1.1 Function Machines Course Administration1a 1b Course Administration Math Placement Syllabus (show...
Transcript of 1.1 Function Machines Course Administration1a 1b Course Administration Math Placement Syllabus (show...
1a 1b
Course Administration
Math Placement
Syllabus (show text)
WebAssign, class keys on syllabus
FDOC_self_enrollment.ppt
Your first homework assignment:
(1) self enroll in WebAssign
(2) Intro to WebAssign
(3) 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.
§1.1 Function Machines
sketch x [f] y = f(x), input, output
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 = 𝑥2, −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 = 𝑥2, 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. 𝑥 = 𝑥, domain not given.
Assume domain= [0, ∞). range = [0,∞). ■
Graphs
Example. 𝑦 = 𝑥2
sketch …− 1 … 0 … 1 …𝑥- (independent variable),
0 …𝑦- (dependent variable), curve■
Example. 𝑦 = 𝑥
sketch 0 … 4 …𝑥-, 0 … 2 …𝑦-, curve■
3a 3b
Example. Graph the set of points that satisfy 𝑦2 = 𝑥.
Table x / 0, 1, 4; y/ 0, ±1, ±2
sketch 0 … 4 …𝑥-, −2 … 0 … 2 …𝑦-, curve
Is this a function?
?? Why? ■
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 𝑥2 + 𝑦2 = 1.
axes, circle, vertical line intersecting circle twice
?? Is this the graph of 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.
Example. 𝑓 𝑥 = 𝑥2,
𝑓 −𝑥 = −𝑥 2 = −𝑥 −𝑥 = 𝑥2 = 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 = 𝑥3,
𝑔 −𝑥 = −𝑥 −𝑥 −𝑥 = −𝑥3 = −𝑔(𝑥) .
Table x/ 0, -1, 1, -2, 2; y/ 0, -1, 1, -8, 8
sketch −2 … 0 … 2 …𝑥-, −8 … 0 … 8 …𝑦-, curve
?? If 𝑔 is odd, its graph is symmetric wrt what? ■
Knowing even or odd symmetry helps us sketch
functions.
5a 5b
§1.2 Catalog Of Functions
Straight lines
slope intercept form
𝑦 = 𝑚𝑥 + 𝑏
sketch … 0 …𝑥- axis, 0 …𝑦- axis, intercept 𝑏,
line,(𝑥, 𝑦),Δ𝑥, Δ𝑦
slope = Δ𝑦
Δ𝑥=
𝑦−𝑏
𝑥−0= 𝑚
Example. 𝑦 =1
2𝑥 + 1
Table
sketch −2 … 0 … 2 …𝑥-, 0 … 1 … 2 …𝑦-, intercept,
line
■
point slope form
𝑦 − 𝑦1 = 𝑚(𝑥 − 𝑥1)
sketch 0 …𝑥-, …𝑦-,(𝑥1, 𝑦1), line, (𝑥, 𝑦),𝑥 − 𝑥1 = 𝛥𝑥,
𝑦 − 𝑦1 = 𝛥𝑦
𝛥𝑦
𝛥𝑥= 𝑚
6a 6b
Example. 𝑦 − 2 =1
2(𝑥 − 2)
convert to slope intercept form
solve for y
𝑦 =
same as previous example ■
Power Functions
general form
𝑦 = 𝑥𝛼 where 𝛼 is a constant
Example 𝑦 = 𝑥 straight line
?? slope ?? y-intercept
Example 𝑦 = 𝑥2 parabola
Example 𝑦 = 𝑥1
2 square root (𝑥1
2 = 𝑥)
Example 𝑦 = 𝑥−1 (𝑥−1 = 1/𝑥)
sketch −2 … 0 … 2 …𝑥-, −2 … 0 … 2 …𝑦-
■
7a 7b
Polynomial Functions
Example 𝑦 = 1 − 𝑥2
sketch −2 … 0 … 2 …𝑥-, −2 … 0 … 1 …𝑦-, parabola
■
A quadratic function is a polynomial of degree 2.
𝑦 = 𝑎2𝑥2 + 𝑎1𝑥 + 𝑎0
𝑎2, 𝑎1 , 𝑎0 are constants add
The degree of a polynomial is the highest power that
it contains.
A polynomial of degree 𝑛 has the form
𝑃 𝑥 = 𝑎𝑛𝑥𝑛 + 𝑎𝑛−1𝑥
𝑛−1 + ⋯ + 𝑎2𝑥2 + 𝑎1𝑥 + 𝑎0.
Piecewise defined functions
Example 𝑓 𝑥 = 𝑥 + 1, 𝑥 ≠ 11, 𝑥 = 1
sketch −1 … 2 …𝑥-, 0 … 3 …𝑦-, line with hole, dot
■
Example Absolute Value Function 𝑓 𝑥 = |𝑥|
gives distance from the origin on the real number line
sketch −2 … 0 … 2, bracket −2 and 0
distance between −2 and 0 is 2, −2 = 2
8a 8b
graph 𝑦 = |𝑥|
sketch −2 … 0 … 2 …𝑥-, 0 … 2 …𝑦-, graph
piecewise definition 𝑥 = 𝑥, 𝑥 > 0−𝑥, 𝑥 < 0
■
Rational Functions
A rational function is a ratio of two polynomials
𝑓 𝑥 =𝑃(𝑥)
𝑄(𝑥)
where 𝑃, 𝑄 are polynomials.
Example (a case of special interest to us)
𝑓 𝑥 =𝑥2−1
𝑥−1
?? domain of f?
Important Algebraic Trick!
𝑥 − 1 𝑥 + 1 =
In general, 𝑥 − 𝑎 𝑥 + 𝑎 = 𝑥2 − 𝑎2. Then
𝑓 𝑥 = 𝑥 − 1 (𝑥 + 1)
𝑥 − 1=
𝑥 + 1, if 𝑥 ≠ 1 undefined, if 𝑥 = 1
sketch −1 … 0 … 2 …𝑥-, 0 … 2 …𝑦-, line with hole
■
9a 9b
Sine and Cosine
sketch – 𝜋, … 0 …3𝜋
2…𝑥-, −1 … 0 … 1.., sin(𝑥),
cos(𝑥)
properties of sine
−1 ≤ sin 𝑥 ≤ 1
sin 0 = 0, sin 𝜋 = 0, sin 2𝜋 = 0, generally
sin 𝑛𝜋 = 0 for 𝑛 an integer
sin 𝜋
2 = 1, sin −
𝜋
2 = −1
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
cos 𝜋
2 = 0, cos
3𝜋
2 = 0, cos
5𝜋
2 = 0, generally
cos 𝑛 +1
2 𝜋 = 0 for 𝑛 an integer
cos 0 = 1, cos 𝜋 = −1
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
sketch 45-45-90 triangle, lengths of sides
Pythagorean theorem: 1
2+
1
2= 1
sin 𝜋
4 = cos
𝜋
4 =
1
2=
2
2≈ 0.71
sketch 30-60-90 triangle, lengths of sides
Pythagorean theorem: 3
4+
1
4= 1
sin 𝜋
6 = cos
𝜋
3 =
1
2
sin 𝜋
3 = cos
𝜋
6 =
3
2≈ 0.87
Tangent
tan 𝑥 =sin 𝑥
cos 𝑥
tan −𝑥 =
?? symmetry
11a 11b
graph tangent
sketch –𝜋
2… 0 …
3𝜋
2…𝑥-, … 0 …𝑦-, vert. asymptotes,
curve
−𝜋
2,
𝜋
2,
3𝜋
2, … not in the domain of tan ?? why
tan 𝑥 + 𝜋 =
period 𝜋
tan 0 = tan 𝜋 = ⋯ = 0
tan 𝜋
4 =
sin 𝜋
4
cos 𝜋
4
= 1
similarly find tan 𝜋
6 =
1
3 and tan
𝜋
3 = 3
§1.3 The Limit of a Function
sketch …𝑎… . , …𝐿…, curve 𝑓, hole at 𝑎
Informal definition of limit
lim𝑥→𝑎 𝑓 𝑥 = 𝐿
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).
Alternate notation: 𝑓 𝑥 → 𝐿 as 𝑥 → 𝑎
12a 12b
Example. 𝑓 𝑥 = 𝑥 + 1
… 1 … , … 1 … 2 …, line 𝑓
imagine a bug approaching 𝑥 = 1 on either side
add arrows towards 𝑥 = 1, arrows towards 𝑦 = 2
lim𝑥→1 𝑓 𝑥 = 2
lim𝑥→𝑎 𝑓(𝑥)has nothing to do with 𝑓(𝑎)
Example 𝑔 𝑥 = 𝑥 + 1, 𝑥 ≠ 11, 𝑥 = 1
sketch … 1 …𝑥-, … 1 … 2 …, line with hole, dot
add arrows towards 𝑥 = 1, arrows towards 𝑦 = 2
lim𝑥→1 𝑔 𝑥 = 2 ■
13a 13b
Limit is a 2-sided concept
Example. Step function
𝐻 𝑥 = 0, 𝑥 < 01, 𝑥 ≥ 0
… 0 …𝑥-, 0 … 1 …, 𝐻(𝑥)
lim𝑥→0 𝐻(𝑥) DNE (does not exist)■
One sided limits
Informal definition
lim𝑥→𝑎− 𝑓 𝑥 = 𝐿1 limit from left or left hand limit
means we can make 𝑓(𝑥) as close as we wish to 𝐿1
by taking 𝑥 sufficiently close to 𝑎 from the left.
lim𝑥→𝑎+ 𝑓 𝑥 = 𝐿2 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)
?? lim𝑥→0+ 𝐻 𝑥 =
?? lim𝑥→0− 𝐻 𝑥 =
■
14a 14b
the (2-sided) limit
lim𝑥→𝑎 𝑓 𝑥 = 𝐿
if and only if
lim𝑥→𝑎+ 𝑓 𝑥 = 𝐿 and lim𝑥→𝑎− 𝑓 𝑥 = 𝐿
?? Practice with limits
Precise definition of a limit
0 …𝑎…𝑥-, 0 …𝐿…𝑦-, 𝑓(𝑥), hole at 𝑥 = 𝑎, dot
lim𝑥→𝑎 𝑓 𝑥 = 𝐿 means
For every 𝜖 > 0 there is a 𝛿 > 0 such that
if 𝑎 − 𝛿 < 𝑥 < 𝑎 + 𝛿 (𝑥 ≠ 𝑎)
then 𝐿 − 𝜖 < 𝑓 𝑥 < 𝐿 + 𝜖
add 𝑎 − 𝛿, 𝑎 + 𝛿, 𝐿 − 𝜖, 𝐿 = 𝜖, segments
§1.4 Calculating Limits
Two Special Limits
A. Let 𝑐 be a constant
lim𝑥→𝑎 𝑐 = 𝑐
0 …𝑎…𝑥-, 0 …𝑦-, line 𝑦 = 𝑐, arrows approaching
𝑥 = 𝑎
15a 15b
B. Consider 𝑓 𝑥 = 𝑥
lim𝑥→𝑎 𝑥 = 𝑎
0 …𝑎…𝑥-, 0 …𝑎…𝑦-, 𝑓 𝑥 = 𝑥
add arrows approaching 𝑥 = 𝑎, add arrows
approaching 𝑦 = 𝑎
Five Limit Laws
Suppose that 𝑐 is a constant and
lim𝑥→𝑎 𝑓(𝑥), lim𝑥→𝑎 𝑔(𝑥) exist.
1. sum law (limit of sum is sum of limits)
lim𝑥→𝑎 𝑓 𝑥 + 𝑔 𝑥 = lim𝑥→𝑎 𝑓 𝑥 + lim𝑥→𝑎 𝑔(𝑥)
2. difference law
lim𝑥→𝑎 𝑓 𝑥 − 𝑔 𝑥 = lim𝑥→𝑎 𝑓 𝑥 − lim𝑥→𝑎 𝑔(𝑥)
3. constant multiple law
lim𝑥→𝑎 𝑐 𝑓 𝑥 = 𝑐 lim𝑥→𝑎 𝑓(𝑥)
4. product law
lim𝑥→𝑎 𝑓 𝑥 𝑔 𝑥 = lim𝑥→𝑎 𝑓 𝑥 ⋅ lim𝑥→𝑎 𝑔(𝑥)
5. quotient law
lim𝑥→𝑎𝑓 𝑥
𝑔 𝑥 =
lim 𝑥→𝑎 𝑓(𝑥)
lim 𝑥→𝑎 𝑔(𝑥) ,
so long as lim𝑥→𝑎 𝑔 𝑥 ≠ 0
16a 16b
Example
lim𝑥→1 3𝑥 + 5 sum law
= lim𝑥→1
3𝑥 + lim𝑥→1 5
constant multiple law
= 3 lim𝑥→1 𝑥 + lim𝑥→1 5
special limits
= 3 ⋅ 1 + 5 = 8 ■
Example
lim𝑥→2𝑥2
𝑥+1 quotient law
=lim 𝑥→2 𝑥2
lim 𝑥→2(𝑥+1)
product and sum laws
= lim 𝑥→2 𝑥⋅lim 𝑥→2 𝑥
lim 𝑥→2 𝑥+lim 𝑥→2 1
special limits
=2⋅2
2+1 =
4
3 ■
Repeated Application of the Product Law
lim𝑥→𝑎 𝑓 𝑥 ⋅ 𝑔 𝑥 ⋅ 𝑥 = lim𝑥→𝑎 𝑓 𝑥 ⋅ 𝑘(𝑥)
where 𝑔 ⋅ = 𝑘
product law
= lim𝑥→𝑎 𝑓 𝑥 ⋅ lim𝑥→𝑎 𝑘(𝑥)
product law
= lim𝑥→𝑎 𝑓 𝑥 ⋅ lim𝑥→𝑎 𝑔 𝑥 ⋅ lim𝑥→𝑎 (𝑥)
Power Law
Suppose 𝑓 𝑥 = 𝑔 𝑥 = (𝑥) then from above
lim𝑥→𝑎 𝑓 𝑥 3 = lim𝑥→𝑎 𝑓(𝑥) 3
Apply the same reasoning to a product of 𝑛 factors of
𝑓(𝑥) to get the Power Law:
lim𝑥→𝑎 𝑓 𝑥 𝑛 = lim𝑥→𝑎 𝑓(𝑥) 𝑛
where 𝑛 is any positive integer
17a 17b
Example. Cubic Polynomial.
lim𝑥→2 𝑥3 − 4𝑥 difference law
= lim𝑥→2
𝑥3 − lim𝑥→2
4𝑥
power, constant multiple laws
= lim𝑥→2 𝑥 3 = 4 lim𝑥→2 𝑥
special limits
= 23 − 4 ⋅ 2 = 0 ■
Recall polynomials of degree 𝑛. Their general form is
𝑃 𝑥 = 𝑐𝑛𝑥𝑛 + 𝑐𝑛−1𝑥
𝑛−1 + ⋯ + 𝑐1𝑥 + 𝑐0
where 𝑐𝑛 , 𝑐𝑛−1, … , 𝑐1, 𝑐0 are constants.
By reasoning similar to the cubic polynomial example
lim𝑥→𝑎 𝑃 𝑥 sum law, const. multiple law
= 𝑐𝑛 lim𝑥→𝑎 𝑥𝑛 + ⋯ + 𝑐1 lim𝑥→𝑎 𝑥 + lim𝑥→𝑎 𝑐0
power law, special limit
= 𝑐𝑛𝑎𝑛 + ⋯𝑐1𝑎 + 𝑐0
= 𝑃(𝑎)
we have discovered the following
Direct Substitution Property for polynomials
If 𝑃(𝑥) is any polynomial and 𝑎 is a real number then
lim𝑥→𝑎 𝑃 𝑥 = 𝑃(𝑎).
This is much easier to apply than the limit laws!
Recall that a rational function is a ratio of two
polynomials.
𝑓 𝑥 =𝑃 𝑥
𝑄 𝑥 , where 𝑃 and 𝑄 are polynomials
Let 𝑓(𝑥) be any rational function
lim𝑥→𝑎 𝑓 𝑥 =lim 𝑥→𝑎 𝑃(𝑥)
lim 𝑥→𝑎 𝑄(𝑥) quotient law
=𝑃(𝑎)
𝑄(𝑎) direct subst. for polys
so long as 𝑄 𝑎 ≠ 0.
we now have a …
18a 18b
Direct Substitution Property for rational functions
If 𝑓(𝑥) is a rational function and 𝑎 is a number in the
domain of 𝑓
lim𝑥→𝑎
𝑓 𝑥 = 𝑓(𝑎)
Example.
lim𝑥→1
𝑥4 + 𝑥2 − 6
𝑥4 − 2𝑥 + 3=
14 + 12 − 6
14 + 2 ⋅ 1 + 3=
−4
6=
−2
3
by Direct Substitution for rational functions! ■
Root Law For Limits
Let 𝑛 be a positive integer
lim𝑥→𝑎
𝑓(𝑥)𝑛
= lim𝑥→𝑎
𝑓(𝑥)𝑛 ,
If 𝑛 is even then require lim𝑥→𝑎 𝑓 𝑥 > 0.
Example. lim𝑥→−2 𝑢4 + 3𝑢 + 6 root law
= lim𝑥→−2
(𝑢4 + 3𝑢 + 6)
direct subst. for polys.
= 16 − 6 + 6 = 4 ■
Indeterminate Forms
Example. Let lim𝑡→2𝑡2+𝑡−6
𝑡2−4= 𝐿
We call this an indeterminate form of type 0
0 since
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:
𝑡2+𝑡−6
𝑡2−4=
𝑡+3 (𝑡−2)
𝑡+2 (𝑡−2) if 𝑡 ≠ 2
=𝑡+3
𝑡−2
19a 19b
Recall that lim𝑡→2 𝑓(𝑡) does not depend on 𝑓(2)!
Then
𝐿 = lim𝑡→2𝑡+3
𝑡−3=
5
4. ■
Rationalization and Cancellation
Example. Consider the following indeterminate form
lim𝑥→−1 𝑥+2−1
𝑥+1= 𝐿 type
0
0.
Rationalize the quotient and simplify as follows:
𝑥+2−1
𝑥+1=
𝑥+2−1
𝑥+1 𝑥+2+1
𝑥+2+1=
(𝑥+1)
𝑥+1 ( 𝑥+2+1)
if 𝑥 ≠ −1
=1
𝑥+2+1
but lim𝑥→−1 𝑓(𝑥) does not depend on 𝑓(−1)! Then
𝐿 = lim𝑥→−11
𝑥+2+1=
1
2. ■
Limits Involving Absolute Values
Recall that a limit exists iff the corresponding left and
right hand limits are equal.
lim𝑥→𝑎 𝑓 𝑥 = 𝐿
⇔ (both lim𝑥→𝑎+
𝑓 𝑥 = 𝐿 and lim𝑥→𝑎−
𝑓 𝑥 = 𝐿)
Recall the piecewise definition of absolute value
𝑧 = 𝑧, 𝑧 > 0−𝑧, 𝑧 < 0
.
Use this when evaluating limits involving absolute
values.
Example. Let 𝐿 = lim𝑥→
3
2
2 𝑥2−3𝑥
|2𝑥−3|
2𝑥 − 3 = 2𝑥 − 3, 𝑥 > 3/23 − 2𝑥, 𝑥 < 3/2
Find 𝐿1 = lim𝑥→
3
2+
2 𝑥2−3𝑥
2𝑥−3= lim
𝑥→3
2+
𝑥(2𝑥−3)
2𝑥−3
= lim𝑥→
3
2+𝑥 =
3
2
20a 20b
and 𝐿2 = lim𝑥→
3
2−
2 𝑥2−3𝑥
3−2𝑥= lim
𝑥→3
2−
𝑥(2𝑥−3)
3−2𝑥=
= lim𝑥→
3
2−−𝑥 = −
3
2
𝐿 does not exist because 𝐿1 ≠ 𝐿2.■
Limits of Trig Functions
−𝜋
2… 0 …
𝜋
2…𝑥-, sin(𝑥)
lim𝑥→0 sin 𝑥 = 0
add 𝑥, slope agrees with sin(𝑥) at the origin
(1)
This limit is a type 0/0 indeterminate form. However,
the ratio sin 𝑥
𝑥→ 1 as 𝑥 → 0. The text proves this
using a geometric argument and the squeeze
theorem.
Example. Find L = lim𝑡→0sin (2𝑡)
𝑡 (*)
𝐿 = lim𝑡→0sin (2𝑡)
2𝑡⋅ 2
const. multiple law
= 2 ⋅ lim𝑡→0sin (2𝑡)
2𝑡
Let 𝑢 = 2𝑡. Notice that 𝑢 → 0 as 𝑡 → 0. Thus
𝐿 = 2 ⋅ lim𝑢→0sin (𝑢)
𝑢 by equation (1)
= 2
lim𝑥→0sin (𝑥)
𝑥= 1
21a 21b
WARNING By a trig identity
sin 2𝑡 = 2 sin 𝑡 cos(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
2… 0 …
𝜋
2…𝑥-, cos(𝑥)
By the graph it is clear that
lim𝑥→0 cos 𝑥 = 1 (2)
Corollary. lim𝜃→0cos 𝜃 −1
𝜃= 0
Proof.
lim𝜃→0cos 𝜃 −1
𝜃 multiply by 1
= lim𝜃→0cos 𝜃 −1
𝜃
cos 𝜃 +1
cos 𝜃 +1
simplify numerator
= lim𝜃→0cos 2 𝜃 −1
𝜃(cos 𝜃 +1)
sin2 𝜃 + cos2(𝜃) = 1
= lim𝜃→0− sin 2(𝜃)
𝜃(cos 𝜃 +1)
algebra
= lim𝜃→0sin (𝜃)
𝜃
− sin 𝜃
cos 𝜃 +1
product law for limits
= lim𝜃→0sin (𝜃)
𝜃⋅ lim𝜃→0
− sin 𝜃
cos 𝜃 +1
using (1), (2) and the quotient law
= 0 ■
22a 22b
One further example.
?? Find L = lim𝑥→0tan (2𝑥)
𝑥.
■
§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
𝑓 is continuous on (𝑏, 𝑐)
𝑔 is discontinuous at 𝑎.
common abbreviations: cts = continuous and dcts =
discontinuous.
23a 23b
Formal definition. A function 𝑓 is continuous at a
number 𝑎 if
lim𝑥→𝑎
𝑓 𝑥 = 𝑓(𝑎)
if not, 𝑓 is discontinuous at 𝑎.
Three conditions required for continuity:
(1) 𝑓(𝑎) exists
(2) lim𝑥→𝑎 𝑓(𝑥) exists
(3) lim𝑥→𝑎 𝑓 𝑥 = 𝑓(𝑎)
Three types of discontinuity
A. Infinite discontinuity
Example. 𝑓 𝑥 = 1/𝑥2 sketch
𝑓 is discontinuous at 𝑥 = 0. (1) and (2) are violated.
B. Jump discontinuity
Example. 𝑔 𝑥 = 1, 𝑥 ≥ 00, 𝑥 < 0
sketch
𝑔 is discontinuous at 𝑥 = 0. (2) is violated.
C. Removable discontinuity
Example. 𝑥 = 𝑥 + 1, 𝑥 ≠ 10, 𝑥 = 1
sketch
is discontinuous at 𝑥 = 1. (3) is violated.
24a 24b
Continuity on an open interval
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
lim𝑥→𝑎
𝑃 𝑥 = 𝑃(𝑎)
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
lim𝑥→𝑎
𝑓 𝑥 = 𝑓(𝑎)
This is the definition of continuity for a function 𝑓 at
point 𝑎. ■
Example. 𝑓 𝑥 =𝑥2−1
𝑥−1 is continuous on (−∞, 1) and
(1,∞). ■
Fact. sin(𝑥) and cos(𝑥) are continuous at every
real no. 𝑥.
−𝜋…−𝜋
2… 0 …
𝜋
2…𝜋…𝑥-, …− 1 … 0 … 1 …𝑦-,
curve of 𝑦 = sin(𝑥), curve of 𝑦 = cos(𝑥)
No formal proof, but notice that these curves can be
drawn without lifting pencil from paper.■
25a 25b
Fact. 𝑦 = tan(𝑥) is continuous at every 𝑥 except
values 𝑥 =𝜋
2+ 𝑛𝜋, where 𝑛 is an integer.
Proof.
lim𝑥→𝑎 tan 𝑥 = lim𝑥→𝑎sin 𝑥
cos 𝑥 quotient law
=lim 𝑥→𝑎 sin 𝑥
lim 𝑥→𝑎 cos 𝑥 cty of sin & cos
= tan(𝑎)
unless cos 𝑎 = 0.
But from graph above, cos 𝑎 = 0 except at
𝑥 =𝜋
2+ 𝑛𝜋, where 𝑛 is an integer. ■
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 constant multiple
4. 𝑓𝑔 product
5. 𝑓/𝑔 provided 𝑔 𝑎 ≠ 0 quotient
Proof. Each part follows from the corresponding limit
law. For example, consider (4)
lim𝑥→𝑎 𝑓 𝑥 ⋅ 𝑔 𝑥 product law for limits
= lim𝑥→𝑎 𝑓 𝑥 ⋅ lim𝑥→𝑎 𝑔(𝑥)
continuity of 𝑓 and 𝑔
= 𝑓 𝑎 ⋅ 𝑔(𝑎)
This is the definition of continuity of 𝑓 𝑥 ⋅ 𝑔(𝑥) ■
26a 26b
Continuity from the left and from the right
A function 𝑓 is continuous from the right at no. 𝑎 if
lim𝑥→𝑎+
𝑓 𝑥 = 𝑓(𝑎)
and 𝑓 is continuous from the left and no. 𝑎 if
lim𝑥→𝑎−
𝑓 𝑥 = 𝑓(𝑎)
Example. Step Function
𝑔 𝑥 = 1, 𝑥 ≥ 00, 𝑥 < 0
… 0 …𝑥-, 0 … 1 …𝑦-, 𝑔(𝑥)
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
𝑦 = 𝑥 is continuous from the right at every integer.
Continuity on an interval (including endpoints)
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.
27a 27b
Example. 𝑓 𝑥 = 𝑥 is continuous on [0, ∞).
Why? For any 𝑎 > 0, by the limit laws for roots
lim𝑥→𝑎
𝑥 = lim𝑥→𝑎
𝑥 = 𝑎
For 𝑎 = 0, lim𝑥→0+ 𝑥 = 0 = 𝑎
which is “obvious” from the graph
0 …𝑥-, 0 …𝑦-, 𝑦 = 𝑥
■
?? Practice with continuity transparency.
Continuity and Compositions
Composition: Function Machine Picture
𝑥 → 𝑔 → 𝑢 = 𝑔(𝑥)
𝑢 → 𝑓 → 𝑦 = 𝑓(𝑢)
𝑦 = 𝑓 𝑢 = 𝑓(𝑔(𝑥)) 𝑔 is the ‘inner function’
𝑓 is the ‘outer function’
Example. 𝐹 𝑥 = cos 𝑥
𝑔 𝑥 = 𝑥
𝑓 𝑢 = cos(𝑢)■
28a 28b
Theorem (Limits of Compositions)
If 𝑓 is continuous at 𝑏 and
lim𝑥→𝑎 𝑔 𝑥 = 𝑏
then
lim𝑥→𝑎 𝑓 𝑔 𝑥 = 𝑓(lim𝑥→𝑎 𝑔(𝑥)) = 𝑓(𝑏)
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).
Example. Evaluate lim𝑥→𝜋2 cos 𝑥 = 𝐿
From the root law, lim𝑥→𝜋2 𝑥 = 𝜋
𝑓 𝑢 = cos(𝑢) is continuous at 𝑢 = 𝜋
By the theorem
𝐿 = cos(lim𝑥→𝜋2 𝑥) = cos 𝜋 = −1 ■
Combine the theorem above with the condition that
𝑏 = 𝑔 𝑎 , in other words the condition that 𝑔 is
continuous at 𝑎, to get:
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
lim𝑥→𝑎 𝑓 𝑔 𝑥 = 𝑓 lim𝑥→𝑎 𝑔 𝑥
= 𝑓 𝑔 𝑎
This is the definition of continuity for 𝑓 𝑔 𝑥 . ■
In words, a continuous function of a continuous
function is continuous.
29a 29b
Example. Where is 𝐹 𝑥 = cos( 𝑥) continuous?
Note that 𝐹 𝑥 = 𝑓 𝑔 𝑥 where
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
Proof. Not given, but the intermediate value
theorem is “obvious”.
Example. Prove that there is an 𝑥 that solves
sin 𝑥 = 1 − 𝑥
on the interval (0,𝜋
2).
Proof. Let 𝑓 𝑥 = 1 − 𝑥 − sin(𝑥)
then 𝑓 0 = 1 − 0 − sin 0 = 1 > 0
𝑓 𝜋
2 = 1 −
𝜋
2− 1 = −
𝜋
2< 0
Since 𝑁 = 0 lies between 𝑓 0 > 0 and 𝑓 𝜋
2 < 0,
and 𝑓 is an arithmetic combination of continuous
functions (and therefore continuous itself), it follows
from the Intermediate Value Theorem that there is a
𝑐 ∈ 0,𝜋
2 such that 𝑓 𝑐 = 0. ■
30a 30b
§1.6 Limits involving infinity
Infinite Limits
Example. 𝑦 = 𝑔 𝑥 = 1/𝑥.
−2 … 0 … 2 …𝑥-, −2 … 0 … 2 …𝑦-, 𝑔(𝑥)
We write lim𝑥→0+ 𝑔 𝑥 = ∞
The symbol ∞ is a version of DNE. It means 𝑔(𝑥)
becomes greater than any fixed value.
We write lim𝑥→0− 𝑔 𝑥 = −∞
The symbol −∞ means that 𝑔(𝑥) becomes less than
any fixed value ■
Example. 𝑥 = 1/ 𝑥 − 3 2
0 … 6 …𝑥-, 0 … 4 …𝑦-, (𝑥)
?? lim𝑥→3+ 𝑥 =
?? lim𝑥→3− 𝑥 =
The one sided limit are “equal” (they DNE in the
same way)
Thus we write lim𝑥→3 𝑥 = ∞.
31a 31b
The vertical line 𝑥 = 𝑎 is a vertical asymptote of
𝑦 = 𝑓(𝑥) if
1) lim𝑥→𝑎+ 𝑓 𝑥 = +∞ or −∞
and / or
2) lim𝑥→𝑎− 𝑓 𝑥 = +∞ or −∞
𝑥 = 𝑎 is a vertical asymptote. add 𝑥 = 3
Example. 𝑦 = log2(𝑥)
domain is ℝ+ = { positive numbers}
Table x/ ¼ =2-2, ½ =2-1, 1=20, 2=21, 4=22 / y …
0 … 4 …𝑥-, −2 … 0 … 2 …𝑦-, dots, curve
The y-axis (or line 𝑥 = 0) is a vertical asymptote. ■
32a 32b
Limits at Infinity
Example. 𝑓 𝑥 = 𝑥2/(𝑥2 + 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
𝑓(𝑥) approaches 1 as |𝑥| increases ■
Definitions
lim𝑥→∞ 𝑔(𝑥) = 𝐿
means that 𝑔(𝑥) can be made as close as we wish to
𝐿 by taking |𝑥| sufficiently large with 𝑥 > 0.
lim𝑥→−∞ 𝑔(𝑥) = 𝐿
means that 𝑔(𝑥) can be made as close as we wish to
𝐿 by taking |𝑥| sufficiently large with 𝑥 < 0.
In the previous example,
lim𝑥→∞ 𝑓 𝑥 = 1, and lim𝑥→−∞ 𝑓(𝑥) = 1
The line 𝑦 = 𝐿 is a horizontal asymptote of 𝑦 = 𝑓(𝑥)
if
lim𝑥→∞ 𝑓 𝑥 = 𝐿 and/or lim𝑥→−∞ 𝑓 𝑥 = 𝐿
?? Transparency: Practice with asymptotes
33a 33b
Consider lim𝑥→∞ 1/𝑥
For 𝑥 > 10, 0 <1
𝑥<
1
10
For 𝑥 > 100, 0 <1
𝑥<
1
100
We can make 1/𝑥 as close as we wish to zero by
taking 𝑥 sufficiently large. Conclude
lim𝑥→∞1
𝑥= 0
Consider lim𝑥→−∞ 1/𝑥
For 𝑥 < −10, −1
10<
1
𝑥< 0
For 𝑥 < −100,−1
100<
1
𝑥< 0
We can make 1/𝑥 as close as we wish to zero by
taking |𝑥| sufficiently large with 𝑥 < 0. Conclude
lim𝑥→−∞1
𝑥= 0
Generalize
If 𝑟 is a positive real number
lim𝑥→∞1
𝑥𝑟= 0 Equation (1)
we won’t prove this
Examples. lim𝑥→∞1
𝑥2= 0
lim𝑥→∞1
𝑥= lim𝑥→∞
1
𝑥12
= 0 ■
Infinite Limit Laws
These are obtained from the limit laws we have
already seen by replacing 𝑎 with ±∞.
Suppose that 𝑐 is a constant and
lim𝑥→±∞ 𝑓(𝑥), lim𝑥→±∞ 𝑔(𝑥)exist.
1. sum law (limit of sum is sum of limits)
lim𝑥→±∞
𝑓 𝑥 + 𝑔 𝑥 = lim𝑥→±∞
𝑓 𝑥 + lim𝑥→±∞
𝑔(𝑥)
34a 34b
2. difference law
lim𝑥→±∞
𝑓 𝑥 − 𝑔 𝑥 = lim𝑥→±∞
𝑓 𝑥 − lim𝑥→±∞
𝑔(𝑥)
3. constant multiple law
lim𝑥→±∞
𝑐 𝑓 𝑥 = 𝑐 lim𝑥→±∞
𝑓(𝑥)
4. product law
lim𝑥→±∞
𝑓 𝑥 𝑔 𝑥 = lim𝑥→±∞
𝑓 𝑥 ⋅ lim𝑥→±∞
𝑔(𝑥)
5. quotient law
lim𝑥→±∞
𝑓 𝑥
𝑔 𝑥 =
lim𝑥→±∞
𝑓 𝑥
lim𝑥→±∞
𝑔 𝑥
so long as lim𝑥→±∞ 𝑔 𝑥 ≠ 0
Example. Find lim𝑥→∞2𝑥+3
16𝑥2+5= 𝐿.
𝐿 = lim𝑥→∞2𝑥+3
16𝑥2+5 multiply by 1
= lim𝑥→∞2𝑥+3
16𝑥2+5
1/𝑥
1/𝑥 (note 𝑥 ≠ 0)
multiply numerator and denominator by 1/𝑥
= lim𝑥→∞2+3/𝑥
16+5/𝑥2 quotient law
=lim 𝑥→∞ ( 2+3/𝑥)
lim 𝑥→∞ 16+5/𝑥2 (denominator is not zero)
=2+0
16+0 limit laws and equation 1
=1
2 ■
Example. Find lim𝑥→−5−𝑥−3
𝑥+5
Cannot use quotient law because 𝑥 + 5 → 0− as
𝑥 → −5−
Notice that as 𝑥 → −5− we have 𝑥 − 3 → −8−
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
lim𝑥→−5−𝑥−3
𝑥+5= ∞ ■