means −1 ≤ 𝑥 ≤ 1
𝑥 ∈ (−1,1) means −1 < 𝑥 < 1
𝑥 ∈ (−∞, ∞) means 𝑥 is any real no.
Example. 𝑔(𝑥) = 𝑥 2 , domain = (−∞, ∞). range=
[0, ∞). ■
If the domain of a function is not given explicitly,
assume it is the largest set of numbers that makes
sense.
Example. 𝑦 = √𝑥
Example. ℎ(𝑥) = √𝑥, domain not given.
sketch 0 … 4 … 𝑥-, 0 … 2 … 𝑦-, curve■
Assume domain= [0, ∞). range = [0, ∞). ■
Graphs
Example. 𝑦 = 𝑥 2
sketch … − 1 … 0 … 1 … 𝑥- (independent variable),
0 … 𝑦- (dependent variable), curve■
3a
Example. Graph the set of points that satisfy 𝑦 2 = 𝑥.
Table x / 0, 1, 4; y/ 0, ±1, ±2
3b
Vertical Line Test
A curve in the xy-plane is the graph of a function iff
no vertical line intersects the curve more than once.
Example. Draw the graph of 𝑥 2 + 𝑦 2 = 1.
axes, circle, vertical line intersecting circle twice
sketch 0 … 4 … 𝑥-, −2 … 0 … 2 … 𝑦-, curve
?? Is this the graph of a function?
?? Why?
Is this a function?
?? Why? ■
4a
4b
Even and odd symmetry
If 𝑓(𝑥) satisfies
𝑓(−𝑥) = 𝑓(𝑥)
for every 𝑥 in its domain, then 𝑓 is called an even
function.
sketch −2 … 0 … 2 … 𝑥-, −8 … 0 … 8 … 𝑦-, curve
2
Example. 𝑓(𝑥) = 𝑥 ,
𝑓(−𝑥) = (−𝑥)2 = (−𝑥)(−𝑥) = 𝑥 2 = 𝑓(𝑥).
?? If 𝑓 is even, its graph is symmetric wrt which axis?
■
If 𝑔(𝑥) satisfies
𝑔(−𝑥) = −𝑔(𝑥)
for every 𝑥 in its domain, then 𝑔 is an odd function.
Example. 𝑔(𝑥) = 𝑥 3 ,
?? If 𝑔 is odd, its graph is symmetric wrt what? ■
𝑔(−𝑥) = (−𝑥)(−𝑥)(−𝑥) = −𝑥 3 = −𝑔(𝑥) .
Table x/ 0, -1, 1, -2, 2; y/ 0, -1, 1, -8, 8
Knowing even or odd symmetry helps us sketch
functions.
5a
5b
§1.2 Catalog Of Functions
sketch −2 … 0 … 2 … 𝑥-, 0 … 1 … 2 … 𝑦-, intercept,
line
Straight lines
slope intercept form
𝑦 = 𝑚𝑥 + 𝑏
sketch … 0 … 𝑥- axis, 0 … 𝑦- axis, intercept 𝑏,
line,(𝑥, 𝑦),Δ𝑥, Δ𝑦
■
point slope form
𝑦 − 𝑦1 = 𝑚(𝑥 − 𝑥1 )
sketch 0 … 𝑥-, … 𝑦-,(𝑥1 , 𝑦1 ), line, (𝑥, 𝑦),𝑥 − 𝑥1 = 𝛥𝑥,
𝑦 − 𝑦1 = 𝛥𝑦
slope =
Δ𝑦
Δ𝑥
=
𝑦−𝑏
𝑥−0
=𝑚
1
Example. 𝑦 = 𝑥 + 1
2
Table
𝛥𝑦
𝛥𝑥
=𝑚
6a
6b
1
Example. 𝑦 − 2 = (𝑥 − 2)
Example 𝑦 = 𝑥 2 parabola
convert to slope intercept form
Example 𝑦 = 𝑥 2 square root (𝑥 2 = √𝑥)
solve for y
Example 𝑦 = 𝑥 −1 (𝑥 −1 = 1/𝑥)
𝑦=
sketch −2 … 0 … 2 … 𝑥-, −2 … 0 … 2 … 𝑦-
2
1
same as previous example ■
Power Functions
general form
𝑦 = 𝑥 𝛼 where 𝛼 is a constant
Example 𝑦 = 𝑥 straight line
?? slope ?? y-intercept
■
1
7a
7b
Polynomial Functions
Piecewise defined functions
Example 𝑦 = 1 − 𝑥 2
Example 𝑓(𝑥) = {
sketch −2 … 0 … 2 … 𝑥-, −2 … 0 … 1 … 𝑦-, parabola
𝑥 + 1, 𝑥 ≠ 1
1,
𝑥=1
sketch −1 … 2 … 𝑥-, 0 … 3 … 𝑦-, line with hole, dot
■
■
Example Absolute Value Function 𝑓(𝑥) = |𝑥|
A quadratic function is a polynomial of degree 2.
gives distance from the origin on the real number line
2
𝑦 = 𝑎2 𝑥 + 𝑎1 𝑥 + 𝑎0
𝑎2 , 𝑎1 , 𝑎0 are constants add
sketch −2 … 0 … 2, bracket −2 and 0
The degree of a polynomial is the highest power that
it contains.
A polynomial of degree 𝑛 has the form
𝑃(𝑥) = 𝑎𝑛 𝑥 𝑛 + 𝑎𝑛−1 𝑥 𝑛−1 + ⋯ + 𝑎2 𝑥 2 + 𝑎1 𝑥 + 𝑎0 .
distance between −2 and 0 is 2, |−2| = 2
8a
8b
graph 𝑦 = |𝑥|
Example (a case of special interest to us)
sketch −2 … 0 … 2 … 𝑥-, 0 … 2 … 𝑦-, graph
𝑓(𝑥) =
𝑥 2 −1
𝑥−1
?? domain of f?
Important Algebraic Trick!
(𝑥 − 1)(𝑥 + 1) =
In general, (𝑥 − 𝑎)(𝑥 + 𝑎) = 𝑥 2 − 𝑎2 . Then
𝑥, 𝑥 > 0
piecewise definition |𝑥| = {
■
−𝑥, 𝑥 < 0
𝑓(𝑥) =
sketch −1 … 0 … 2 … 𝑥-, 0 … 2 … 𝑦-, line with hole
Rational Functions
A rational function is a ratio of two polynomials
𝑓(𝑥) =
𝑃(𝑥)
𝑄(𝑥)
where 𝑃, 𝑄 are polynomials.
(𝑥 − 1)(𝑥 + 1)
𝑥 + 1,
if 𝑥 ≠ 1
={
undefined, if 𝑥 = 1
𝑥−1
■
9a
9b
𝜋
Sine and Cosine
sketch – 𝜋, … 0 …
sin ( ) = 1,
2
3𝜋
2
… 𝑥-, −1 … 0 … 1.., sin(𝑥),
cos(𝑥)
𝜋
sin (− ) = −1
2
sin(𝑥 + 2𝜋) = sin(𝑥) periodicity with period 2𝜋
sin(𝑥 + 𝜋) = −sin(𝑥) advance by half a period
?? sin(−𝑥) = ⋯, ?? symmetry?
properties of cosine
−1 ≤ cos(𝑥) ≤ 1
𝜋
3𝜋
5𝜋
cos ( ) = 0, cos ( ) = 0, cos ( ) = 0, generally
2
2
2
1
cos ((𝑛 + ) 𝜋) = 0 for 𝑛 an integer
2
cos(0) = 1, cos(𝜋) = −1
properties of sine
−1 ≤ sin(𝑥) ≤ 1
sin(0) = 0, sin(𝜋) = 0, sin(2𝜋) = 0, generally
sin(𝑛𝜋) = 0 for 𝑛 an integer
cos(𝑥 + 2𝜋) = cos(𝑥) periodicity with period 2𝜋
cos(𝑥 + 𝜋) = −cos(𝑥) advance by half a period
?? cos(−𝑥) = ⋯, ?? symmetry?
10a
10b
Two Important Triangles
Pythagorean theorem:
sketch 45-45-90 triangle, lengths of sides
𝜋
𝜋
1
𝜋
𝜋
√3
2
Tangent
tan(𝑥) =
𝜋
𝜋
sin ( ) = cos ( ) =
4
4
1
√2
1
2
=
1
+ =1
2
√2
2
4
1
+ =1
4
sin ( ) = cos ( ) =
6
3
2
sin ( ) = cos ( ) =
3
6
Pythagorean theorem:
3
sin(𝑥)
cos(𝑥)
tan(−𝑥) =
≈ 0.71
sketch 30-60-90 triangle, lengths of sides
?? symmetry
≈ 0.87
11a
11b
graph tangent
§1.3 The Limit of a Function
𝜋
3𝜋
2
2
sketch – … 0 …
… 𝑥-, … 0 … 𝑦-, vert. asymptotes,
sketch … 𝑎 … . , … 𝐿 …, curve 𝑓, hole at 𝑎
curve
𝜋
𝜋
2
2
− ,
,
3𝜋
2
, … not in the domain of tan ?? why
Informal definition of limit
lim 𝑓(𝑥) = 𝐿
tan(𝑥 + 𝜋) =
𝑥→𝑎
period 𝜋
means that we can make 𝑓(𝑥) as close as we wish to
𝐿 by taking 𝑥 sufficiently close to 𝑎 (but not equal to
𝑎).
tan(0) = tan(𝜋) = ⋯ = 0
𝜋
tan ( ) =
4
𝜋
4
𝜋
cos( 4 )
sin( )
=1
𝜋
similarly find tan ( ) =
6
1
√3
𝜋
and tan ( ) = √3
3
Alternate notation: 𝑓(𝑥) → 𝐿 as 𝑥 → 𝑎
12a
12b
lim 𝑓(𝑥) = 2
𝑥→1
Example. 𝑓(𝑥) = 𝑥 + 1
… 1 … , … 1 … 2 …, line 𝑓
lim 𝑓(𝑥)has nothing to do with 𝑓(𝑎)
𝑥→𝑎
Example 𝑔(𝑥) = {
𝑥 + 1, 𝑥 ≠ 1
1,
𝑥=1
sketch … 1 … 𝑥-, … 1 … 2 …, line with hole, dot
imagine a bug approaching 𝑥 = 1 on either side
add arrows towards 𝑥 = 1, arrows towards 𝑦 = 2
add arrows towards 𝑥 = 1, arrows towards 𝑦 = 2
13a
13b
lim 𝑔(𝑥) = 2 ■
𝑥→1
Limit is a 2-sided concept
One sided limits
Example. Step function
Informal definition
0, 𝑥 < 0
𝐻(𝑥) = {
1, 𝑥 ≥ 0
𝑥→𝑎−
… 0 … 𝑥-, 0 … 1 …, 𝐻(𝑥)
means we can make 𝑓(𝑥) as close as we wish to 𝐿1
by taking 𝑥 sufficiently close to 𝑎 from the left.
lim 𝑓(𝑥) = 𝐿1
lim 𝑓(𝑥) = 𝐿2
𝑥→𝑎+
limit from left or left hand limit
limit from right or …
means we can make 𝑓(𝑥) as close as we wish to 𝐿2
by taking 𝑥 sufficiently close to 𝑎 from the right.
Example. (step function again)
lim 𝐻(𝑥) DNE (does not exist)■
𝑥→0
?? lim 𝐻(𝑥) =
𝑥→0+
14a
14b
■
?? lim 𝐻(𝑥) =
𝑥→0−
the (2-sided) limit
lim 𝑓(𝑥) = 𝐿 means
𝑥→𝑎
lim 𝑓(𝑥) = 𝐿
𝑥→𝑎
if and only if
For every 𝜖 > 0 there is a 𝛿 > 0 such that
lim 𝑓(𝑥) = 𝐿 and lim 𝑓(𝑥) = 𝐿
𝑥→𝑎+
if
𝑎−𝛿 <𝑥 <𝑎+𝛿
then
𝐿 − 𝜖 < 𝑓(𝑥) < 𝐿 + 𝜖
(𝑥 ≠ 𝑎)
𝑥→𝑎−
?? Practice with limits
add 𝑎 − 𝛿, 𝑎 + 𝛿, 𝐿 − 𝜖, 𝐿 = 𝜖, segments
§1.4 Calculating Limits
Precise definition of a limit
0 … 𝑎 … 𝑥-, 0 … 𝐿 … 𝑦-, 𝑓(𝑥), hole at 𝑥 = 𝑎, dot
Two Special Limits
A. Let 𝑐 be a constant
lim 𝑐 = 𝑐
𝑥→𝑎
0 … 𝑎 … 𝑥-, 0 … 𝑦-, line 𝑦 = 𝑐, arrows approaching
𝑥 = 𝑎
15a
15b
Five Limit Laws
Suppose that 𝑐 is a constant and
lim 𝑓(𝑥),
𝑥→𝑎
B. Consider 𝑓(𝑥) = 𝑥
lim 𝑥 = 𝑎
𝑥→𝑎
1.
lim 𝑔(𝑥) exist.
𝑥→𝑎
sum law
(limit of sum is sum of limits)
lim [𝑓(𝑥) + 𝑔(𝑥)] = lim 𝑓(𝑥) + lim 𝑔(𝑥)
𝑥→𝑎
𝑥→𝑎
𝑥→𝑎
0 … 𝑎 … 𝑥-, 0 … 𝑎 … 𝑦-, 𝑓(𝑥) = 𝑥
2.
difference law
lim [𝑓(𝑥) − 𝑔(𝑥)] = lim 𝑓(𝑥) − lim 𝑔(𝑥)
𝑥→𝑎
3.
𝑥→𝑎
𝑥→𝑎
constant multiple law
lim 𝑐 𝑓(𝑥) = 𝑐 lim 𝑓(𝑥)
𝑥→𝑎
add arrows approaching 𝑥 = 𝑎, add arrows
approaching 𝑦 = 𝑎
4.
𝑥→𝑎
product law
lim 𝑓(𝑥)𝑔(𝑥) = lim 𝑓(𝑥) ⋅ lim 𝑔(𝑥)
𝑥→𝑎
𝑥→𝑎
𝑥→𝑎
16a
5.
lim
16b
quotient law
𝑓(𝑥)
𝑥→𝑎 𝑔(𝑥)
lim 𝑓(𝑥)
= 𝑥→𝑎
lim 𝑔(𝑥)
product and sum laws
=
,
lim 𝑥⋅lim 𝑥
𝑥→2
𝑥→2
𝑥→𝑎
𝑥→2
lim 𝑥+lim 1
𝑥→2
so long as lim 𝑔(𝑥) ≠ 0
special limits
𝑥→𝑎
=
Example
2⋅2
2+1
=
4
■
3
Repeated Application of the Product Law
lim 3𝑥 + 5
sum law
𝑥→1
lim 𝑓(𝑥) ⋅ 𝑔(𝑥) ⋅ ℎ(𝑥) = lim 𝑓(𝑥) ⋅ 𝑘(𝑥)
𝑥→𝑎
𝑥→1
product law
constant multiple law
= lim 𝑓(𝑥) ⋅ lim 𝑘(𝑥)
𝑥→𝑎
= 3 lim 𝑥 + lim 5
𝑥→1
𝑥→𝑎
𝑥→1
product law
special limits
=3⋅1+5
=8
where
𝑔⋅ℎ =𝑘
= lim 3𝑥 + lim 5
𝑥→1
𝑥→𝑎
■
= lim 𝑓(𝑥) ⋅ lim 𝑔(𝑥) ⋅ lim ℎ(𝑥)
𝑥→𝑎
𝑥→𝑎
𝑥→𝑎
Example
lim
𝑥2
quotient law
𝑥→2 𝑥+1
=
Power Law
Suppose 𝑓(𝑥) = 𝑔(𝑥) = ℎ(𝑥) then from above
lim 𝑥 2
𝑥→2
lim (𝑥+1)
𝑥→2
17a
17b
3
lim 𝑓(𝑥)3 = (lim 𝑓(𝑥))
𝑥→𝑎
𝑥→𝑎
Apply the same reasoning to a product of 𝑛 factors of
𝑓(𝑥) to get the Power Law:
Recall polynomials of degree 𝑛. Their general form is
𝑃(𝑥) = 𝑐𝑛 𝑥 𝑛 + 𝑐𝑛−1 𝑥 𝑛−1 + ⋯ + 𝑐1 𝑥 + 𝑐0
where 𝑐𝑛 , 𝑐𝑛−1 , … , 𝑐1 , 𝑐0 are constants.
By reasoning similar to the cubic polynomial example
lim 𝑃(𝑥)
𝑛
lim 𝑓(𝑥)𝑛 = (lim 𝑓(𝑥))
𝑥→𝑎
sum law, const. multiple law
𝑥→𝑎
𝑥→𝑎
= 𝑐𝑛 lim 𝑥 𝑛 + ⋯ + 𝑐1 lim 𝑥 + lim 𝑐0
𝑥→𝑎
𝑥→𝑎
where 𝑛 is any positive integer
power law, special limit
Example. Cubic Polynomial.
= 𝑐𝑛 𝑎𝑛 + ⋯ 𝑐1 𝑎 + 𝑐0
lim (𝑥 3 − 4𝑥)
difference law
𝑥→2
= lim 𝑥 3 − lim 4𝑥
𝑥→2
Direct Substitution Property for polynomials
If 𝑃(𝑥) is any polynomial and 𝑎 is a real number then
3
= (lim 𝑥) = 4 lim 𝑥
𝑥→2
lim 𝑃(𝑥) = 𝑃(𝑎).
special limits
= 23 − 4 ⋅ 2 = 0
= 𝑃(𝑎)
we have discovered the following
𝑥→2
power, constant multiple laws
𝑥→2
𝑥→𝑎
■
𝑥→𝑎
This is much easier to apply than the limit laws!
18a
18b
Recall that a rational function is a ratio of two
polynomials.
𝑓(𝑥) =
𝑃(𝑥)
𝑄(𝑥)
where 𝑃 and 𝑄 are polynomials
,
𝑥4 + 𝑥2 − 6
14 + 1 2 − 6
−4 −2
=
=
=
𝑥→1 𝑥 4 − 2𝑥 + 3
14 + 2 ⋅ 1 + 3
6
3
lim
by Direct Substitution for rational functions! ■
Let 𝑓(𝑥) be any rational function
Root Law For Limits
lim 𝑃(𝑥)
lim 𝑓(𝑥) = 𝑥→𝑎
𝑥→𝑎
quotient law
lim 𝑄(𝑥)
𝑥→𝑎
=
𝑃(𝑎)
Let 𝑛 be a positive integer
𝑛
lim √𝑓(𝑥) = 𝑛√ lim 𝑓(𝑥),
𝑥→𝑎
direct subst. for polys
𝑄(𝑎)
𝑥→𝑎
so long as 𝑄(𝑎) ≠ 0.
we now have a …
If 𝑛 is even then require lim 𝑓(𝑥) > 0.
𝑥→𝑎
Direct Substitution Property for rational functions
If 𝑓(𝑥) is a rational function and 𝑎 is a number in the
domain of 𝑓
lim 𝑓(𝑥) = 𝑓(𝑎)
𝑥→𝑎
Example. lim √𝑢4 + 3𝑢 + 6
root law
𝑥→−2
= √ lim (𝑢4 + 3𝑢 + 6)
𝑥→−2
direct subst. for polys.
Example.
= √16 − 6 + 6 = 4
■
19a
19b
Example. Consider the following indeterminate form
Indeterminate Forms
Example. Let lim
𝑡 2 +𝑡−6
𝑡→2 𝑡 2 −4
√𝑥+2−1
𝑥→−1 𝑥+1
lim
=𝐿
0
=𝐿
type .
0
Rationalize the quotient and simplify as follows:
We call this an indeterminate form of type
0
0
since
√𝑥+2−1
𝑥+1
=
√𝑥+2−1 √𝑥+2+1
𝑥+1
√𝑥+2+1
direct substitution of 𝑡 = 2 into the rational function
gives that quotient, which is not defined. We cannot
use the quotient law!
Factor the numerator and denominator:
𝑡 2 +𝑡−6
𝑡 2 −4
= (𝑡+2)(𝑡−2)
if 𝑡 ≠ 2
if 𝑥 ≠ −1
=
1
√𝑥+2+1
but lim 𝑓(𝑥) does not depend on 𝑓(−1)! Then
𝐿 = lim
1
𝑥→−1 √𝑥+2+1
𝑡+3
1
= .
2
■
𝑡−2
Recall that lim 𝑓(𝑡) does not depend on 𝑓(2)! Then
𝑡→2
𝐿 = lim
√𝑥+2+1)
𝑥→−1
(𝑡+3)(𝑡−2)
=
(𝑥+1)
= (𝑥+1)(
𝑡+3
𝑡→2 𝑡−3
5
= .
4
■
Limits Involving Absolute Values
Recall that a limit exists iff the corresponding left and
right hand limits are equal.
lim 𝑓(𝑥) = 𝐿
Rationalization and Cancellation
𝑥→𝑎
⇔ (both lim 𝑓(𝑥) = 𝐿 and lim 𝑓(𝑥) = 𝐿)
𝑥→𝑎+
𝑥→𝑎−
20a
20b
Recall the piecewise definition of absolute value
|𝑧| = {
𝑧,
𝑧>0
.
−𝑧, 𝑧 < 0
Limits of Trig Functions
Use this when evaluating limits involving absolute
values.
Example. Let 𝐿 = lim3
𝑥→
𝜋
𝜋
2
2
− … 0 … … 𝑥-, sin(𝑥)
2 𝑥 2 −3𝑥
|2𝑥−3|
2
2𝑥 − 3, 𝑥 > 3/2
|2𝑥 − 3| = {
3 − 2𝑥, 𝑥 < 3/2
Find 𝐿1 = lim
3
2 𝑥 2 −3𝑥
𝑥→ + 2𝑥−3
2
= lim
3
𝑥(2𝑥−3)
𝑥→ + 2𝑥−3
2
lim sin(𝑥) = 0
= lim
𝑥=
3
𝑥→ +
𝑥→0
3
add 𝑥, slope agrees with sin(𝑥) at the origin
2
2
and 𝐿2 = lim
3
2 𝑥 2 −3𝑥
𝑥→ − 3−2𝑥
2
= lim
3
𝑥(2𝑥−3)
𝑥→ − 3−2𝑥
=
2
= lim
−𝑥 = −
3
𝑥→ −
2
𝐿 does not exist because 𝐿1 ≠ 𝐿2 .■
3
2
lim
𝑥→0
sin(𝑥)
𝑥
=1
(1)
21a
21b
This limit is a type 0/0 indeterminate form. However,
sin(𝑥)
the ratio
→ 1 as 𝑥 → 0. The text proves this
𝑥
using a geometric argument and the squeeze
theorem.
Example. Find L = lim
𝑡→0
𝐿 = lim
sin(2𝑡)
𝑡→0
2𝑡
sin(2𝑡)
(*)
𝑡
Thus, simplifying Equation (*) by writing “sin(2𝑡) =
2 sin(𝑡)” shows incorrect reasoning, even though it
leads to the correct answer. This would likely lead to
a loss of points on an exam. For full credit multiply
and divide by 2 as shown. ■
𝜋
𝜋
2
2
− … 0 … … 𝑥-, cos(𝑥)
⋅2
const. multiple law
= 2 ⋅ lim
𝑡→0
sin(2𝑡)
2𝑡
Let 𝑢 = 2𝑡. Notice that 𝑢 → 0 as 𝑡 → 0. Thus
𝐿 = 2 ⋅ lim
𝑢→0
sin(𝑢)
𝑢
by equation (1)
=2
By the graph it is clear that
lim cos(𝑥) = 1
WARNING By a trig identity
sin(2𝑡) = 2 sin(𝑡) cos(𝑡).
(2)
𝑥→0
Corollary. lim
𝜃→0
cos(𝜃)−1
𝜃
=0
22a
22b
Proof.
lim
𝜃→0
cos(𝜃)−1
?? Find L = lim
multiply by 1
𝜃
= lim
𝑥→0
cos(𝜃)−1 cos(𝜃)+1
𝜃
𝜃→0
cos(𝜃)+1
simplify numerator
= lim
cos2(𝜃)−1
𝜃→0 𝜃(cos(𝜃)+1)
sin2 (𝜃) + cos 2 (𝜃) = 1
= lim
− sin2 (𝜃)
𝜃→0 𝜃(cos(𝜃)+1)
algebra
= lim
𝜃→0
sin(𝜃) − sin(𝜃)
𝜃
cos(𝜃)+1
product law for limits
= lim
𝜃→0
sin(𝜃)
𝜃
⋅ lim
− sin(𝜃)
𝜃→0 cos(𝜃)+1
using (1), (2) and the quotient law
=0
One further example.
■
■
tan(2𝑥)
𝑥
.
23a
§1.5 Continuity
Informal definition A function is continuous is it can
be drawn without removing pencil from the paper.
… 𝑏 … 𝑎 … 𝑐 … 𝑥-. ,… 𝑦-, 𝑓 continuous on(𝑏, 𝑐),
𝑔 with hole at 𝑎
23b
Formal definition. A function 𝑓 is continuous at a
number 𝑎 if
lim 𝑓(𝑥) = 𝑓(𝑎)
𝑥→𝑎
if not, 𝑓 is discontinuous at 𝑎.
Three conditions required for continuity:
(1) 𝑓(𝑎) exists
(2)
lim 𝑓(𝑥) exists
𝑥→𝑎
(3)
lim 𝑓(𝑥) = 𝑓(𝑎)
𝑥→𝑎
Three types of discontinuity
A.
Infinite discontinuity
Example. 𝑓(𝑥) = 1/𝑥 2
sketch
𝑓 is continuous on (𝑏, 𝑐)
𝑔 is discontinuous at 𝑎.
common abbreviations: cts = continuous and dcts =
discontinuous.
𝑓 is discontinuous at 𝑥 = 0. (1) and (2) are violated.
24a
B.
24b
Jump discontinuity
1, 𝑥 ≥ 0
Example. 𝑔(𝑥) = {
0, 𝑥 < 0
Continuity on an open interval
sketch
If 𝑓 is continuous at each point of an open interval 𝐼,
we say 𝑓 is continuous on 𝐼.
Fact. Every polynomial is continuous at every real
number
Proof. Let 𝑃 be a polynomial. For any real no. 𝑎, by
the direct substitution property for polynomials
𝑔 is discontinuous at 𝑥 = 0. (2) is violated.
lim 𝑃(𝑥) = 𝑃(𝑎)
𝑥→𝑎
C.
Removable discontinuity
𝑥 + 1, 𝑥 ≠ 1
Example. ℎ(𝑥) = {
0,
𝑥=1
sketch
This is also the definition of continuity for a function
𝑃 at a point! Thus 𝑃 is continuous on (−∞, ∞). ■
Fact. Every rational function is continuous at every
point of its domain.
Proof. Let 𝑓 be a rational function and let 𝑎 ∈
domain of 𝑓. By the direct substitution property for
rational functions
ℎ is discontinuous at 𝑥 = 1. (3) is violated.
lim 𝑓(𝑥) = 𝑓(𝑎)
𝑥→𝑎
25a
25b
This is the definition of continuity for a function 𝑓 at
point 𝑎. ■
Example. 𝑓(𝑥) =
𝑥 2 −1
𝑥−1
is continuous on (−∞, 1) and
(1, ∞). ■
Fact. 𝑦 = tan(𝑥) is continuous at every 𝑥 except
𝜋
values 𝑥 = + 𝑛𝜋, where 𝑛 is an integer.
2
Proof.
lim tan(𝑥) = lim
𝑥→𝑎
Fact. sin(𝑥) and cos(𝑥) are continuous at every real
no. 𝑥.
𝜋
𝜋
2
2
−𝜋 … − … 0 … … 𝜋 … 𝑥-, … − 1 … 0 … 1 … 𝑦-,
curve of 𝑦 = sin(𝑥), curve of 𝑦 = cos(𝑥)
sin(𝑥)
quotient law
𝑥→𝑎 cos(𝑥)
=
lim sin(𝑥)
𝑥→𝑎
lim cos(𝑥)
cty of sin & cos
𝑥→𝑎
= tan(𝑎)
unless cos(𝑎) = 0.
𝜋
But from graph above, cos(𝑎) = 0 except at 𝑥 = +
2
𝑛𝜋, where 𝑛 is an integer. ■
No formal proof, but notice that these curves can be
drawn without lifting pencil from paper.■
26a
26b
Theorem (Arithmetic combinations of continuous
functions)
If 𝑓 and 𝑔 are continuous at no. 𝑎 and 𝑐 is a constant,
then the following combinations are continuous at 𝑎.
1.
𝑓+𝑔
sum
2.
𝑓−𝑔
difference
3.
𝑐𝑓
4.
𝑓𝑔
5.
𝑓/𝑔 provided 𝑔(𝑎) ≠ 0 quotient
constant multiple
This is the definition of continuity of 𝑓(𝑥) ⋅ 𝑔(𝑥)
Continuity from the left and from the right
A function 𝑓 is continuous from the right at no. 𝑎 if
lim 𝑓(𝑥) = 𝑓(𝑎)
𝑥→𝑎+
and 𝑓 is continuous from the left and no. 𝑎 if
lim 𝑓(𝑥) = 𝑓(𝑎)
𝑥→𝑎−
product
Example. Step Function
𝑔(𝑥) = {
Proof. Each part follows from the corresponding limit
law. For example, consider (4)
lim 𝑓(𝑥) ⋅ 𝑔(𝑥)
product law for limits
𝑥→𝑎
= lim 𝑓(𝑥) ⋅ lim 𝑔(𝑥)
𝑥→𝑎
𝑥→𝑎
continuity of 𝑓 and 𝑔
= 𝑓(𝑎) ⋅ 𝑔(𝑎)
1,
0,
… 0 … 𝑥-, 0 … 1 … 𝑦-, 𝑔(𝑥)
𝑥≥0
𝑥<0
■
27a
𝑔(𝑥) is continuous from the right at 𝑥 = 0 and
continuous at every other 𝑥. ■
Example. Greatest integer function
⟦𝑥⟧ is the largest integer less than or equal to 𝑥
−1 … 0 … 1 … 𝑥-, −1 … 2 … 𝑦-, 𝑦 = ⟦𝑥 ⟧
27b
A function 𝑓 is continuous on an interval if it is
continuous at every no. on the interval. Continuity
at an endpoint means continuity from the right or
from the left.
Example. 𝑓(𝑥) = √𝑥 is continuous on [0, ∞).
Why? For any 𝑎 > 0, by the limit laws for roots
lim √𝑥 = √ lim 𝑥 = √𝑎
𝑥→𝑎
𝑥→𝑎
For 𝑎 = 0, lim √𝑥 = 0 = √𝑎
𝑥→0+
which is “obvious” from the graph
0 … 𝑥-, 0 … 𝑦-, 𝑦 = √𝑥
𝑦 = ⟦𝑥⟧ is continuous from the right at every integer.
Continuity on an interval (including endpoints)
28a
28b
■
Example. 𝐹(𝑥) = cos(√𝑥)
𝑔(𝑥) = √𝑥
?? Practice with continuity transparency.
𝑓(𝑢) = cos(𝑢)■
Continuity and Compositions
Theorem (Limits of Compositions)
Composition: Function Machine Picture
If 𝑓 is continuous at 𝑏 and
𝑥 → [𝑔] → 𝑢 = 𝑔(𝑥)
lim 𝑔(𝑥) = 𝑏
𝑥→𝑎
then
lim 𝑓(𝑔(𝑥)) = 𝑓(lim 𝑔(𝑥)) = 𝑓(𝑏)
𝑥→𝑎
𝑥→𝑎
Intuitively, if 𝑥 is close to 𝑎 then 𝑔(𝑥) is close to 𝑏.
Since 𝑓 is continuous at 𝑏, if 𝑔(𝑥) is close to 𝑏 then
𝑓(𝑔(𝑥)) is close to 𝑓(𝑏).
𝑢 → [𝑓] → 𝑦 = 𝑓(𝑢)
Example. Evaluate lim2 cos(√𝑥) = 𝐿
𝑥→𝜋
𝑦 = 𝑓(𝑢) = 𝑓(𝑔(𝑥))
𝑔 is the ‘inner function’
𝑓 is the ‘outer function’
From the root law,
lim √𝑥 = 𝜋
𝑥→𝜋2
𝑓(𝑢) = cos(𝑢) is continuous at 𝑢 = 𝜋
29a
29b
By the theorem
𝐿 = cos( lim2 √𝑥) = cos(𝜋) = −1
𝑥→𝜋
■
Combine the theorem above with the condition that
𝑏 = 𝑔(𝑎), in other words the condition that 𝑔 is
continuous at 𝑎, to get:
In words, a continuous function of a continuous
function is continuous.
Example. Where is 𝐹(𝑥) = cos(√𝑥) continuous?
Note that 𝐹(𝑥) = 𝑓(𝑔(𝑥)) where
Theorem (Compositions of continuous functions)
If 𝑔 is continuous at a no. 𝑎 and 𝑓 is continuous at
𝑔(𝑎), then 𝑓(𝑔(𝑥)) is continuous at 𝑎.
Proof. From the theorem above
lim 𝑓(𝑔(𝑥)) = 𝑓 (lim 𝑔(𝑥))
𝑥→𝑎
𝑥→𝑎
= 𝑓(𝑔(𝑎))
This is the definition of continuity for 𝑓(𝑔(𝑥)). ■
𝑔(𝑥) = √𝑥
is continuous for 𝑥 > 0
𝑓(𝑢) = cos(𝑢)
is continuous for any real no. 𝑢
By the theorem, 𝐹 is continuous for 𝑥 ≥ 0. ■
Intermediate Value Theorem
Let 𝑓 be continuous on the closed interval [𝑎, 𝑏] with
𝑓(𝑎) ≠ 𝑓(𝑏). If 𝑁 is any number between 𝑓(𝑎) and
𝑓(𝑏), then there is a no. 𝑐 in (𝑎, 𝑏) such that 𝑓(𝑐) =
𝑁.
Idea
… 𝑎 … 𝑏 … 𝑥-, 0 … 𝑓(𝑏) … 𝑓(𝑎) … 𝑦-, 𝑓, 𝑐, 𝑁
30a
30b
from the Intermediate Value Theorem that there is a
𝜋
𝑐 ∈ (0, ) such that 𝑓(𝑐) = 0. ■
2
§1.6 Limits involving infinity
Proof. Not given, but the intermediate value
theorem is “obvious”.
Infinite Limits
Example. 𝑦 = 𝑔(𝑥) = 1/𝑥.
−2 … 0 … 2 … 𝑥-, −2 … 0 … 2 … 𝑦-, 𝑔(𝑥)
Example. Prove that there is an 𝑥 that solves
sin(𝑥) = 1 − 𝑥
𝜋
on the interval (0, ).
2
Proof. Let 𝑓(𝑥) = 1 − 𝑥 − sin(𝑥)
then 𝑓(0) = 1 − 0 − sin(0) = 1 > 0
𝜋
𝜋
𝜋
𝑓( ) = 1− −1 = − < 0
2
2
2
𝜋
Since 𝑁 = 0 lies between 𝑓(0) > 0 and 𝑓 ( ) < 0,
2
and 𝑓 is an arithmetic combination of continuous
functions (and therefore continuous itself), it follows
We write
lim 𝑔(𝑥) = ∞
𝑥→0+
31a
31b
The symbol ∞ is a version of DNE. It means 𝑔(𝑥)
becomes greater than any fixed value.
?? lim ℎ(𝑥) =
𝑥→3+
?? lim ℎ(𝑥) =
𝑥→3−
We write lim 𝑔(𝑥) = −∞
𝑥→0−
The symbol −∞ means that 𝑔(𝑥) becomes less than
any fixed value ■
Example. ℎ(𝑥) = 1/(𝑥 − 3)
2
0 … 6 … 𝑥-, 0 … 4 … 𝑦-, ℎ(𝑥)
The one sided limit are “equal” (they DNE in the
same way)
Thus we write lim ℎ(𝑥) = ∞.
𝑥→3
The vertical line 𝑥 = 𝑎 is a vertical asymptote of 𝑦 =
𝑓(𝑥) if
1)
lim 𝑓(𝑥) = +∞ or −∞
𝑥→𝑎+
and / or
2)
lim 𝑓(𝑥) = +∞ or −∞
𝑥→𝑎−
𝑥 = 𝑎 is a vertical asymptote. add 𝑥 = 3
Example. 𝑦 = log 2 (𝑥)
domain is ℝ+ = { positive numbers}
Table x/ ¼ =2-2, ½ =2-1, 1=20, 2=21, 4=22 / y …
32a
32b
The y-axis (or line 𝑥 = 0) is a vertical asymptote. ■
0 … 4 … 𝑥-, −2 … 0 … 2 … 𝑦-, dots, curve
Limits at Infinity
Example. 𝑓(𝑥) = 𝑥 2 /(𝑥 2 + 1)
Note that 𝑓 has even symmetry, so we only need to
tabulate values of 𝑥 ≥ 0.
Table: x/ 0, 1, 2, 3 / f(x) …
−3 … 0 … 3 … 𝑥-, 0 … 1 … 𝑦-, dots, curve
33a
33b
The line 𝑦 = 𝐿 is a horizontal asymptote of 𝑦 = 𝑓(𝑥)
if
lim 𝑓(𝑥) = 𝐿
𝑥→∞
𝑓(𝑥) approaches 1 as |𝑥| increases ■
and/or
lim 𝑓(𝑥) = 𝐿
𝑥→−∞
?? Transparency: Practice with asymptotes
Definitions
lim 𝑔(𝑥) = 𝐿
Consider lim 1/𝑥
means that 𝑔(𝑥) can be made as close as we wish to
𝐿 by taking |𝑥| sufficiently large with 𝑥 > 0.
For 𝑥 > 10, 0 < <
𝑥→∞
𝑥→∞
1
1
𝑥
10
1
1
𝑥
100
For 𝑥 > 100, 0 < <
lim 𝑔(𝑥) = 𝐿
𝑥→−∞
means that 𝑔(𝑥) can be made as close as we wish to
𝐿 by taking |𝑥| sufficiently large with 𝑥 < 0.
We can make 1/𝑥 as close as we wish to zero by
taking 𝑥 sufficiently large. Conclude
lim
1
𝑥→∞ 𝑥
=0
In the previous example,
lim 𝑓(𝑥) = 1,
𝑥→∞
and
lim 𝑓(𝑥) = 1
𝑥→−∞
Consider
lim 1/𝑥
𝑥→−∞
For 𝑥 < −10,
−
1
10
1
< <0
𝑥
34a
34b
For 𝑥 < −100, −
1
1
100
These are obtained from the limit laws we have
already seen by replacing 𝑎 with ±∞.
< <0
𝑥
We can make 1/𝑥 as close as we wish to zero by
taking |𝑥| sufficiently large with 𝑥 < 0. Conclude
lim
1
𝑥→−∞ 𝑥
Suppose that 𝑐 is a constant and
lim 𝑓(𝑥),
=0
𝑥→±∞
1.
lim 𝑔(𝑥)exist.
𝑥→±∞
sum law
(limit of sum is sum of limits)
lim [𝑓(𝑥) + 𝑔(𝑥)] = lim 𝑓(𝑥) + lim 𝑔(𝑥)
𝑥→±∞
2.
Generalize
1
𝑥→∞ 𝑥 𝑟
difference law
𝑥→±∞
=0
Equation (1)
3.
lim
1
𝑥→∞ 𝑥 2
lim
1
𝑥→∞ √𝑥
𝑥→±∞
𝑥→±∞
constant multiple law
lim 𝑐 𝑓(𝑥) = 𝑐 lim 𝑓(𝑥)
we won’t prove this
Examples.
𝑥→±∞
lim [𝑓(𝑥) − 𝑔(𝑥)] = lim 𝑓(𝑥) − lim 𝑔(𝑥)
If 𝑟 is a positive real number
lim
𝑥→±∞
𝑥→±∞
4.
=0
= lim
1
1
𝑥→∞ 𝑥 2
𝑥→±∞
product law
lim 𝑓(𝑥)𝑔(𝑥) = lim 𝑓(𝑥) ⋅ lim 𝑔(𝑥)
=0 ■
𝑥→±∞
5.
𝑥→±∞
𝑥→±∞
quotient law
lim 𝑓(𝑥)
𝑓(𝑥) 𝑥→±∞
=
𝑥→±∞ 𝑔(𝑥)
lim 𝑔(𝑥)
lim
Infinite Limit Laws
𝑥→±∞
35a
35b
Cannot use quotient law because 𝑥 + 5 → 0− as 𝑥 →
−5−
so long as lim 𝑔(𝑥) ≠ 0
𝑥→±∞
Notice that as 𝑥 → −5− we have 𝑥 − 3 → −8−
2𝑥+3
Example. Find lim
𝑥→∞ √16𝑥 2 +5
𝐿 = lim
= 𝐿.
2𝑥+3
multiply by 1
𝑥→∞ √16𝑥 2 +5
= lim
2𝑥+3
1/𝑥
𝑥→∞ √16𝑥 2 +5 1/𝑥
(note 𝑥 ≠ 0)
Since both the numerator and denominator are
negative as 𝑥 → −5− , the ratio must be positive.
Since the denominator is approaching 0 and the
numerator is not
lim −
𝑥−3
𝑥→−5 𝑥+5
multiply numerator and denominator by 1/𝑥
= lim
2+3/𝑥
quotient law
𝑥→∞ √16+5/𝑥 2
=
lim ( 2+3/𝑥)
𝑥→∞
lim √16+5/𝑥 2
(denominator is not zero)
𝑥→∞
=
=
2+0
limit laws and equation 1
√16+0
1
2
■
Example. Find lim −
𝑥−3
𝑥→−5 𝑥+5
=∞■