1.1 function machines course administration1a 1b course administration math placement syllabus (show...
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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= ∞ ■