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MATH 31 LESSONS
Chapter 1: Limits
1. Linear Functions and Tangents
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Section 1.1: Linear Functions and
The Tangent Problem
Read Textbook pp. 5 - 9
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A. Linear Functions
Recall
y = m x + b
where
m = slope
b = y-intercept
x
y
b
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x
y
(x1, y1)
(x2, y2)
x = x2 - x1
y = y2 - y1
12
12
Δ
Δslope
xx
yy
x
y
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Note:
is called the rate of change of y with respect to x
i.e. how quickly y changes relative to changes in
x
x
y
Δ
Δ
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is called the rate of change of y with respect to x
i.e. how quickly y changes relative to changes in
x
If y is positive, then y is increasing
If y is negative, then y is decreasing
x
y
Δ
Δ
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e.g.
How can this be interpreted?
x
y
x = 2
y = 52
5
Δ
Δ
x
y
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When x increases by 2 units, y increases by 5 units.
x
y
x = 2
y = 52
5
Δ
Δ
x
y
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Clearly,
the greater the slope
the greater the rate of change of y (with respect to x)
the “faster” y changes
Low rate of changeHigh rate of change
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Note:
x
y
Δ
Δ
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Ex. 1 Find the equation of the linear function that passes
through A(2, 1) and B(-4, -9).
Answer in general form.
Try this example on your own first.Then, check out the solution.
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A(2, 1) and B(-4, -9)
Find slope
x1 y1 x2 y2
24
19
12
12
xx
yym
6
10
3
5
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A(2, 1)
Use point-slope formula:
x1 y1
11 xxmyy
23
51 xy
3
5m
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11 xxmyy
23
51 xy
2513 xy Multiply both sides by 3 to remove the denominator
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2513 xy
10533 xy
0735 yx
Ax + By + C = 0
A > 0
A, B, and C are integers
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Ex. 2 For the linear function 32x + 4y - 19 = 0,
if x decreases by 2, how does y change?
Try this example on your own first.Then, check out the solution.
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Find slope
019432 yx
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019432 yx
19324 xy
4
198 xy
Put in slope-intercept form: y = mx + b
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The slope is -8
019432 yx
19324 xy
4
198 xy
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Find the change in y (y) :
8m8Δ
Δm
x
y
2x
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8m8Δ
Δm
x
y
xy 8
2x
28
16
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So, y increases by 16
8Δ
Δm
x
y
xy 8
28
16
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B. The Tangent Problem
What is a tangent line?
The question is more challenging than it appears,
and as we will see, only calculus can truly answer it.
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For circles, the tangent line is readily defined.
A tangent is a straight line
that touches the circle
only once. Tangent, t
Not tangent
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A tangent line is a straight line that touches the
circle only once
However, the definition above is not adequate for
general curves.
To show why, consider the next two illustrations.
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Consider the following parabola:
y = f (x)
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Both l and t touch the curve only once.
However, only t is a tangent.
l
t
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This line t is a tangent.
t
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For an even more interesting example, consider
the cubic function below:
y = f (x)
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The line t is a tangent line, even though it crosses
the curve twice!
t
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So, the statement ...
“a tangent line is a straight line that touches a curve
only once”
is clearly inadequate.
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C. An Approach to the Tangent Problem
Much of this unit is devoted to solving the
tangent problem.
To illustrate the approach we will take, consider
the following problem.
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Question: Find the slope of the tangent line to the
parabola y = x2 at the point P(2, 4).
y = x2
y
x2
4
t
P(2, 4)
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Problem:
In order to find the slope
of the tangent line, we need
to know two points on
the tangent line.
However, we only know one.
y = x2
y
x2
4
t
P(2, 4)
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Solution:
We introduce a second point
on the curve y = x2,
somewhere close to P.
We will call this point Q,
and it will have the
coordinates (x, x2).
y = x2
x 2
4
t
P(2, 4)
Q(x, x2)
x2
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The line l that connects
P and Q is called a
secant line.
It can be thought of as
an approximation of
the tangent line.
y = x2
x 2
4
t
P(2, 4)
Q(x, x2)
x2
l
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The slope of the secant
line l is given byy = x2
x 2
4
t
P(2, 4)
Q(x, x2)
x2
l
12
12
xx
yyml
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The slope of the secant
line l is given byy = x2
x 2
4
t
P(2, 4)
Q(x, x2)
x2
l
12
12
xx
yyml
2
42
x
x
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The slope of the secant
line l is given byy = x2
x 2
4
t
P(2, 4)
Q(x, x2)
x2
l
12
12
xx
yyml
2
42
x
x
2
22
x
xx
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The slope of the secant
line l is given byy = x2
x 2
4
t
P(2, 4)
Q(x, x2)
x2
l
12
12
xx
yyml
2
42
x
x
2,2 xx
2
22
x
xx
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How can we use the secant
slope to find the tangent
slope?
The main idea is to
start moving Q closer
and closer to the point P.
y = x2
x 2
4
t
P(2, 4)
Q(x, x2)
x2
l
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Notice that as Q
approaches P ...
x approaches 2
l approaches t
mPQ approaches mt
y = x2
x 2
4 P(2, 4)
Q(x, x2)x2
t l
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In fact, if Q gets really close
to P, the secant line l is almost
identical to the tangent line t
It follows that the slope
of the secant line would
be almost identical
to the slope of the tangent.
y = x2
x2
4 P(2, 4)Q(x, x2)
x2
t l
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In calculus, we attempt to
bring Q right to P, so that
the line l actually becomes
the tangent line t.
In this way, the slope is
exactly the same.
But how do we do this?
We use the idea of limits.
y = x2
2
4 P(2, 4)Q
t l
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We will now find the
pattern of slopes as Q
approaches P
(i.e. as x approaches 2
“from the left”)y = x2
x 2
4
t
P(2, 4)
Q(x, x2)
x2
l
x mt = x + 2
1
1.5
1.9
1.99
1.999
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Slope of the secant as x approaches 2 from the left:
x mt = x + 2
1
1.5
1.9
1.99
1.999
3
3.5
3.9
3.99
3.999
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Slope of the secant as x approaches 2 from the left:
x mt = x + 2
1
1.5
1.9
1.99
1.999
3
3.5
3.9
3.99
3.999
2 4
As x approaches 2 from the left, the slopes appear to be approaching 4
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We should also find the slope of the secants as
x approaches 2 from the right as well:
x mt = x + 2
3
2.5
2.1
2.01
2.001
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So, as x gets extremely close to 2,
the slope of the secant lines get extremely close to 4.
In fact, in calculus we state that if x gets infinitely close to 2
(e.g. x = 1.9 ), the slopes would become so close to 4
(e.g. slope = 3.9 ) as to become actually equal to 4.
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So, as x gets extremely close to 2,
the slope of the secant lines get extremely close to 4.
In fact, in calculus we state that if x gets infinitely close to 2
(e.g. x = 1.9 ), the slopes would become so close to 4
(e.g. slope = 3.9 ) as to become actually equal to 4.
This value of 4 would represent the true slope of the tangent.
We say that the slope of the tangent is the limit
of the secant slopes, because it represents the limit (endpoint)
of slopes as x gets infinitely close to 2.
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Notation:
If the tangent slope is the limit of the secant slopes,
then we write:
tPQPQ
mm
lim
We say, “The limit as Q approaches P of the slopes of secant line PQ is the slope of the tangent.”
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If the tangent slope is the limit of the secant slopes,
then we write:
or
tPQPQ
mm
lim
42
4lim
2
2
x
xx
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Note:
We could also find the
equation of the tangent,
since we now know the
slope and one point.
y = x2
2
4 P(2, 4)
t
mt = 4
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Equation of tangent: P(2, 4) mt = 4
11 xxmyy
244 xy
844 xy
44 xy or 044 yx
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Ex. 3 For the function y = x2 - 5x, determine the equation
of the tangent line at the point P(2, -6).
Be certain to:
- include a sketch
- find the formula for the secant slope
- see what the slopes approach as x -2
from both sides to get the tangent slope
Try this example on your own first.Then, check out the solution.
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Sketch: y = x2 - 5xy
x2
-6
t
P(2, -6)
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We introduce
a second point
Q (x, x2 - 5x)
on the curve.
The point PQ is the
secant line.
As Q approaches P,
the secant approaches the tangent.
y = x2 - 5xy
x2
-6
t
Q(x, x2 - 5x)
P(2, -6)
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Slope of secant:
y = x2 - 5xy
x2
-6
t
P(2, -6)
Q(x, x2 - 5x)
12
12
xx
yyml
2
652
x
xx
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Slope of secant:
12
12
xx
yyml
2
652
x
xx
2
652
x
xx 2
23
x
xx
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Slope of secant:
12
12
xx
yyml
2
652
x
xx
2
652
x
xx 2
23
x
xx
2,3 xx
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Slope of the secant as x approaches 2 from the left:
x mt = x - 3
1
1.5
1.9
1.99
1.999
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Slope of the secant as x approaches 2 from the left:
x mt = x - 3
1
1.5
1.9
1.99
1.999
-2
-1.5
-1.1
-1.01
-1.001
2 -1
As x approaches 2 from the left, the slopes appear to be approaching -1
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Slope of the secant as x approaches 2 from the right:
x mt = x - 3
3
2.5
2.1
2.01
2.001
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Slope of the secant as x approaches 2 from the left:
x mt = x - 3
3
2.5
2.1
2.01
2.001
0
-0.5
-0.9
-0.99
-0.999
2 -1
As x approaches 2 from the right, the slopes appear to be approaching -1 as well
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The tangent slope is the limit of the secant slopes.
So, it follows that
PQPQ
t mm
lim
2
65lim
2
2
x
xxx
1
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Equation of tangent: P(2, -6) mt = -1
11 xxmyy
216 xy
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Equation of tangent: P(2, -6) mt = -1
11 xxmyy
216 xy
26 xy
4 xy or 04 yx