1a 1b
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(3) Math 171 week #1A
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ยง1.1 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. ๐ ๐ฅ = ๐ฅ2, โ1 โค ๐ฅ โค 1. domain
= [โ1,1]. range = [0,1]. โ
2a 2b
Interval Notation
๐ฅ โ [โ1,1] means โ1 โค ๐ฅ โค 1
๐ฅ โ (โ1,1) means โ1 < ๐ฅ < 1
๐ฅ โ (โโ,โ) means ๐ฅ is any real no.
Example. ๐ ๐ฅ = ๐ฅ2, domain = (โโ,โ).
range= [0, โ). โ
If the domain of a function is not given explicitly,
assume it is the largest set of numbers that makes
sense.
Example. ๐ ๐ฅ = ๐ฅ, domain not given.
Assume domain= [0, โ). range = [0,โ). โ
Graphs
Example. ๐ฆ = ๐ฅ2
sketch โฆโ 1 โฆ 0 โฆ 1 โฆ๐ฅ- (independent variable),
0 โฆ๐ฆ- (dependent variable), curveโ
Example. ๐ฆ = ๐ฅ
sketch 0 โฆ 4 โฆ๐ฅ-, 0 โฆ 2 โฆ๐ฆ-, curveโ
3a 3b
Example. Graph the set of points that satisfy ๐ฆ2 = ๐ฅ.
Table x / 0, 1, 4; y/ 0, ยฑ1, ยฑ2
sketch 0 โฆ 4 โฆ๐ฅ-, โ2 โฆ 0 โฆ 2 โฆ๐ฆ-, curve
Is this a function?
?? Why? โ
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 ๐ฅ2 + ๐ฆ2 = 1.
axes, circle, vertical line intersecting circle twice
?? Is this the graph of a function?
?? Why?
4a 4b
Even and odd symmetry
If ๐(๐ฅ) satisfies
๐ โ๐ฅ = ๐(๐ฅ)
for every ๐ฅ in its domain, then ๐ is called an even
function.
Example. ๐ ๐ฅ = ๐ฅ2,
๐ โ๐ฅ = โ๐ฅ 2 = โ๐ฅ โ๐ฅ = ๐ฅ2 = ๐(๐ฅ).
?? If ๐ is even, its graph is symmetric wrt which axis?
โ
If ๐(๐ฅ) satisfies
๐ โ๐ฅ = โ๐(๐ฅ)
for every ๐ฅ in its domain, then ๐ is an odd function.
Example. ๐ ๐ฅ = ๐ฅ3,
๐ โ๐ฅ = โ๐ฅ โ๐ฅ โ๐ฅ = โ๐ฅ3 = โ๐(๐ฅ) .
Table x/ 0, -1, 1, -2, 2; y/ 0, -1, 1, -8, 8
sketch โ2 โฆ 0 โฆ 2 โฆ๐ฅ-, โ8 โฆ 0 โฆ 8 โฆ๐ฆ-, curve
?? If ๐ is odd, its graph is symmetric wrt what? โ
Knowing even or odd symmetry helps us sketch
functions.
5a 5b
ยง1.2 Catalog Of Functions
Straight lines
slope intercept form
๐ฆ = ๐๐ฅ + ๐
sketch โฆ 0 โฆ๐ฅ- axis, 0 โฆ๐ฆ- axis, intercept ๐,
line,(๐ฅ, ๐ฆ),ฮ๐ฅ, ฮ๐ฆ
slope = ฮ๐ฆ
ฮ๐ฅ=
๐ฆโ๐
๐ฅโ0= ๐
Example. ๐ฆ =1
2๐ฅ + 1
Table
sketch โ2 โฆ 0 โฆ 2 โฆ๐ฅ-, 0 โฆ 1 โฆ 2 โฆ๐ฆ-, intercept,
line
โ
point slope form
๐ฆ โ ๐ฆ1 = ๐(๐ฅ โ ๐ฅ1)
sketch 0 โฆ๐ฅ-, โฆ๐ฆ-,(๐ฅ1, ๐ฆ1), line, (๐ฅ, ๐ฆ),๐ฅ โ ๐ฅ1 = ๐ฅ๐ฅ,
๐ฆ โ ๐ฆ1 = ๐ฅ๐ฆ
๐ฅ๐ฆ
๐ฅ๐ฅ= ๐
6a 6b
Example. ๐ฆ โ 2 =1
2(๐ฅ โ 2)
convert to slope intercept form
solve for y
๐ฆ =
same as previous example โ
Power Functions
general form
๐ฆ = ๐ฅ๐ผ where ๐ผ is a constant
Example ๐ฆ = ๐ฅ straight line
?? slope ?? y-intercept
Example ๐ฆ = ๐ฅ2 parabola
Example ๐ฆ = ๐ฅ1
2 square root (๐ฅ1
2 = ๐ฅ)
Example ๐ฆ = ๐ฅโ1 (๐ฅโ1 = 1/๐ฅ)
sketch โ2 โฆ 0 โฆ 2 โฆ๐ฅ-, โ2 โฆ 0 โฆ 2 โฆ๐ฆ-
โ
7a 7b
Polynomial Functions
Example ๐ฆ = 1 โ ๐ฅ2
sketch โ2 โฆ 0 โฆ 2 โฆ๐ฅ-, โ2 โฆ 0 โฆ 1 โฆ๐ฆ-, parabola
โ
A quadratic function is a polynomial of degree 2.
๐ฆ = ๐2๐ฅ2 + ๐1๐ฅ + ๐0
๐2, ๐1 , ๐0 are constants add
The degree of a polynomial is the highest power that
it contains.
A polynomial of degree ๐ has the form
๐ ๐ฅ = ๐๐๐ฅ๐ + ๐๐โ1๐ฅ
๐โ1 + โฏ + ๐2๐ฅ2 + ๐1๐ฅ + ๐0.
Piecewise defined functions
Example ๐ ๐ฅ = ๐ฅ + 1, ๐ฅ โ 11, ๐ฅ = 1
sketch โ1 โฆ 2 โฆ๐ฅ-, 0 โฆ 3 โฆ๐ฆ-, line with hole, dot
โ
Example Absolute Value Function ๐ ๐ฅ = |๐ฅ|
gives distance from the origin on the real number line
sketch โ2 โฆ 0 โฆ 2, bracket โ2 and 0
distance between โ2 and 0 is 2, โ2 = 2
8a 8b
graph ๐ฆ = |๐ฅ|
sketch โ2 โฆ 0 โฆ 2 โฆ๐ฅ-, 0 โฆ 2 โฆ๐ฆ-, graph
piecewise definition ๐ฅ = ๐ฅ, ๐ฅ > 0โ๐ฅ, ๐ฅ < 0
โ
Rational Functions
A rational function is a ratio of two polynomials
๐ ๐ฅ =๐(๐ฅ)
๐(๐ฅ)
where ๐, ๐ are polynomials.
Example (a case of special interest to us)
๐ ๐ฅ =๐ฅ2โ1
๐ฅโ1
?? domain of f?
Important Algebraic Trick!
๐ฅ โ 1 ๐ฅ + 1 =
In general, ๐ฅ โ ๐ ๐ฅ + ๐ = ๐ฅ2 โ ๐2. Then
๐ ๐ฅ = ๐ฅ โ 1 (๐ฅ + 1)
๐ฅ โ 1=
๐ฅ + 1, if ๐ฅ โ 1 undefined, if ๐ฅ = 1
sketch โ1 โฆ 0 โฆ 2 โฆ๐ฅ-, 0 โฆ 2 โฆ๐ฆ-, line with hole
โ
9a 9b
Sine and Cosine
sketch โ ๐, โฆ 0 โฆ3๐
2โฆ๐ฅ-, โ1 โฆ 0 โฆ 1.., sin(๐ฅ),
cos(๐ฅ)
properties of sine
โ1 โค sin ๐ฅ โค 1
sin 0 = 0, sin ๐ = 0, sin 2๐ = 0, generally
sin ๐๐ = 0 for ๐ an integer
sin ๐
2 = 1, sin โ
๐
2 = โ1
sin ๐ฅ + 2๐ = sin(๐ฅ) periodicity with period 2๐
sin ๐ฅ + ๐ = โsin(๐ฅ) advance by half a period
?? sin โ๐ฅ = โฏ, ?? symmetry?
properties of cosine
โ1 โค cos ๐ฅ โค 1
cos ๐
2 = 0, cos
3๐
2 = 0, cos
5๐
2 = 0, generally
cos ๐ +1
2 ๐ = 0 for ๐ an integer
cos 0 = 1, cos ๐ = โ1
cos ๐ฅ + 2๐ = cos(๐ฅ) periodicity with period 2๐
cos ๐ฅ + ๐ = โcos(๐ฅ) advance by half a period
?? cos โ๐ฅ = โฏ, ?? symmetry?
10a 10b
Two Important Triangles
sketch 45-45-90 triangle, lengths of sides
Pythagorean theorem: 1
2+
1
2= 1
sin ๐
4 = cos
๐
4 =
1
2=
2
2โ 0.71
sketch 30-60-90 triangle, lengths of sides
Pythagorean theorem: 3
4+
1
4= 1
sin ๐
6 = cos
๐
3 =
1
2
sin ๐
3 = cos
๐
6 =
3
2โ 0.87
Tangent
tan ๐ฅ =sin ๐ฅ
cos ๐ฅ
tan โ๐ฅ =
?? symmetry
11a 11b
graph tangent
sketch โ๐
2โฆ 0 โฆ
3๐
2โฆ๐ฅ-, โฆ 0 โฆ๐ฆ-, vert. asymptotes,
curve
โ๐
2,
๐
2,
3๐
2, โฆ not in the domain of tan ?? why
tan ๐ฅ + ๐ =
period ๐
tan 0 = tan ๐ = โฏ = 0
tan ๐
4 =
sin ๐
4
cos ๐
4
= 1
similarly find tan ๐
6 =
1
3 and tan
๐
3 = 3
ยง1.3 The Limit of a Function
sketch โฆ๐โฆ . , โฆ๐ฟโฆ, curve ๐, hole at ๐
Informal definition of limit
lim๐ฅโ๐ ๐ ๐ฅ = ๐ฟ
means that we can make ๐(๐ฅ) as close as we wish to
๐ฟ by taking ๐ฅ sufficiently close to ๐ (but not equal to
๐).
Alternate notation: ๐ ๐ฅ โ ๐ฟ as ๐ฅ โ ๐
12a 12b
Example. ๐ ๐ฅ = ๐ฅ + 1
โฆ 1 โฆ , โฆ 1 โฆ 2 โฆ, line ๐
imagine a bug approaching ๐ฅ = 1 on either side
add arrows towards ๐ฅ = 1, arrows towards ๐ฆ = 2
lim๐ฅโ1 ๐ ๐ฅ = 2
lim๐ฅโ๐ ๐(๐ฅ)has nothing to do with ๐(๐)
Example ๐ ๐ฅ = ๐ฅ + 1, ๐ฅ โ 11, ๐ฅ = 1
sketch โฆ 1 โฆ๐ฅ-, โฆ 1 โฆ 2 โฆ, line with hole, dot
add arrows towards ๐ฅ = 1, arrows towards ๐ฆ = 2
lim๐ฅโ1 ๐ ๐ฅ = 2 โ
13a 13b
Limit is a 2-sided concept
Example. Step function
๐ป ๐ฅ = 0, ๐ฅ < 01, ๐ฅ โฅ 0
โฆ 0 โฆ๐ฅ-, 0 โฆ 1 โฆ, ๐ป(๐ฅ)
lim๐ฅโ0 ๐ป(๐ฅ) DNE (does not exist)โ
One sided limits
Informal definition
lim๐ฅโ๐โ ๐ ๐ฅ = ๐ฟ1 limit from left or left hand limit
means we can make ๐(๐ฅ) as close as we wish to ๐ฟ1
by taking ๐ฅ sufficiently close to ๐ from the left.
lim๐ฅโ๐+ ๐ ๐ฅ = ๐ฟ2 limit from right or โฆ
means we can make ๐(๐ฅ) as close as we wish to ๐ฟ2
by taking ๐ฅ sufficiently close to ๐ from the right.
Example. (step function again)
?? lim๐ฅโ0+ ๐ป ๐ฅ =
?? lim๐ฅโ0โ ๐ป ๐ฅ =
โ
14a 14b
the (2-sided) limit
lim๐ฅโ๐ ๐ ๐ฅ = ๐ฟ
if and only if
lim๐ฅโ๐+ ๐ ๐ฅ = ๐ฟ and lim๐ฅโ๐โ ๐ ๐ฅ = ๐ฟ
?? Practice with limits
Precise definition of a limit
0 โฆ๐โฆ๐ฅ-, 0 โฆ๐ฟโฆ๐ฆ-, ๐(๐ฅ), hole at ๐ฅ = ๐, dot
lim๐ฅโ๐ ๐ ๐ฅ = ๐ฟ means
For every ๐ > 0 there is a ๐ฟ > 0 such that
if ๐ โ ๐ฟ < ๐ฅ < ๐ + ๐ฟ (๐ฅ โ ๐)
then ๐ฟ โ ๐ < ๐ ๐ฅ < ๐ฟ + ๐
add ๐ โ ๐ฟ, ๐ + ๐ฟ, ๐ฟ โ ๐, ๐ฟ = ๐, segments
ยง1.4 Calculating Limits
Two Special Limits
A. Let ๐ be a constant
lim๐ฅโ๐ ๐ = ๐
0 โฆ๐โฆ๐ฅ-, 0 โฆ๐ฆ-, line ๐ฆ = ๐, arrows approaching
๐ฅ = ๐
15a 15b
B. Consider ๐ ๐ฅ = ๐ฅ
lim๐ฅโ๐ ๐ฅ = ๐
0 โฆ๐โฆ๐ฅ-, 0 โฆ๐โฆ๐ฆ-, ๐ ๐ฅ = ๐ฅ
add arrows approaching ๐ฅ = ๐, add arrows
approaching ๐ฆ = ๐
Five Limit Laws
Suppose that ๐ is a constant and
lim๐ฅโ๐ ๐(๐ฅ), lim๐ฅโ๐ ๐(๐ฅ) exist.
1. sum law (limit of sum is sum of limits)
lim๐ฅโ๐ ๐ ๐ฅ + ๐ ๐ฅ = lim๐ฅโ๐ ๐ ๐ฅ + lim๐ฅโ๐ ๐(๐ฅ)
2. difference law
lim๐ฅโ๐ ๐ ๐ฅ โ ๐ ๐ฅ = lim๐ฅโ๐ ๐ ๐ฅ โ lim๐ฅโ๐ ๐(๐ฅ)
3. constant multiple law
lim๐ฅโ๐ ๐ ๐ ๐ฅ = ๐ lim๐ฅโ๐ ๐(๐ฅ)
4. product law
lim๐ฅโ๐ ๐ ๐ฅ ๐ ๐ฅ = lim๐ฅโ๐ ๐ ๐ฅ โ lim๐ฅโ๐ ๐(๐ฅ)
5. quotient law
lim๐ฅโ๐๐ ๐ฅ
๐ ๐ฅ =
lim ๐ฅโ๐ ๐(๐ฅ)
lim ๐ฅโ๐ ๐(๐ฅ) ,
so long as lim๐ฅโ๐ ๐ ๐ฅ โ 0
16a 16b
Example
lim๐ฅโ1 3๐ฅ + 5 sum law
= lim๐ฅโ1
3๐ฅ + lim๐ฅโ1 5
constant multiple law
= 3 lim๐ฅโ1 ๐ฅ + lim๐ฅโ1 5
special limits
= 3 โ 1 + 5 = 8 โ
Example
lim๐ฅโ2๐ฅ2
๐ฅ+1 quotient law
=lim ๐ฅโ2 ๐ฅ2
lim ๐ฅโ2(๐ฅ+1)
product and sum laws
= lim ๐ฅโ2 ๐ฅโ lim ๐ฅโ2 ๐ฅ
lim ๐ฅโ2 ๐ฅ+lim ๐ฅโ2 1
special limits
=2โ 2
2+1 =
4
3 โ
Repeated Application of the Product Law
lim๐ฅโ๐ ๐ ๐ฅ โ ๐ ๐ฅ โ ๐ ๐ฅ = lim๐ฅโ๐ ๐ ๐ฅ โ ๐(๐ฅ)
where ๐ โ ๐ = ๐
product law
= lim๐ฅโ๐ ๐ ๐ฅ โ lim๐ฅโ๐ ๐(๐ฅ)
product law
= lim๐ฅโ๐ ๐ ๐ฅ โ lim๐ฅโ๐ ๐ ๐ฅ โ lim๐ฅโ๐ ๐(๐ฅ)
Power Law
Suppose ๐ ๐ฅ = ๐ ๐ฅ = ๐(๐ฅ) then from above
lim๐ฅโ๐ ๐ ๐ฅ 3 = lim๐ฅโ๐ ๐(๐ฅ) 3
Apply the same reasoning to a product of ๐ factors of
๐(๐ฅ) to get the Power Law:
lim๐ฅโ๐ ๐ ๐ฅ ๐ = lim๐ฅโ๐ ๐(๐ฅ) ๐
where ๐ is any positive integer
17a 17b
Example. Cubic Polynomial.
lim๐ฅโ2 ๐ฅ3 โ 4๐ฅ difference law
= lim๐ฅโ2
๐ฅ3 โ lim๐ฅโ2
4๐ฅ
power, constant multiple laws
= lim๐ฅโ2 ๐ฅ 3 = 4 lim๐ฅโ2 ๐ฅ
special limits
= 23 โ 4 โ 2 = 0 โ
Recall polynomials of degree ๐. Their general form is
๐ ๐ฅ = ๐๐๐ฅ๐ + ๐๐โ1๐ฅ
๐โ1 + โฏ + ๐1๐ฅ + ๐0
where ๐๐ , ๐๐โ1, โฆ , ๐1, ๐0 are constants.
By reasoning similar to the cubic polynomial example
lim๐ฅโ๐ ๐ ๐ฅ sum law, const. multiple law
= ๐๐ lim๐ฅโ๐ ๐ฅ๐ + โฏ + ๐1 lim๐ฅโ๐ ๐ฅ + lim๐ฅโ๐ ๐0
power law, special limit
= ๐๐๐๐ + โฏ๐1๐ + ๐0
= ๐(๐)
we have discovered the following
Direct Substitution Property for polynomials
If ๐(๐ฅ) is any polynomial and ๐ is a real number then
lim๐ฅโ๐ ๐ ๐ฅ = ๐(๐).
This is much easier to apply than the limit laws!
Recall that a rational function is a ratio of two
polynomials.
๐ ๐ฅ =๐ ๐ฅ
๐ ๐ฅ , where ๐ and ๐ are polynomials
Let ๐(๐ฅ) be any rational function
lim๐ฅโ๐ ๐ ๐ฅ =lim ๐ฅโ๐ ๐(๐ฅ)
lim ๐ฅโ๐ ๐(๐ฅ) quotient law
=๐(๐)
๐(๐) direct subst. for polys
so long as ๐ ๐ โ 0.
we now have a โฆ
18a 18b
Direct Substitution Property for rational functions
If ๐(๐ฅ) is a rational function and ๐ is a number in the
domain of ๐
lim๐ฅโ๐
๐ ๐ฅ = ๐(๐)
Example.
lim๐ฅโ1
๐ฅ4 + ๐ฅ2 โ 6
๐ฅ4 โ 2๐ฅ + 3=
14 + 12 โ 6
14 + 2 โ 1 + 3=
โ4
6=
โ2
3
by Direct Substitution for rational functions! โ
Root Law For Limits
Let ๐ be a positive integer
lim๐ฅโ๐
๐(๐ฅ)๐
= lim๐ฅโ๐
๐(๐ฅ)๐ ,
If ๐ is even then require lim๐ฅโ๐ ๐ ๐ฅ > 0.
Example. lim๐ฅโโ2 ๐ข4 + 3๐ข + 6 root law
= lim๐ฅโโ2
(๐ข4 + 3๐ข + 6)
direct subst. for polys.
= 16 โ 6 + 6 = 4 โ
Indeterminate Forms
Example. Let lim๐กโ2๐ก2+๐กโ6
๐ก2โ4= ๐ฟ
We call this an indeterminate form of type 0
0 since
direct substitution of ๐ก = 2 into the rational function
gives that quotient, which is not defined. We cannot
use the quotient law!
Factor the numerator and denominator:
๐ก2+๐กโ6
๐ก2โ4=
๐ก+3 (๐กโ2)
๐ก+2 (๐กโ2) if ๐ก โ 2
=๐ก+3
๐กโ2
19a 19b
Recall that lim๐กโ2 ๐(๐ก) does not depend on ๐(2)!
Then
๐ฟ = lim๐กโ2๐ก+3
๐กโ3=
5
4. โ
Rationalization and Cancellation
Example. Consider the following indeterminate form
lim๐ฅโโ1 ๐ฅ+2โ1
๐ฅ+1= ๐ฟ type
0
0.
Rationalize the quotient and simplify as follows:
๐ฅ+2โ1
๐ฅ+1=
๐ฅ+2โ1
๐ฅ+1 ๐ฅ+2+1
๐ฅ+2+1=
(๐ฅ+1)
๐ฅ+1 ( ๐ฅ+2+1)
if ๐ฅ โ โ1
=1
๐ฅ+2+1
but lim๐ฅโโ1 ๐(๐ฅ) does not depend on ๐(โ1)! Then
๐ฟ = lim๐ฅโโ11
๐ฅ+2+1=
1
2. โ
Limits Involving Absolute Values
Recall that a limit exists iff the corresponding left and
right hand limits are equal.
lim๐ฅโ๐ ๐ ๐ฅ = ๐ฟ
โ (both lim๐ฅโ๐+
๐ ๐ฅ = ๐ฟ and lim๐ฅโ๐โ
๐ ๐ฅ = ๐ฟ)
Recall the piecewise definition of absolute value
๐ง = ๐ง, ๐ง > 0โ๐ง, ๐ง < 0
.
Use this when evaluating limits involving absolute
values.
Example. Let ๐ฟ = lim๐ฅโ
3
2
2 ๐ฅ2โ3๐ฅ
|2๐ฅโ3|
2๐ฅ โ 3 = 2๐ฅ โ 3, ๐ฅ > 3/23 โ 2๐ฅ, ๐ฅ < 3/2
Find ๐ฟ1 = lim๐ฅโ
3
2+
2 ๐ฅ2โ3๐ฅ
2๐ฅโ3= lim
๐ฅโ3
2+
๐ฅ(2๐ฅโ3)
2๐ฅโ3
= lim๐ฅโ
3
2+๐ฅ =
3
2
20a 20b
and ๐ฟ2 = lim๐ฅโ
3
2โ
2 ๐ฅ2โ3๐ฅ
3โ2๐ฅ= lim
๐ฅโ3
2โ
๐ฅ(2๐ฅโ3)
3โ2๐ฅ=
= lim๐ฅโ
3
2โโ๐ฅ = โ
3
2
๐ฟ does not exist because ๐ฟ1 โ ๐ฟ2.โ
Limits of Trig Functions
โ๐
2โฆ 0 โฆ
๐
2โฆ๐ฅ-, sin(๐ฅ)
lim๐ฅโ0 sin ๐ฅ = 0
add ๐ฅ, slope agrees with sin(๐ฅ) at the origin
(1)
This limit is a type 0/0 indeterminate form. However,
the ratio sin ๐ฅ
๐ฅโ 1 as ๐ฅ โ 0. The text proves this
using a geometric argument and the squeeze
theorem.
Example. Find L = lim๐กโ0sin (2๐ก)
๐ก (*)
๐ฟ = lim๐กโ0sin (2๐ก)
2๐กโ 2
const. multiple law
= 2 โ lim๐กโ0sin (2๐ก)
2๐ก
Let ๐ข = 2๐ก. Notice that ๐ข โ 0 as ๐ก โ 0. Thus
๐ฟ = 2 โ lim๐ขโ0sin (๐ข)
๐ข by equation (1)
= 2
lim๐ฅโ0sin (๐ฅ)
๐ฅ= 1
21a 21b
WARNING By a trig identity
sin 2๐ก = 2 sin ๐ก cos(๐ก).
Thus, simplifying Equation (*) by writing โsin 2๐ก =
2 sin(๐ก)โ 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. โ
โ๐
2โฆ 0 โฆ
๐
2โฆ๐ฅ-, cos(๐ฅ)
By the graph it is clear that
lim๐ฅโ0 cos ๐ฅ = 1 (2)
Corollary. lim๐โ0cos ๐ โ1
๐= 0
Proof.
lim๐โ0cos ๐ โ1
๐ multiply by 1
= lim๐โ0cos ๐ โ1
๐
cos ๐ +1
cos ๐ +1
simplify numerator
= lim๐โ0cos 2 ๐ โ1
๐(cos ๐ +1)
sin2 ๐ + cos2(๐) = 1
= lim๐โ0โ sin 2(๐)
๐(cos ๐ +1)
algebra
= lim๐โ0sin (๐)
๐
โ sin ๐
cos ๐ +1
product law for limits
= lim๐โ0sin (๐)
๐โ lim๐โ0
โ sin ๐
cos ๐ +1
using (1), (2) and the quotient law
= 0 โ
22a 22b
One further example.
?? Find L = lim๐ฅโ0tan (2๐ฅ)
๐ฅ.
โ
ยง1.5 Continuity
Informal definition A function is continuous is it can
be drawn without removing pencil from the paper.
โฆ๐โฆ๐โฆ๐โฆ๐ฅ-. ,โฆ๐ฆ-, ๐ continuous on ๐, ๐ ,
๐ with hole at ๐
๐ is continuous on (๐, ๐)
๐ is discontinuous at ๐.
common abbreviations: cts = continuous and dcts =
discontinuous.
23a 23b
Formal definition. A function ๐ is continuous at a
number ๐ if
lim๐ฅโ๐
๐ ๐ฅ = ๐(๐)
if not, ๐ is discontinuous at ๐.
Three conditions required for continuity:
(1) ๐(๐) exists
(2) lim๐ฅโ๐ ๐(๐ฅ) exists
(3) lim๐ฅโ๐ ๐ ๐ฅ = ๐(๐)
Three types of discontinuity
A. Infinite discontinuity
Example. ๐ ๐ฅ = 1/๐ฅ2 sketch
๐ is discontinuous at ๐ฅ = 0. (1) and (2) are violated.
B. Jump discontinuity
Example. ๐ ๐ฅ = 1, ๐ฅ โฅ 00, ๐ฅ < 0
sketch
๐ is discontinuous at ๐ฅ = 0. (2) is violated.
C. Removable discontinuity
Example. ๐ ๐ฅ = ๐ฅ + 1, ๐ฅ โ 10, ๐ฅ = 1
sketch
๐ is discontinuous at ๐ฅ = 1. (3) is violated.
24a 24b
Continuity on an open interval
If ๐ is continuous at each point of an open interval ๐ผ,
we say ๐ is continuous on ๐ผ.
Fact. Every polynomial is continuous at every real
number
Proof. Let ๐ be a polynomial. For any real no. ๐, by
the direct substitution property for polynomials
lim๐ฅโ๐
๐ ๐ฅ = ๐(๐)
This is also the definition of continuity for a function
๐ at a point! Thus ๐ is continuous on (โโ, โ). โ
Fact. Every rational function is continuous at every
point of its domain.
Proof. Let ๐ be a rational function and let
๐ โ domain of ๐. By the direct substitution property
for rational functions
lim๐ฅโ๐
๐ ๐ฅ = ๐(๐)
This is the definition of continuity for a function ๐ at
point ๐. โ
Example. ๐ ๐ฅ =๐ฅ2โ1
๐ฅโ1 is continuous on (โโ, 1) and
(1,โ). โ
Fact. sin(๐ฅ) and cos(๐ฅ) are continuous at every
real no. ๐ฅ.
โ๐โฆโ๐
2โฆ 0 โฆ
๐
2โฆ๐โฆ๐ฅ-, โฆโ 1 โฆ 0 โฆ 1 โฆ๐ฆ-,
curve of ๐ฆ = sin(๐ฅ), curve of ๐ฆ = cos(๐ฅ)
No formal proof, but notice that these curves can be
drawn without lifting pencil from paper.โ
25a 25b
Fact. ๐ฆ = tan(๐ฅ) is continuous at every ๐ฅ except
values ๐ฅ =๐
2+ ๐๐, where ๐ is an integer.
Proof.
lim๐ฅโ๐ tan ๐ฅ = lim๐ฅโ๐sin ๐ฅ
cos ๐ฅ quotient law
=lim ๐ฅโ๐ sin ๐ฅ
lim ๐ฅโ๐ cos ๐ฅ cty of sin & cos
= tan(๐)
unless cos ๐ = 0.
But from graph above, cos ๐ = 0 except at
๐ฅ =๐
2+ ๐๐, where ๐ is an integer. โ
Theorem (Arithmetic combinations of continuous
functions)
If ๐ and ๐ are continuous at no. ๐ and ๐ is a constant,
then the following combinations are continuous at ๐.
1. ๐ + ๐ sum
2. ๐ โ ๐ difference
3. ๐๐ constant multiple
4. ๐๐ product
5. ๐/๐ provided ๐ ๐ โ 0 quotient
Proof. Each part follows from the corresponding limit
law. For example, consider (4)
lim๐ฅโ๐ ๐ ๐ฅ โ ๐ ๐ฅ product law for limits
= lim๐ฅโ๐ ๐ ๐ฅ โ lim๐ฅโ๐ ๐(๐ฅ)
continuity of ๐ and ๐
= ๐ ๐ โ ๐(๐)
This is the definition of continuity of ๐ ๐ฅ โ ๐(๐ฅ) โ
26a 26b
Continuity from the left and from the right
A function ๐ is continuous from the right at no. ๐ if
lim๐ฅโ๐+
๐ ๐ฅ = ๐(๐)
and ๐ is continuous from the left and no. ๐ if
lim๐ฅโ๐โ
๐ ๐ฅ = ๐(๐)
Example. Step Function
๐ ๐ฅ = 1, ๐ฅ โฅ 00, ๐ฅ < 0
โฆ 0 โฆ๐ฅ-, 0 โฆ 1 โฆ๐ฆ-, ๐(๐ฅ)
๐(๐ฅ) is continuous from the right at ๐ฅ = 0 and
continuous at every other ๐ฅ. โ
Example. Greatest integer function
๐ฅ is the largest integer less than or equal to ๐ฅ
โ1 โฆ 0 โฆ 1 โฆ๐ฅ-, โ1 โฆ 2 โฆ๐ฆ-, ๐ฆ = ๐ฅ
๐ฆ = ๐ฅ is continuous from the right at every integer.
Continuity on an interval (including endpoints)
A function ๐ 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. ๐ ๐ฅ = ๐ฅ is continuous on [0, โ).
Why? For any ๐ > 0, by the limit laws for roots
lim๐ฅโ๐
๐ฅ = lim๐ฅโ๐
๐ฅ = ๐
For ๐ = 0, lim๐ฅโ0+ ๐ฅ = 0 = ๐
which is โobviousโ from the graph
0 โฆ๐ฅ-, 0 โฆ๐ฆ-, ๐ฆ = ๐ฅ
โ
?? Practice with continuity transparency.
Continuity and Compositions
Composition: Function Machine Picture
๐ฅ โ ๐ โ ๐ข = ๐(๐ฅ)
๐ข โ ๐ โ ๐ฆ = ๐(๐ข)
๐ฆ = ๐ ๐ข = ๐(๐(๐ฅ)) ๐ is the โinner functionโ
๐ is the โouter functionโ
Example. ๐น ๐ฅ = cos ๐ฅ
๐ ๐ฅ = ๐ฅ
๐ ๐ข = cos(๐ข)โ
28a 28b
Theorem (Limits of Compositions)
If ๐ is continuous at ๐ and
lim๐ฅโ๐ ๐ ๐ฅ = ๐
then
lim๐ฅโ๐ ๐ ๐ ๐ฅ = ๐(lim๐ฅโ๐ ๐(๐ฅ)) = ๐(๐)
Intuitively, if ๐ฅ is close to ๐ then ๐(๐ฅ) is close to ๐.
Since ๐ is continuous at ๐, if ๐(๐ฅ) is close to ๐ then
๐ ๐ ๐ฅ is close to ๐(๐).
Example. Evaluate lim๐ฅโ๐2 cos ๐ฅ = ๐ฟ
From the root law, lim๐ฅโ๐2 ๐ฅ = ๐
๐ ๐ข = cos(๐ข) is continuous at ๐ข = ๐
By the theorem
๐ฟ = cos(lim๐ฅโ๐2 ๐ฅ) = cos ๐ = โ1 โ
Combine the theorem above with the condition that
๐ = ๐ ๐ , in other words the condition that ๐ is
continuous at ๐, to get:
Theorem (Compositions of continuous functions)
If ๐ is continuous at a no. ๐ and ๐ is continuous at
๐(๐), then ๐ ๐ ๐ฅ is continuous at ๐.
Proof. From the theorem above
lim๐ฅโ๐ ๐ ๐ ๐ฅ = ๐ lim๐ฅโ๐ ๐ ๐ฅ
= ๐ ๐ ๐
This is the definition of continuity for ๐ ๐ ๐ฅ . โ
In words, a continuous function of a continuous
function is continuous.
29a 29b
Example. Where is ๐น ๐ฅ = cos( ๐ฅ) continuous?
Note that ๐น ๐ฅ = ๐ ๐ ๐ฅ where
๐ ๐ฅ = ๐ฅ is continuous for ๐ฅ > 0
๐ ๐ข = cos(๐ข) is continuous for any real no. ๐ข
By the theorem, ๐น is continuous for ๐ฅ โฅ 0. โ
Intermediate Value Theorem
Let ๐ be continuous on the closed interval [๐, ๐] with
๐ ๐ โ ๐(๐). If ๐ is any number between ๐(๐) and
๐(๐), then there is a no. ๐ in (๐, ๐) such that
๐ ๐ = ๐.
Idea โฆ๐โฆ๐โฆ๐ฅ-, 0 โฆ๐ ๐ โฆ๐ ๐ โฆ๐ฆ-, ๐, ๐, ๐
Proof. Not given, but the intermediate value
theorem is โobviousโ.
Example. Prove that there is an ๐ฅ that solves
sin ๐ฅ = 1 โ ๐ฅ
on the interval (0,๐
2).
Proof. Let ๐ ๐ฅ = 1 โ ๐ฅ โ sin(๐ฅ)
then ๐ 0 = 1 โ 0 โ sin 0 = 1 > 0
๐ ๐
2 = 1 โ
๐
2โ 1 = โ
๐
2< 0
Since ๐ = 0 lies between ๐ 0 > 0 and ๐ ๐
2 < 0,
and ๐ is an arithmetic combination of continuous
functions (and therefore continuous itself), it follows
from the Intermediate Value Theorem that there is a
๐ โ 0,๐
2 such that ๐ ๐ = 0. โ
30a 30b
ยง1.6 Limits involving infinity
Infinite Limits
Example. ๐ฆ = ๐ ๐ฅ = 1/๐ฅ.
โ2 โฆ 0 โฆ 2 โฆ๐ฅ-, โ2 โฆ 0 โฆ 2 โฆ๐ฆ-, ๐(๐ฅ)
We write lim๐ฅโ0+ ๐ ๐ฅ = โ
The symbol โ is a version of DNE. It means ๐(๐ฅ)
becomes greater than any fixed value.
We write lim๐ฅโ0โ ๐ ๐ฅ = โโ
The symbol โโ means that ๐(๐ฅ) becomes less than
any fixed value โ
Example. ๐ ๐ฅ = 1/ ๐ฅ โ 3 2
0 โฆ 6 โฆ๐ฅ-, 0 โฆ 4 โฆ๐ฆ-, ๐(๐ฅ)
?? lim๐ฅโ3+ ๐ ๐ฅ =
?? lim๐ฅโ3โ ๐ ๐ฅ =
The one sided limit are โequalโ (they DNE in the
same way)
Thus we write lim๐ฅโ3 ๐ ๐ฅ = โ.
31a 31b
The vertical line ๐ฅ = ๐ is a vertical asymptote of
๐ฆ = ๐(๐ฅ) if
1) lim๐ฅโ๐+ ๐ ๐ฅ = +โ or โโ
and / or
2) lim๐ฅโ๐โ ๐ ๐ฅ = +โ or โโ
๐ฅ = ๐ is a vertical asymptote. add ๐ฅ = 3
Example. ๐ฆ = log2(๐ฅ)
domain is โ+ = { positive numbers}
Table x/ ยผ =2-2, ยฝ =2-1, 1=20, 2=21, 4=22 / y โฆ
0 โฆ 4 โฆ๐ฅ-, โ2 โฆ 0 โฆ 2 โฆ๐ฆ-, dots, curve
The y-axis (or line ๐ฅ = 0) is a vertical asymptote. โ
32a 32b
Limits at Infinity
Example. ๐ ๐ฅ = ๐ฅ2/(๐ฅ2 + 1)
Note that ๐ has even symmetry, so we only need to
tabulate values of ๐ฅ โฅ 0.
Table: x/ 0, 1, 2, 3 / f(x) โฆ
โ3 โฆ 0 โฆ 3 โฆ๐ฅ-, 0 โฆ 1 โฆ๐ฆ-, dots, curve
๐(๐ฅ) approaches 1 as |๐ฅ| increases โ
Definitions
lim๐ฅโโ ๐(๐ฅ) = ๐ฟ
means that ๐(๐ฅ) can be made as close as we wish to
๐ฟ by taking |๐ฅ| sufficiently large with ๐ฅ > 0.
lim๐ฅโโโ ๐(๐ฅ) = ๐ฟ
means that ๐(๐ฅ) can be made as close as we wish to
๐ฟ by taking |๐ฅ| sufficiently large with ๐ฅ < 0.
In the previous example,
lim๐ฅโโ ๐ ๐ฅ = 1, and lim๐ฅโโโ ๐(๐ฅ) = 1
The line ๐ฆ = ๐ฟ is a horizontal asymptote of ๐ฆ = ๐(๐ฅ)
if
lim๐ฅโโ ๐ ๐ฅ = ๐ฟ and/or lim๐ฅโโโ ๐ ๐ฅ = ๐ฟ
?? Transparency: Practice with asymptotes
33a 33b
Consider lim๐ฅโโ 1/๐ฅ
For ๐ฅ > 10, 0 <1
๐ฅ<
1
10
For ๐ฅ > 100, 0 <1
๐ฅ<
1
100
We can make 1/๐ฅ as close as we wish to zero by
taking ๐ฅ sufficiently large. Conclude
lim๐ฅโโ1
๐ฅ= 0
Consider lim๐ฅโโโ 1/๐ฅ
For ๐ฅ < โ10, โ1
10<
1
๐ฅ< 0
For ๐ฅ < โ100,โ1
100<
1
๐ฅ< 0
We can make 1/๐ฅ as close as we wish to zero by
taking |๐ฅ| sufficiently large with ๐ฅ < 0. Conclude
lim๐ฅโโโ1
๐ฅ= 0
Generalize
If ๐ is a positive real number
lim๐ฅโโ1
๐ฅ๐= 0 Equation (1)
we wonโt prove this
Examples. lim๐ฅโโ1
๐ฅ2= 0
lim๐ฅโโ1
๐ฅ= lim๐ฅโโ
1
๐ฅ12
= 0 โ
Infinite Limit Laws
These are obtained from the limit laws we have
already seen by replacing ๐ with ยฑโ.
Suppose that ๐ is a constant and
lim๐ฅโยฑโ ๐(๐ฅ), lim๐ฅโยฑโ ๐(๐ฅ)exist.
1. sum law (limit of sum is sum of limits)
lim๐ฅโยฑโ
๐ ๐ฅ + ๐ ๐ฅ = lim๐ฅโยฑโ
๐ ๐ฅ + lim๐ฅโยฑโ
๐(๐ฅ)
34a 34b
2. difference law
lim๐ฅโยฑโ
๐ ๐ฅ โ ๐ ๐ฅ = lim๐ฅโยฑโ
๐ ๐ฅ โ lim๐ฅโยฑโ
๐(๐ฅ)
3. constant multiple law
lim๐ฅโยฑโ
๐ ๐ ๐ฅ = ๐ lim๐ฅโยฑโ
๐(๐ฅ)
4. product law
lim๐ฅโยฑโ
๐ ๐ฅ ๐ ๐ฅ = lim๐ฅโยฑโ
๐ ๐ฅ โ lim๐ฅโยฑโ
๐(๐ฅ)
5. quotient law
lim๐ฅโยฑโ
๐ ๐ฅ
๐ ๐ฅ =
lim๐ฅโยฑโ
๐ ๐ฅ
lim๐ฅโยฑโ
๐ ๐ฅ
so long as lim๐ฅโยฑโ ๐ ๐ฅ โ 0
Example. Find lim๐ฅโโ2๐ฅ+3
16๐ฅ2+5= ๐ฟ.
๐ฟ = lim๐ฅโโ2๐ฅ+3
16๐ฅ2+5 multiply by 1
= lim๐ฅโโ2๐ฅ+3
16๐ฅ2+5
1/๐ฅ
1/๐ฅ (note ๐ฅ โ 0)
multiply numerator and denominator by 1/๐ฅ
= lim๐ฅโโ2+3/๐ฅ
16+5/๐ฅ2 quotient law
=lim ๐ฅโโ ( 2+3/๐ฅ)
lim ๐ฅโโ 16+5/๐ฅ2 (denominator is not zero)
=2+0
16+0 limit laws and equation 1
=1
2 โ
Example. Find lim๐ฅโโ5โ๐ฅโ3
๐ฅ+5
Cannot use quotient law because ๐ฅ + 5 โ 0โ as
๐ฅ โ โ5โ
Notice that as ๐ฅ โ โ5โ we have ๐ฅ โ 3 โ โ8โ
Since both the numerator and denominator are
negative as ๐ฅ โ โ5โ, the ratio must be positive.
Since the denominator is approaching 0 and the
numerator is not
lim๐ฅโโ5โ๐ฅโ3
๐ฅ+5= โ โ