relativeresourcemanager
TRANSCRIPT
Math 1190 – Calculus I
Ken Keating
Kennesaw State University
Chapter 1
Ken Keating (Kennesaw State University) Math 1190 – Calculus I Chapter 1 1 / 33
Table of Contents
1 Preface
2 Section 1.1
3 Section 1.2
4 Section 1.3
5 Section 1.4
6 Section 1.6
7 Section 1.7
8 Section 1.8
Ken Keating (Kennesaw State University) Math 1190 – Calculus I Chapter 1 2 / 33
What is Calculus?
According to Wikipedia:
Calculus is a branch of mathematics focused on limits, func-tions, derivatives, integrals, and infinite series. . . . It hastwo major branches, differential calculus and integral calcu-lus, which are related by the fundamental theorem of calculus.Calculus is the study of change, in the same way that geome-try is the study of shape and algebra is the study of operations andtheir application to solving equations. . . . Calculus has widespreadapplications in science, economics, and engineering and can solvemany problems for which algebra alone is insufficient.
Ken Keating (Kennesaw State University) Math 1190 – Calculus I Chapter 1 3 / 33
What is Calculus?
According to Wikipedia:
Calculus is a branch of mathematics focused on limits, func-tions, derivatives, integrals, and infinite series. . . . It hastwo major branches, differential calculus and integral calcu-lus, which are related by the fundamental theorem of calculus.Calculus is the study of change, in the same way that geome-try is the study of shape and algebra is the study of operations andtheir application to solving equations. . . . Calculus has widespreadapplications in science, economics, and engineering and can solvemany problems for which algebra alone is insufficient.
Ken Keating (Kennesaw State University) Math 1190 – Calculus I Chapter 1 3 / 33
Section 1.1 – Functions and Their Graphs
Definition
A function f from a set D to a set Y is a rule that assigns a unique(single) element f (x) ∈ Y to each element x ∈ D.
Ken Keating (Kennesaw State University) Math 1190 – Calculus I Section 1.1 4 / 33
Section 1.1 – Functions and Their Graphs
Definition
A function f from a set D to a set Y is a rule that assigns a unique(single) element f (x) ∈ Y to each element x ∈ D.
Ken Keating (Kennesaw State University) Math 1190 – Calculus I Section 1.1 4 / 33
Section 1.1 – Functions and Their Graphs
Definition
A function f from a set D to a set Y is a rule that assigns a unique(single) element f (x) ∈ Y to each element x ∈ D.
Ken Keating (Kennesaw State University) Math 1190 – Calculus I Section 1.1 4 / 33
Section 1.1 – Functions and Their Graphs
Functions pass the vertical line test
Relations that fail the vertical line test can be analyzed piecewiseI e.g. a circle
Types of functionsI piecewise-definedI increasing/decreasingI odd/evenI linearI powerI polynomialI rationalI algebraicI trigonometricI exponentialI logarithmicI transcendental
Ken Keating (Kennesaw State University) Math 1190 – Calculus I Section 1.1 5 / 33
Section 1.1 – Functions and Their Graphs
Functions pass the vertical line test
Relations that fail the vertical line test can be analyzed piecewise
I e.g. a circle
Types of functionsI piecewise-definedI increasing/decreasingI odd/evenI linearI powerI polynomialI rationalI algebraicI trigonometricI exponentialI logarithmicI transcendental
Ken Keating (Kennesaw State University) Math 1190 – Calculus I Section 1.1 5 / 33
Section 1.1 – Functions and Their Graphs
Functions pass the vertical line test
Relations that fail the vertical line test can be analyzed piecewiseI e.g. a circle
Types of functionsI piecewise-definedI increasing/decreasingI odd/evenI linearI powerI polynomialI rationalI algebraicI trigonometricI exponentialI logarithmicI transcendental
Ken Keating (Kennesaw State University) Math 1190 – Calculus I Section 1.1 5 / 33
Section 1.1 – Functions and Their Graphs
Functions pass the vertical line test
Relations that fail the vertical line test can be analyzed piecewiseI e.g. a circle
Types of functions
I piecewise-definedI increasing/decreasingI odd/evenI linearI powerI polynomialI rationalI algebraicI trigonometricI exponentialI logarithmicI transcendental
Ken Keating (Kennesaw State University) Math 1190 – Calculus I Section 1.1 5 / 33
Section 1.1 – Functions and Their Graphs
Functions pass the vertical line test
Relations that fail the vertical line test can be analyzed piecewiseI e.g. a circle
Types of functionsI piecewise-definedI increasing/decreasingI odd/evenI linearI powerI polynomialI rationalI algebraicI trigonometricI exponentialI logarithmicI transcendental
Ken Keating (Kennesaw State University) Math 1190 – Calculus I Section 1.1 5 / 33
Section 1.2 – Combining Functions; Shifting and ScalingGraphs
(f + g)(x), (f − g)(x), (fg)(x),
(f
g
)(x), including domain
(f ◦ g)(x) = f (g(x)), including domain
shifts: y = f (x) + k (vertical) , y = f (x + h) (horizontal)
scaling: y = cf (x) (vertical) , y = f (cx) (horizontal)
reflecting: y = −f (x) (x-axis) , y = f (−x) (y-axis)
Ken Keating (Kennesaw State University) Math 1190 – Calculus I Section 1.2 6 / 33
Section 1.2 – Combining Functions; Shifting and ScalingGraphs
(f + g)(x), (f − g)(x), (fg)(x),
(f
g
)(x), including domain
(f ◦ g)(x) = f (g(x)), including domain
shifts: y = f (x) + k (vertical) , y = f (x + h) (horizontal)
scaling: y = cf (x) (vertical) , y = f (cx) (horizontal)
reflecting: y = −f (x) (x-axis) , y = f (−x) (y-axis)
Ken Keating (Kennesaw State University) Math 1190 – Calculus I Section 1.2 6 / 33
Section 1.2 – Combining Functions; Shifting and ScalingGraphs
(f + g)(x), (f − g)(x), (fg)(x),
(f
g
)(x), including domain
(f ◦ g)(x) = f (g(x)), including domain
shifts: y = f (x) + k (vertical) , y = f (x + h) (horizontal)
scaling: y = cf (x) (vertical) , y = f (cx) (horizontal)
reflecting: y = −f (x) (x-axis) , y = f (−x) (y-axis)
Ken Keating (Kennesaw State University) Math 1190 – Calculus I Section 1.2 6 / 33
Section 1.2 – Combining Functions; Shifting and ScalingGraphs
(f + g)(x), (f − g)(x), (fg)(x),
(f
g
)(x), including domain
(f ◦ g)(x) = f (g(x)), including domain
shifts: y = f (x) + k (vertical) , y = f (x + h) (horizontal)
scaling: y = cf (x) (vertical) , y = f (cx) (horizontal)
reflecting: y = −f (x) (x-axis) , y = f (−x) (y-axis)
Ken Keating (Kennesaw State University) Math 1190 – Calculus I Section 1.2 6 / 33
Section 1.2 – Combining Functions; Shifting and ScalingGraphs
(f + g)(x), (f − g)(x), (fg)(x),
(f
g
)(x), including domain
(f ◦ g)(x) = f (g(x)), including domain
shifts: y = f (x) + k (vertical) , y = f (x + h) (horizontal)
scaling: y = cf (x) (vertical) , y = f (cx) (horizontal)
reflecting: y = −f (x) (x-axis) , y = f (−x) (y-axis)
Ken Keating (Kennesaw State University) Math 1190 – Calculus I Section 1.2 6 / 33
Section 1.2 – Combining Functions; Shifting and ScalingGraphs
(f + g)(x), (f − g)(x), (fg)(x),
(f
g
)(x), including domain
(f ◦ g)(x) = f (g(x)), including domain
shifts: y = f (x) + k (vertical) , y = f (x + h) (horizontal)
scaling: y = cf (x) (vertical) , y = f (cx) (horizontal)
reflecting: y = −f (x) (x-axis) , y = f (−x) (y-axis)
Ken Keating (Kennesaw State University) Math 1190 – Calculus I Section 1.2 6 / 33
Section 1.3 – Rates of Change and Tangents to Curves
Recall slope =rise
run=
change in y
change in x=
y2 − y1x2 − x1
=∆y
∆x
ex: linear function – note ∆y and ∆x
Ken Keating (Kennesaw State University) Math 1190 – Calculus I Section 1.3 7 / 33
Section 1.3 – Rates of Change and Tangents to Curves
Recall slope =rise
run=
change in y
change in x=
y2 − y1x2 − x1
=∆y
∆x
ex: linear function – note ∆y and ∆x
Ken Keating (Kennesaw State University) Math 1190 – Calculus I Section 1.3 7 / 33
Section 1.3 – Rates of Change and Tangents to Curves
Recall slope =rise
run=
change in y
change in x=
y2 − y1x2 − x1
=∆y
∆x
ex: linear function – note ∆y and ∆x
Ken Keating (Kennesaw State University) Math 1190 – Calculus I Section 1.3 7 / 33
Section 1.3 – Rates of Change and Tangents to Curves
ex: quadratic function
Note: slope of secant line betweenP and Q = average rate of changebetween P and Q
Ken Keating (Kennesaw State University) Math 1190 – Calculus I Section 1.3 8 / 33
Section 1.3 – Rates of Change and Tangents to Curves
ex: quadratic function
Note: slope of secant line betweenP and Q = average rate of changebetween P and Q
Ken Keating (Kennesaw State University) Math 1190 – Calculus I Section 1.3 8 / 33
Section 1.3 – Rates of Change and Tangents to Curves
ex: quadratic function
Note: slope of secant line betweenP and Q = average rate of changebetween P and Q
Ken Keating (Kennesaw State University) Math 1190 – Calculus I Section 1.3 8 / 33
Section 1.3 – Rates of Change and Tangents to Curves
Definition
The average rate of change of y = f (x) with respect to x over the interval[x1, x2] is
∆y
∆x=
f (x2)− f (x1)
x2 − x1=
f (x1 + h)− f (x1)
h, h 6= 0
So we know we can find the slope of the line between any two points Pand Q on a curve. What about the slope of the curve at any givenpoint?
It’s reasonable (and correct, as we’ll see later this semester) to associatethe slope of the line tangent to a point on a curve with the slope of thecurve at that point.
Ken Keating (Kennesaw State University) Math 1190 – Calculus I Section 1.3 9 / 33
Section 1.3 – Rates of Change and Tangents to Curves
Definition
The average rate of change of y = f (x) with respect to x over the interval[x1, x2] is
∆y
∆x=
f (x2)− f (x1)
x2 − x1=
f (x1 + h)− f (x1)
h, h 6= 0
So we know we can find the slope of the line between any two points Pand Q on a curve. What about the slope of the curve at any givenpoint?
It’s reasonable (and correct, as we’ll see later this semester) to associatethe slope of the line tangent to a point on a curve with the slope of thecurve at that point.
Ken Keating (Kennesaw State University) Math 1190 – Calculus I Section 1.3 9 / 33
Section 1.3 – Rates of Change and Tangents to Curves
Definition
The average rate of change of y = f (x) with respect to x over the interval[x1, x2] is
∆y
∆x=
f (x2)− f (x1)
x2 − x1=
f (x1 + h)− f (x1)
h, h 6= 0
So we know we can find the slope of the line between any two points Pand Q on a curve. What about the slope of the curve at any givenpoint?
It’s reasonable (and correct, as we’ll see later this semester) to associatethe slope of the line tangent to a point on a curve with the slope of thecurve at that point.
Ken Keating (Kennesaw State University) Math 1190 – Calculus I Section 1.3 9 / 33
Section 1.3 – Rates of Change and Tangents to Curves
As Q moves closer to P,the slope of the secant line between P and Q approaches the slope ofthe tangent line at P
the length of ∆x = h gets closer to 0
the average rate of change between P and Q approaches theinstantaneous rate of change at P
Ken Keating (Kennesaw State University) Math 1190 – Calculus I Section 1.3 10 / 33
Section 1.3 – Rates of Change and Tangents to Curves
As Q moves closer to P,
the slope of the secant line between P and Q approaches the slope ofthe tangent line at P
the length of ∆x = h gets closer to 0
the average rate of change between P and Q approaches theinstantaneous rate of change at P
Ken Keating (Kennesaw State University) Math 1190 – Calculus I Section 1.3 10 / 33
Section 1.3 – Rates of Change and Tangents to Curves
As Q moves closer to P,the slope of the secant line between P and Q approaches the slope ofthe tangent line at P
the length of ∆x = h gets closer to 0
the average rate of change between P and Q approaches theinstantaneous rate of change at P
Ken Keating (Kennesaw State University) Math 1190 – Calculus I Section 1.3 10 / 33
Section 1.3 – Rates of Change and Tangents to Curves
As Q moves closer to P,the slope of the secant line between P and Q approaches the slope ofthe tangent line at P
the length of ∆x = h gets closer to 0
the average rate of change between P and Q approaches theinstantaneous rate of change at P
Ken Keating (Kennesaw State University) Math 1190 – Calculus I Section 1.3 10 / 33
Section 1.3 – Rates of Change and Tangents to Curves
As Q moves closer to P,the slope of the secant line between P and Q approaches the slope ofthe tangent line at P
the length of ∆x = h gets closer to 0
the average rate of change between P and Q approaches theinstantaneous rate of change at P
Ken Keating (Kennesaw State University) Math 1190 – Calculus I Section 1.3 10 / 33
Section 1.3 – Rates of Change and Tangents to Curves
ex: (p.23 #10) Find the slope of the curve at the given point P and findthe equation of the tangent line at P.
y = 5− x2, P(1, 4)
Ken Keating (Kennesaw State University) Math 1190 – Calculus I Section 1.3 11 / 33
Section 1.4 – Limit of a Function and Limit Laws
Before defining what a limit is, let’s look at an example and try todetermine what the function values are approaching as x → 1.
f (−1) = 0 f (−1) = 0 f (−1) = 0
f (0) = 1 f (0) = 1 f (0) = 1
f (1) = undefined f (1) = 1 f (1) = 2
as x → 1, y → 2 as x → 1, y → 2 as x → 1, y → 2
Ken Keating (Kennesaw State University) Math 1190 – Calculus I Section 1.4 12 / 33
Section 1.4 – Limit of a Function and Limit Laws
Before defining what a limit is, let’s look at an example and try todetermine what the function values are approaching as x → 1.
f (−1) = 0 f (−1) = 0 f (−1) = 0
f (0) = 1 f (0) = 1 f (0) = 1
f (1) = undefined f (1) = 1 f (1) = 2
as x → 1, y → 2 as x → 1, y → 2 as x → 1, y → 2
Ken Keating (Kennesaw State University) Math 1190 – Calculus I Section 1.4 12 / 33
Section 1.4 – Limit of a Function and Limit Laws
Before defining what a limit is, let’s look at an example and try todetermine what the function values are approaching as x → 1.
f (−1) = 0 f (−1) = 0 f (−1) = 0
f (0) = 1 f (0) = 1 f (0) = 1
f (1) = undefined f (1) = 1 f (1) = 2
as x → 1, y → 2 as x → 1, y → 2 as x → 1, y → 2
Ken Keating (Kennesaw State University) Math 1190 – Calculus I Section 1.4 12 / 33
Section 1.4 – Limit of a Function and Limit Laws
Before defining what a limit is, let’s look at an example and try todetermine what the function values are approaching as x → 1.
f (−1) = 0 f (−1) = 0 f (−1) = 0
f (0) = 1 f (0) = 1 f (0) = 1
f (1) = undefined f (1) = 1 f (1) = 2
as x → 1, y → 2 as x → 1, y → 2 as x → 1, y → 2
Ken Keating (Kennesaw State University) Math 1190 – Calculus I Section 1.4 12 / 33
Section 1.4 – Limit of a Function and Limit Laws
Before defining what a limit is, let’s look at an example and try todetermine what the function values are approaching as x → 1.
f (−1) = 0 f (−1) = 0 f (−1) = 0
f (0) = 1 f (0) = 1 f (0) = 1
f (1) = undefined f (1) = 1 f (1) = 2
as x → 1, y → 2 as x → 1, y → 2 as x → 1, y → 2
Ken Keating (Kennesaw State University) Math 1190 – Calculus I Section 1.4 12 / 33
Section 1.4 – Limit of a Function and Limit Laws
Before defining what a limit is, let’s look at an example and try todetermine what the function values are approaching as x → 1.
f (−1) = 0 f (−1) = 0 f (−1) = 0
f (0) = 1 f (0) = 1 f (0) = 1
f (1) = undefined f (1) = 1 f (1) = 2
as x → 1, y → 2 as x → 1, y → 2 as x → 1, y → 2
Ken Keating (Kennesaw State University) Math 1190 – Calculus I Section 1.4 12 / 33
Section 1.4 – Limit of a Function and Limit Laws
Before defining what a limit is, let’s look at an example and try todetermine what the function values are approaching as x → 1.
f (−1) = 0 f (−1) = 0 f (−1) = 0
f (0) = 1 f (0) = 1 f (0) = 1
f (1) = undefined f (1) = 1 f (1) = 2
as x → 1, y → 2 as x → 1, y → 2 as x → 1, y → 2
Ken Keating (Kennesaw State University) Math 1190 – Calculus I Section 1.4 12 / 33
Section 1.4 – Limit of a Function and Limit Laws
Definition
Let f (x) be defined on an open interval about x0, except possibly at x0itself. If f (x) is arbitrarily close to L for all x sufficiently close to x0, wesay that f approaches the limit L as x approaches x0, and we write
limx→x0
f (x) = L.
Definition (formal – as given in Sec. 1.5)
Let f (x) be defined on an open interval about x0, except possibly at x0itself. We say that the limit of f(x) as x approaches x0 is the numberL, and write
limx→x0
f (x) = L,
if, for every number ε > 0, there exists a corresponding number δ > 0 suchthat for all x ,
0 < |x − x0| < δ ⇒ |f (x)− L| < ε.
Ken Keating (Kennesaw State University) Math 1190 – Calculus I Section 1.4 13 / 33
Section 1.4 – Limit of a Function and Limit Laws
Definition
Let f (x) be defined on an open interval about x0, except possibly at x0itself. If f (x) is arbitrarily close to L for all x sufficiently close to x0, wesay that f approaches the limit L as x approaches x0, and we write
limx→x0
f (x) = L.
Definition (formal – as given in Sec. 1.5)
Let f (x) be defined on an open interval about x0, except possibly at x0itself. We say that the limit of f(x) as x approaches x0 is the numberL, and write
limx→x0
f (x) = L,
if, for every number ε > 0, there exists a corresponding number δ > 0 suchthat for all x ,
0 < |x − x0| < δ ⇒ |f (x)− L| < ε.Ken Keating (Kennesaw State University) Math 1190 – Calculus I Section 1.4 13 / 33
Section 1.4 – Limit of a Function and Limit Laws
Note, functions may have point(s) where the limit does NOT exist:
(a) jumps
(b) grows too large to have a limit
(c) oscillates too much to have a limit
Ken Keating (Kennesaw State University) Math 1190 – Calculus I Section 1.4 14 / 33
Section 1.4 – Limit of a Function and Limit Laws
Note, functions may have point(s) where the limit does NOT exist:
(a) jumps
(b) grows too large to have a limit
(c) oscillates too much to have a limit
Ken Keating (Kennesaw State University) Math 1190 – Calculus I Section 1.4 14 / 33
Section 1.4 – Limit of a Function and Limit Laws
Note, functions may have point(s) where the limit does NOT exist:
(a) jumps
(b) grows too large to have a limit
(c) oscillates too much to have a limit
Ken Keating (Kennesaw State University) Math 1190 – Calculus I Section 1.4 14 / 33
Section 1.4 – Limit of a Function and Limit Laws
Note, functions may have point(s) where the limit does NOT exist:
(a) jumps
(b) grows too large to have a limit
(c) oscillates too much to have a limit
Ken Keating (Kennesaw State University) Math 1190 – Calculus I Section 1.4 14 / 33
Section 1.4 – Limit of a Function and Limit Laws
Note, functions may have point(s) where the limit does NOT exist:
(a) jumps
(b) grows too large to have a limit
(c) oscillates too much to have a limit
Ken Keating (Kennesaw State University) Math 1190 – Calculus I Section 1.4 14 / 33
Section 1.4 – Limit of a Function and Limit Laws
One common task we’ll have is to find the value of the limit (assuming itexists). To do this we’ll need to make use of ”limit laws”. First, however,we present a few special cases.
f is a constant function, i.e., f (x) = klimx→x0
f (x) = limx→x0
k = k
f is the identity function, i.e., f (x) = xlimx→x0
f (x) = limx→x0
x = x0
Now, our ”limit laws”:
Ken Keating (Kennesaw State University) Math 1190 – Calculus I Section 1.4 15 / 33
Section 1.4 – Limit of a Function and Limit Laws
One common task we’ll have is to find the value of the limit (assuming itexists). To do this we’ll need to make use of ”limit laws”. First, however,we present a few special cases.
f is a constant function, i.e., f (x) = k
limx→x0
f (x) = limx→x0
k = k
f is the identity function, i.e., f (x) = xlimx→x0
f (x) = limx→x0
x = x0
Now, our ”limit laws”:
Ken Keating (Kennesaw State University) Math 1190 – Calculus I Section 1.4 15 / 33
Section 1.4 – Limit of a Function and Limit Laws
One common task we’ll have is to find the value of the limit (assuming itexists). To do this we’ll need to make use of ”limit laws”. First, however,we present a few special cases.
f is a constant function, i.e., f (x) = klimx→x0
f (x) = limx→x0
k = k
f is the identity function, i.e., f (x) = xlimx→x0
f (x) = limx→x0
x = x0
Now, our ”limit laws”:
Ken Keating (Kennesaw State University) Math 1190 – Calculus I Section 1.4 15 / 33
Section 1.4 – Limit of a Function and Limit Laws
One common task we’ll have is to find the value of the limit (assuming itexists). To do this we’ll need to make use of ”limit laws”. First, however,we present a few special cases.
f is a constant function, i.e., f (x) = klimx→x0
f (x) = limx→x0
k = k
f is the identity function, i.e., f (x) = x
limx→x0
f (x) = limx→x0
x = x0
Now, our ”limit laws”:
Ken Keating (Kennesaw State University) Math 1190 – Calculus I Section 1.4 15 / 33
Section 1.4 – Limit of a Function and Limit Laws
One common task we’ll have is to find the value of the limit (assuming itexists). To do this we’ll need to make use of ”limit laws”. First, however,we present a few special cases.
f is a constant function, i.e., f (x) = klimx→x0
f (x) = limx→x0
k = k
f is the identity function, i.e., f (x) = xlimx→x0
f (x) = limx→x0
x = x0
Now, our ”limit laws”:
Ken Keating (Kennesaw State University) Math 1190 – Calculus I Section 1.4 15 / 33
Section 1.4 – Limit of a Function and Limit Laws
One common task we’ll have is to find the value of the limit (assuming itexists). To do this we’ll need to make use of ”limit laws”. First, however,we present a few special cases.
f is a constant function, i.e., f (x) = klimx→x0
f (x) = limx→x0
k = k
f is the identity function, i.e., f (x) = xlimx→x0
f (x) = limx→x0
x = x0
Now, our ”limit laws”:
Ken Keating (Kennesaw State University) Math 1190 – Calculus I Section 1.4 15 / 33
Section 1.4 – Limit of a Function and Limit Laws
Ken Keating (Kennesaw State University) Math 1190 – Calculus I Section 1.4 16 / 33
Section 1.4 – Limit of a Function and Limit Laws
In addition to our ”limit laws”, there are some theorems that will help usquickly compute limits.
Theorem (Limits of Polynomials)
If P(x) = anxn + an−1xn−1 + · · ·+ a0, then
limx→c
P(x) = P(c) = ancn + an−1cn−1 + · · ·+ a0.
In other words, to compute the limit as x approaches c of a polynomial,simply plug c in for x in the polynomial!
Ken Keating (Kennesaw State University) Math 1190 – Calculus I Section 1.4 17 / 33
Section 1.4 – Limit of a Function and Limit Laws
In addition to our ”limit laws”, there are some theorems that will help usquickly compute limits.
Theorem (Limits of Polynomials)
If P(x) = anxn + an−1xn−1 + · · ·+ a0, then
limx→c
P(x) = P(c) = ancn + an−1cn−1 + · · ·+ a0.
In other words, to compute the limit as x approaches c of a polynomial,simply plug c in for x in the polynomial!
Ken Keating (Kennesaw State University) Math 1190 – Calculus I Section 1.4 17 / 33
Section 1.4 – Limit of a Function and Limit Laws
In addition to our ”limit laws”, there are some theorems that will help usquickly compute limits.
Theorem (Limits of Polynomials)
If P(x) = anxn + an−1xn−1 + · · ·+ a0, then
limx→c
P(x) = P(c) = ancn + an−1cn−1 + · · ·+ a0.
In other words, to compute the limit as x approaches c of a polynomial,simply plug c in for x in the polynomial!
Ken Keating (Kennesaw State University) Math 1190 – Calculus I Section 1.4 17 / 33
Section 1.4 – Limit of a Function and Limit Laws
Theorem (Limits of Rational Functions)
If P(x) and Q(x) are polynomials and Q(c) 6= 0, then
limx→c
P(x)
Q(x)=
P(c)
Q(c).
Like we did for polynomials, if we need to compute the limit as xapproaches c of a rational function, we just plug c in for x in both thenumerator and denominator.
Note in the statement of the theorem the phrase ”and Q(c) 6= 0”. If, infact, we DO have Q(c) = 0, there are some techniques we can employ totry to compute the limit. These will be presented by example after wepresent the next two theorems.
Ken Keating (Kennesaw State University) Math 1190 – Calculus I Section 1.4 18 / 33
Section 1.4 – Limit of a Function and Limit Laws
Theorem (Limits of Rational Functions)
If P(x) and Q(x) are polynomials and Q(c) 6= 0, then
limx→c
P(x)
Q(x)=
P(c)
Q(c).
Like we did for polynomials, if we need to compute the limit as xapproaches c of a rational function, we just plug c in for x in both thenumerator and denominator.
Note in the statement of the theorem the phrase ”and Q(c) 6= 0”. If, infact, we DO have Q(c) = 0, there are some techniques we can employ totry to compute the limit. These will be presented by example after wepresent the next two theorems.
Ken Keating (Kennesaw State University) Math 1190 – Calculus I Section 1.4 18 / 33
Section 1.4 – Limit of a Function and Limit Laws
Theorem (Limits of Rational Functions)
If P(x) and Q(x) are polynomials and Q(c) 6= 0, then
limx→c
P(x)
Q(x)=
P(c)
Q(c).
Like we did for polynomials, if we need to compute the limit as xapproaches c of a rational function, we just plug c in for x in both thenumerator and denominator.
Note in the statement of the theorem the phrase ”and Q(c) 6= 0”. If, infact, we DO have Q(c) = 0, there are some techniques we can employ totry to compute the limit. These will be presented by example after wepresent the next two theorems.
Ken Keating (Kennesaw State University) Math 1190 – Calculus I Section 1.4 18 / 33
Section 1.4 – Limit of a Function and Limit Laws
Theorem (The Sandwich, or Squeeze, Theorem)
Suppose that g(x) ≤ f (x) ≤ h(x) for all x in some open intervalcontaining c, except possibly at x = c itself. Suppose also that
limx→c
g(x) = limx→c
h(x) = L.
Then limx→c f (x) = L.
Suppose we don’t know to compute the limit as x approaches c of thefunction f . If we know that f is always between two other functions, gand h, whose limits as x approaches c both equal L, then we areguaranteed that the limit as x approaches c of the function f is also L.
Ken Keating (Kennesaw State University) Math 1190 – Calculus I Section 1.4 19 / 33
Section 1.4 – Limit of a Function and Limit Laws
Theorem (The Sandwich, or Squeeze, Theorem)
Suppose that g(x) ≤ f (x) ≤ h(x) for all x in some open intervalcontaining c, except possibly at x = c itself. Suppose also that
limx→c
g(x) = limx→c
h(x) = L.
Then limx→c f (x) = L.
Suppose we don’t know to compute the limit as x approaches c of thefunction f . If we know that f is always between two other functions, gand h, whose limits as x approaches c both equal L, then we areguaranteed that the limit as x approaches c of the function f is also L.
Ken Keating (Kennesaw State University) Math 1190 – Calculus I Section 1.4 19 / 33
Section 1.4 – Limit of a Function and Limit Laws
Ken Keating (Kennesaw State University) Math 1190 – Calculus I Section 1.4 20 / 33
Section 1.4 – Limit of a Function and Limit Laws
Theorem
If f (x) ≤ g(x) for all x in some open interval containing c, except possiblyat x = c itself, and the limits of f and g both exist as x approaches c, then
limx→c
f (x) ≤ limx→c
g(x).
Ken Keating (Kennesaw State University) Math 1190 – Calculus I Section 1.4 21 / 33
Section 1.4 – Limit of a Function and Limit Laws
ex: p.31 #2
ex: p.32 #10
ex: p.32 #14
ex: p.32 #20
ex: p.32 #32
ex: p.32 #42
ex: p.33 #56
ex: p.33 #66a
ex: p.33 #73
ex: p.34 #75
Ken Keating (Kennesaw State University) Math 1190 – Calculus I Section 1.4 22 / 33
Section 1.4 – Limit of a Function and Limit Laws
ex: p.31 #2
ex: p.32 #10
ex: p.32 #14
ex: p.32 #20
ex: p.32 #32
ex: p.32 #42
ex: p.33 #56
ex: p.33 #66a
ex: p.33 #73
ex: p.34 #75
Ken Keating (Kennesaw State University) Math 1190 – Calculus I Section 1.4 22 / 33
Section 1.4 – Limit of a Function and Limit Laws
ex: p.31 #2
ex: p.32 #10
ex: p.32 #14
ex: p.32 #20
ex: p.32 #32
ex: p.32 #42
ex: p.33 #56
ex: p.33 #66a
ex: p.33 #73
ex: p.34 #75
Ken Keating (Kennesaw State University) Math 1190 – Calculus I Section 1.4 22 / 33
Section 1.4 – Limit of a Function and Limit Laws
ex: p.31 #2
ex: p.32 #10
ex: p.32 #14
ex: p.32 #20
ex: p.32 #32
ex: p.32 #42
ex: p.33 #56
ex: p.33 #66a
ex: p.33 #73
ex: p.34 #75
Ken Keating (Kennesaw State University) Math 1190 – Calculus I Section 1.4 22 / 33
Section 1.4 – Limit of a Function and Limit Laws
ex: p.31 #2
ex: p.32 #10
ex: p.32 #14
ex: p.32 #20
ex: p.32 #32
ex: p.32 #42
ex: p.33 #56
ex: p.33 #66a
ex: p.33 #73
ex: p.34 #75
Ken Keating (Kennesaw State University) Math 1190 – Calculus I Section 1.4 22 / 33
Section 1.4 – Limit of a Function and Limit Laws
ex: p.31 #2
ex: p.32 #10
ex: p.32 #14
ex: p.32 #20
ex: p.32 #32
ex: p.32 #42
ex: p.33 #56
ex: p.33 #66a
ex: p.33 #73
ex: p.34 #75
Ken Keating (Kennesaw State University) Math 1190 – Calculus I Section 1.4 22 / 33
Section 1.4 – Limit of a Function and Limit Laws
ex: p.31 #2
ex: p.32 #10
ex: p.32 #14
ex: p.32 #20
ex: p.32 #32
ex: p.32 #42
ex: p.33 #56
ex: p.33 #66a
ex: p.33 #73
ex: p.34 #75
Ken Keating (Kennesaw State University) Math 1190 – Calculus I Section 1.4 22 / 33
Section 1.4 – Limit of a Function and Limit Laws
ex: p.31 #2
ex: p.32 #10
ex: p.32 #14
ex: p.32 #20
ex: p.32 #32
ex: p.32 #42
ex: p.33 #56
ex: p.33 #66a
ex: p.33 #73
ex: p.34 #75
Ken Keating (Kennesaw State University) Math 1190 – Calculus I Section 1.4 22 / 33
Section 1.4 – Limit of a Function and Limit Laws
ex: p.31 #2
ex: p.32 #10
ex: p.32 #14
ex: p.32 #20
ex: p.32 #32
ex: p.32 #42
ex: p.33 #56
ex: p.33 #66a
ex: p.33 #73
ex: p.34 #75
Ken Keating (Kennesaw State University) Math 1190 – Calculus I Section 1.4 22 / 33
Section 1.4 – Limit of a Function and Limit Laws
ex: p.31 #2
ex: p.32 #10
ex: p.32 #14
ex: p.32 #20
ex: p.32 #32
ex: p.32 #42
ex: p.33 #56
ex: p.33 #66a
ex: p.33 #73
ex: p.34 #75
Ken Keating (Kennesaw State University) Math 1190 – Calculus I Section 1.4 22 / 33
Section 1.4 – Limit of a Function and Limit Laws
”Toolbox” Summary – When computing limx→c
f (x):
first try plugging c in for x
simplify f (x) by canceling common factors, then take the limit
multiply numerator and denominator of f (x) by the ”conjugate”,simplify f (x), then take the limit
when all else fails, make a table of values letting x get closer andcloser to c (from both sides) and see if you can determine if f (x) isapproaching a real number L
Ken Keating (Kennesaw State University) Math 1190 – Calculus I Section 1.4 23 / 33
Section 1.6 – One-Sided Limits
limx→c
f (x) = L means f (x)→ L as x → c from both sides of c
thus, when we say limx→c
f (x) = L, we mean a two-sided limit
a function that fails to have a two-sided limit at c may still haveone-sided limits at c
right-hand limit: limx→c+
f (x) = L if x ∈ (c , b) and f (x)→ L as x → c
left-hand limit: limx→c−
f (x) = L if x ∈ (a, c) and f (x)→ L as x → c
all laws/properties/rules/theorems from Section 1.4 hold for one-sidedlimits
let’s revisit the graphs depicting functions that failed to have a limitas x → 0
Ken Keating (Kennesaw State University) Math 1190 – Calculus I Section 1.6 24 / 33
Section 1.6 – One-Sided Limits
limx→c
f (x) = L means f (x)→ L as x → c from both sides of c
thus, when we say limx→c
f (x) = L, we mean a two-sided limit
a function that fails to have a two-sided limit at c may still haveone-sided limits at c
right-hand limit: limx→c+
f (x) = L if x ∈ (c , b) and f (x)→ L as x → c
left-hand limit: limx→c−
f (x) = L if x ∈ (a, c) and f (x)→ L as x → c
all laws/properties/rules/theorems from Section 1.4 hold for one-sidedlimits
let’s revisit the graphs depicting functions that failed to have a limitas x → 0
Ken Keating (Kennesaw State University) Math 1190 – Calculus I Section 1.6 24 / 33
Section 1.6 – One-Sided Limits
limx→c
f (x) = L means f (x)→ L as x → c from both sides of c
thus, when we say limx→c
f (x) = L, we mean a two-sided limit
a function that fails to have a two-sided limit at c may still haveone-sided limits at c
right-hand limit: limx→c+
f (x) = L if x ∈ (c , b) and f (x)→ L as x → c
left-hand limit: limx→c−
f (x) = L if x ∈ (a, c) and f (x)→ L as x → c
all laws/properties/rules/theorems from Section 1.4 hold for one-sidedlimits
let’s revisit the graphs depicting functions that failed to have a limitas x → 0
Ken Keating (Kennesaw State University) Math 1190 – Calculus I Section 1.6 24 / 33
Section 1.6 – One-Sided Limits
limx→c
f (x) = L means f (x)→ L as x → c from both sides of c
thus, when we say limx→c
f (x) = L, we mean a two-sided limit
a function that fails to have a two-sided limit at c may still haveone-sided limits at c
right-hand limit: limx→c+
f (x) = L if x ∈ (c , b) and f (x)→ L as x → c
left-hand limit: limx→c−
f (x) = L if x ∈ (a, c) and f (x)→ L as x → c
all laws/properties/rules/theorems from Section 1.4 hold for one-sidedlimits
let’s revisit the graphs depicting functions that failed to have a limitas x → 0
Ken Keating (Kennesaw State University) Math 1190 – Calculus I Section 1.6 24 / 33
Section 1.6 – One-Sided Limits
limx→c
f (x) = L means f (x)→ L as x → c from both sides of c
thus, when we say limx→c
f (x) = L, we mean a two-sided limit
a function that fails to have a two-sided limit at c may still haveone-sided limits at c
right-hand limit: limx→c+
f (x) = L if x ∈ (c , b) and f (x)→ L as x → c
left-hand limit: limx→c−
f (x) = L if x ∈ (a, c) and f (x)→ L as x → c
all laws/properties/rules/theorems from Section 1.4 hold for one-sidedlimits
let’s revisit the graphs depicting functions that failed to have a limitas x → 0
Ken Keating (Kennesaw State University) Math 1190 – Calculus I Section 1.6 24 / 33
Section 1.6 – One-Sided Limits
limx→c
f (x) = L means f (x)→ L as x → c from both sides of c
thus, when we say limx→c
f (x) = L, we mean a two-sided limit
a function that fails to have a two-sided limit at c may still haveone-sided limits at c
right-hand limit: limx→c+
f (x) = L if x ∈ (c , b) and f (x)→ L as x → c
left-hand limit: limx→c−
f (x) = L if x ∈ (a, c) and f (x)→ L as x → c
all laws/properties/rules/theorems from Section 1.4 hold for one-sidedlimits
let’s revisit the graphs depicting functions that failed to have a limitas x → 0
Ken Keating (Kennesaw State University) Math 1190 – Calculus I Section 1.6 24 / 33
Section 1.6 – One-Sided Limits
limx→c
f (x) = L means f (x)→ L as x → c from both sides of c
thus, when we say limx→c
f (x) = L, we mean a two-sided limit
a function that fails to have a two-sided limit at c may still haveone-sided limits at c
right-hand limit: limx→c+
f (x) = L if x ∈ (c , b) and f (x)→ L as x → c
left-hand limit: limx→c−
f (x) = L if x ∈ (a, c) and f (x)→ L as x → c
all laws/properties/rules/theorems from Section 1.4 hold for one-sidedlimits
let’s revisit the graphs depicting functions that failed to have a limitas x → 0
Ken Keating (Kennesaw State University) Math 1190 – Calculus I Section 1.6 24 / 33
Section 1.6 – One-Sided Limits
limx→c
f (x) = L means f (x)→ L as x → c from both sides of c
thus, when we say limx→c
f (x) = L, we mean a two-sided limit
a function that fails to have a two-sided limit at c may still haveone-sided limits at c
right-hand limit: limx→c+
f (x) = L if x ∈ (c , b) and f (x)→ L as x → c
left-hand limit: limx→c−
f (x) = L if x ∈ (a, c) and f (x)→ L as x → c
all laws/properties/rules/theorems from Section 1.4 hold for one-sidedlimits
let’s revisit the graphs depicting functions that failed to have a limitas x → 0
Ken Keating (Kennesaw State University) Math 1190 – Calculus I Section 1.6 24 / 33
Section 1.6 – One-Sided Limits
limx→0
f (x) = DNE limx→0
f (x) = DNE limx→0
f (x) = DNE
limx→0−
f (x) = 0 limx→0−
f (x) = DNE limx→0−
f (x) = 0
limx→0+
f (x) = 1 limx→0+
f (x) = DNE limx→0+
f (x) = DNE
Ken Keating (Kennesaw State University) Math 1190 – Calculus I Section 1.6 25 / 33
Section 1.6 – One-Sided Limits
limx→0
f (x) = DNE
limx→0
f (x) = DNE limx→0
f (x) = DNE
limx→0−
f (x) = 0 limx→0−
f (x) = DNE limx→0−
f (x) = 0
limx→0+
f (x) = 1 limx→0+
f (x) = DNE limx→0+
f (x) = DNE
Ken Keating (Kennesaw State University) Math 1190 – Calculus I Section 1.6 25 / 33
Section 1.6 – One-Sided Limits
limx→0
f (x) = DNE
limx→0
f (x) = DNE limx→0
f (x) = DNE
limx→0−
f (x) =
0 limx→0−
f (x) = DNE limx→0−
f (x) = 0
limx→0+
f (x) = 1 limx→0+
f (x) = DNE limx→0+
f (x) = DNE
Ken Keating (Kennesaw State University) Math 1190 – Calculus I Section 1.6 25 / 33
Section 1.6 – One-Sided Limits
limx→0
f (x) = DNE
limx→0
f (x) = DNE limx→0
f (x) = DNE
limx→0−
f (x) = 0
limx→0−
f (x) = DNE limx→0−
f (x) = 0
limx→0+
f (x) = 1 limx→0+
f (x) = DNE limx→0+
f (x) = DNE
Ken Keating (Kennesaw State University) Math 1190 – Calculus I Section 1.6 25 / 33
Section 1.6 – One-Sided Limits
limx→0
f (x) = DNE
limx→0
f (x) = DNE limx→0
f (x) = DNE
limx→0−
f (x) = 0
limx→0−
f (x) = DNE limx→0−
f (x) = 0
limx→0+
f (x) =
1 limx→0+
f (x) = DNE limx→0+
f (x) = DNE
Ken Keating (Kennesaw State University) Math 1190 – Calculus I Section 1.6 25 / 33
Section 1.6 – One-Sided Limits
limx→0
f (x) = DNE
limx→0
f (x) = DNE limx→0
f (x) = DNE
limx→0−
f (x) = 0
limx→0−
f (x) = DNE limx→0−
f (x) = 0
limx→0+
f (x) = 1
limx→0+
f (x) = DNE limx→0+
f (x) = DNE
Ken Keating (Kennesaw State University) Math 1190 – Calculus I Section 1.6 25 / 33
Section 1.6 – One-Sided Limits
limx→0
f (x) = DNE limx→0
f (x) = DNE
limx→0
f (x) = DNE
limx→0−
f (x) = 0
limx→0−
f (x) = DNE limx→0−
f (x) = 0
limx→0+
f (x) = 1
limx→0+
f (x) = DNE limx→0+
f (x) = DNE
Ken Keating (Kennesaw State University) Math 1190 – Calculus I Section 1.6 25 / 33
Section 1.6 – One-Sided Limits
limx→0
f (x) = DNE limx→0
f (x) = DNE
limx→0
f (x) = DNE
limx→0−
f (x) = 0 limx→0−
f (x) =
DNE limx→0−
f (x) = 0
limx→0+
f (x) = 1
limx→0+
f (x) = DNE limx→0+
f (x) = DNE
Ken Keating (Kennesaw State University) Math 1190 – Calculus I Section 1.6 25 / 33
Section 1.6 – One-Sided Limits
limx→0
f (x) = DNE limx→0
f (x) = DNE
limx→0
f (x) = DNE
limx→0−
f (x) = 0 limx→0−
f (x) = DNE
limx→0−
f (x) = 0
limx→0+
f (x) = 1
limx→0+
f (x) = DNE limx→0+
f (x) = DNE
Ken Keating (Kennesaw State University) Math 1190 – Calculus I Section 1.6 25 / 33
Section 1.6 – One-Sided Limits
limx→0
f (x) = DNE limx→0
f (x) = DNE
limx→0
f (x) = DNE
limx→0−
f (x) = 0 limx→0−
f (x) = DNE
limx→0−
f (x) = 0
limx→0+
f (x) = 1 limx→0+
f (x) =
DNE limx→0+
f (x) = DNE
Ken Keating (Kennesaw State University) Math 1190 – Calculus I Section 1.6 25 / 33
Section 1.6 – One-Sided Limits
limx→0
f (x) = DNE limx→0
f (x) = DNE
limx→0
f (x) = DNE
limx→0−
f (x) = 0 limx→0−
f (x) = DNE
limx→0−
f (x) = 0
limx→0+
f (x) = 1 limx→0+
f (x) = DNE
limx→0+
f (x) = DNE
Ken Keating (Kennesaw State University) Math 1190 – Calculus I Section 1.6 25 / 33
Section 1.6 – One-Sided Limits
limx→0
f (x) = DNE limx→0
f (x) = DNE limx→0
f (x) = DNE
limx→0−
f (x) = 0 limx→0−
f (x) = DNE
limx→0−
f (x) = 0
limx→0+
f (x) = 1 limx→0+
f (x) = DNE
limx→0+
f (x) = DNE
Ken Keating (Kennesaw State University) Math 1190 – Calculus I Section 1.6 25 / 33
Section 1.6 – One-Sided Limits
limx→0
f (x) = DNE limx→0
f (x) = DNE limx→0
f (x) = DNE
limx→0−
f (x) = 0 limx→0−
f (x) = DNE limx→0−
f (x) =
0
limx→0+
f (x) = 1 limx→0+
f (x) = DNE
limx→0+
f (x) = DNE
Ken Keating (Kennesaw State University) Math 1190 – Calculus I Section 1.6 25 / 33
Section 1.6 – One-Sided Limits
limx→0
f (x) = DNE limx→0
f (x) = DNE limx→0
f (x) = DNE
limx→0−
f (x) = 0 limx→0−
f (x) = DNE limx→0−
f (x) = 0
limx→0+
f (x) = 1 limx→0+
f (x) = DNE
limx→0+
f (x) = DNE
Ken Keating (Kennesaw State University) Math 1190 – Calculus I Section 1.6 25 / 33
Section 1.6 – One-Sided Limits
limx→0
f (x) = DNE limx→0
f (x) = DNE limx→0
f (x) = DNE
limx→0−
f (x) = 0 limx→0−
f (x) = DNE limx→0−
f (x) = 0
limx→0+
f (x) = 1 limx→0+
f (x) = DNE limx→0+
f (x) =
DNE
Ken Keating (Kennesaw State University) Math 1190 – Calculus I Section 1.6 25 / 33
Section 1.6 – One-Sided Limits
limx→0
f (x) = DNE limx→0
f (x) = DNE limx→0
f (x) = DNE
limx→0−
f (x) = 0 limx→0−
f (x) = DNE limx→0−
f (x) = 0
limx→0+
f (x) = 1 limx→0+
f (x) = DNE limx→0+
f (x) = DNE
Ken Keating (Kennesaw State University) Math 1190 – Calculus I Section 1.6 25 / 33
Section 1.6 – One-Sided Limits
One-sided limits are related to (two-sided) limits in the following way:
Theorem
A function f (x) has a limit as x approaches c if and only if it has left-handand right-hand limits there and these one-sided limits are equal:
limx→c
f (x) = L ⇐⇒ limx→c−
f (x) = L and limx→c+
f (x) = L.
The ”if and only if” (⇐⇒) in this theorem means:
if limx→c
f (x) = L, then both one-sided limits exist and are equal to L
if both one-sided limits exist and are equal to L, then limx→c
f (x) = L
The power of this theorem is in the second bullet. Let’s see it action.
Ken Keating (Kennesaw State University) Math 1190 – Calculus I Section 1.6 26 / 33
Section 1.6 – One-Sided Limits
One-sided limits are related to (two-sided) limits in the following way:
Theorem
A function f (x) has a limit as x approaches c if and only if it has left-handand right-hand limits there and these one-sided limits are equal:
limx→c
f (x) = L ⇐⇒ limx→c−
f (x) = L and limx→c+
f (x) = L.
The ”if and only if” (⇐⇒) in this theorem means:
if limx→c
f (x) = L, then both one-sided limits exist and are equal to L
if both one-sided limits exist and are equal to L, then limx→c
f (x) = L
The power of this theorem is in the second bullet. Let’s see it action.
Ken Keating (Kennesaw State University) Math 1190 – Calculus I Section 1.6 26 / 33
Section 1.6 – One-Sided Limits
Theorem
limθ→0
sin θ
θ= 1 (θin radians)
Ken Keating (Kennesaw State University) Math 1190 – Calculus I Section 1.6 27 / 33
Section 1.6 – One-Sided Limits
To prove the theorem we are going to show that both the left-handand right-hand limits are equal to 1.
We’ll employ concepts from algebra, geometry, and trigonometry, aswell as calculus, in the proof.
Recall:
I area of a triangle =1
2bh
I area of a circle = πr2
I radians in a circle = 2π
I area of a sector =θ
2π· πr2 =
1
2r2θ
I tan θ =opposite
adjacent
I measured in radians, sin θ =x
r
Ken Keating (Kennesaw State University) Math 1190 – Calculus I Section 1.6 28 / 33
Section 1.6 – One-Sided Limits
Notes:the circle is a unit circle
0 < θ <π
2
area ∆OAP =1
2sin θ
area sector OAP =1
2θ
area ∆OAT =1
2tan θ
Ken Keating (Kennesaw State University) Math 1190 – Calculus I Section 1.6 29 / 33
Section 1.6 – One-Sided Limits
Proof.
area ∆OAP < area sector OAP < area ∆OAT
12 sin θ < 1
2θ <12 tan θ
divide each term by 12 sin θ (a positive #, so inequality signs unchanged)
1 < θsin θ <
1cos θ
take the reciprocal of each term (reverses the inequality signs)
1 > sin θθ > cos θ
Since limθ→0+ 1 = 1 and limθ→0+ cos θ = 1, by the Sandwich Theoremlimθ→0+
sin θθ = 1. Also, since sin θ and θ are both odd functions, sin θ
θ is an
even function, so limθ→0−sin θθ = limθ→0+
sin θθ . Thus, limθ→0
sin θθ = 1.
Ken Keating (Kennesaw State University) Math 1190 – Calculus I Section 1.6 30 / 33
Section 1.6 – One-Sided Limits
ex: p.48 #6
ex: p.48 #8
ex: p.48 #12
ex: p.48 #16
ex: p.48 #22
ex: p.48 #32
Ken Keating (Kennesaw State University) Math 1190 – Calculus I Section 1.6 31 / 33
Section 1.6 – One-Sided Limits
ex: p.48 #6
ex: p.48 #8
ex: p.48 #12
ex: p.48 #16
ex: p.48 #22
ex: p.48 #32
Ken Keating (Kennesaw State University) Math 1190 – Calculus I Section 1.6 31 / 33
Section 1.6 – One-Sided Limits
ex: p.48 #6
ex: p.48 #8
ex: p.48 #12
ex: p.48 #16
ex: p.48 #22
ex: p.48 #32
Ken Keating (Kennesaw State University) Math 1190 – Calculus I Section 1.6 31 / 33
Section 1.6 – One-Sided Limits
ex: p.48 #6
ex: p.48 #8
ex: p.48 #12
ex: p.48 #16
ex: p.48 #22
ex: p.48 #32
Ken Keating (Kennesaw State University) Math 1190 – Calculus I Section 1.6 31 / 33
Section 1.6 – One-Sided Limits
ex: p.48 #6
ex: p.48 #8
ex: p.48 #12
ex: p.48 #16
ex: p.48 #22
ex: p.48 #32
Ken Keating (Kennesaw State University) Math 1190 – Calculus I Section 1.6 31 / 33
Section 1.6 – One-Sided Limits
ex: p.48 #6
ex: p.48 #8
ex: p.48 #12
ex: p.48 #16
ex: p.48 #22
ex: p.48 #32
Ken Keating (Kennesaw State University) Math 1190 – Calculus I Section 1.6 31 / 33
Section 1.7 – Continuity
Ken Keating (Kennesaw State University) Math 1190 – Calculus I Section 1.7 32 / 33
Section 1.8 – Limits Involving Infinity
Ken Keating (Kennesaw State University) Math 1190 – Calculus I Section 1.8 33 / 33