Math 31 Chapter 1 Limits Lessons
Lesson 1: Review/Preview
Review of factoring:
1. GCF:
5𝑥 2 − 25𝑥 + 15𝑥 3
5𝑥(𝑥 − 5 + 3𝑥 2 )
5𝑥(3𝑥 2 + 𝑥 − 5)
2. Difference of Squares:
𝑎2 − 𝑏 2 = (𝑎 − 𝑏)(𝑎 + 𝑏)
3. Trinomials:
a. Easy:
𝑥 2 + 𝑏𝑥 + 𝑐
𝑥 2 + 5𝑥 − 6
(x + 6)(x – 1)
Ex.
b. Lucky:
𝑎𝑥 2 + 𝑏𝑥 + 𝑐
Ex. 2𝑥 2 − 5𝑥 + 2
2x
1x
–
–
1
2
2x(-2x)= -4x
1x(-1) = -1x
-5x
Diagonal multiply to see if it works out to the middle term. Since it does, draw a line
through the middle. Whatever is above the line goes in one bracket and below the line
goes in the second bracket….. and there’s your answer.
Answer: (2x – 1)(x – 2)
c. Decomposition:
2𝑥 2 − 5𝑥 + 2
Multiply first and last numbers together. What multiplies to 4 and adds to -5? Take
those two numbers and split up the middle term.
Factor by grouping.
2𝑥 2 − 4𝑥 − 1𝑥 + 2
2(x – 2) –1(x – 2)
(x – 2)(2x – 1)
Binomials/Trinomials with fractional or negative exponents
**Look for the term with the smallest exponent and use it as a common factor.
5
1
1. 𝑥 2 − 𝑥 2
1
**smallest exponent is ½
4
𝑥 2 (𝑥 2 − 1)
1
𝑥 2 (𝑥 2 − 1)
1
𝑥 2 (𝑥 − 1)(𝑥 + 1)
2. 𝑥 + 5 + 6𝑥 −1
𝑥 −1 (𝑥 2 + 5𝑥 + 6)
𝑥 −1 (𝑥 + 3)(𝑥 + 2)
**smallest exponent is -1
**remember that 5 has an x with a 0 exponent
3. 5𝑥 2 (𝑥 2 + 4)2 − (𝑥 2 + 4)3
(𝑥 2 + 4)2 [5𝑥 2 − (𝑥 2 + 4)1 ]
(𝑥 2 + 4)2 [5𝑥 2 − 𝑥 2 − 4]
(𝑥 2 + 4)2 [4𝑥 2 − 4]
4(𝑥 2 + 4)2 [𝑥 2 − 1]
4(𝑥 2 + 4)2 (𝑥 − 1)(𝑥 + 1)
Homework questions:
p. 3 #1, 2adgh, 4cef
Lesson 2: Sum/Difference of Cubes/Factor Theorem/Rationalizing Num. or Den.
1. Sum of Cubes: 𝑎3 + 𝑏 3 = (𝑎 + 𝑏)(𝑎2 − 𝑎𝑏 + 𝑏 2 )
𝑥 3 + 125 is really 𝑥 3 + 53
Ex.
=(𝑥 + 5)(𝑥 2 − 5𝑥 + 25)
2. Difference of Cubes: 𝑎3 − 𝑏 3 = (𝑎 − 𝑏)(𝑎2 + 𝑎𝑏 + 𝑏 2 )
27𝑎3 − 64𝑏 3 is really (3𝑎)3 − (4𝑏)3
Ex.
= (3𝑎 − 4𝑏)(9𝑎2 + 12𝑎𝑏 + 16𝑏 2 )
3. Factor Theorem: A polynomial P(x) has x – b as a factor if and only if P(b) = 0.
Ex. 𝑥 3 − 𝑥 2 − 16𝑥 + 16
***try the factors of 16 (the constant term). Start with the easiest: 1.
(1)3 − (1)2 − 16(−1) + 16 = 0 so x – 1 is a factor. (Remember 30-1?)
Use synthetic or long division….synthetic is usually easier.
𝑥2
x–1
− 16
𝑥 3 − 𝑥 2 − 16𝑥 + 16
𝑥3 − 𝑥2
−16𝑥 + 16
−16𝑥 + 16
0
1
1
-1
1
1
0
Answer:
-16
0
-16
𝑥 2 − 16
To find the final answer, factor 𝑥 2 − 16
Answer: (x – 1)(x – 4)(x + 4)
16
-16
0
Rationalizing numerator and denominator
**needed skill for limits**
Multiply both top and bottom by the conjugate radical (opposite sign in the middle, same
terms)
Numerator:
√𝑥 − 3
𝑥−9
=
=
√𝑥 − 3 √𝑥 + 3
×
𝑥 − 9 √𝑥 + 3
𝑥−9
(𝑥 − 9)(√𝑥 + 3)
**cancel out the x – 9 and the final answer is:
=
1
√𝑥 + 3
Denominator:
1
√𝑥 + 1 − 1
=
1
𝑥
(√𝑥 + 1 + 1)
√𝑥 + 1 − 1 (√𝑥 + 1 + 1)
=
√𝑥 + 1 + 1
𝑥+1−1
=
√𝑥 + 1 + 1
𝑥
Homework assignment: p. 3 #2bce, 3bdf; p. 4 #1c – f, 2cd
Lesson 3: Linear Functions and Tangent Problems (1.1)
We will show how a limit arises when we try to find a tangent to a curve.
Linear function: y = mx + b
m: slope
OR
f(x) = mx + b
b: y-intercept
∆𝑦
𝑦 −𝑦
slope = ∆𝑥 = 𝑥2 −𝑥1
2
1
Slope is also called the rate of change: the rate of change of y with respect to x.
Positive slopes: up and to the right
Negative slopes: down and to the right
Horizontal lines: slope of 0
Vertical lines: undefined slopes
Ex. Find the equation of a line passing through P(-1, 4) and Q(7, -2).
−2−4
𝑚 = 7−−1 =
−6
8
3
= −4
Plug one of the points and the slope into y = mx + b to solve for b.
3
4 = − 4(-1) + b
3
4=4+𝑏
13
4
=𝑏
3
Equation: 𝑦 = − 4 𝑥 +
13
4
If the question asks for general form, multiply everything by 4 to get rid of the common
denominator:
4y = -3x + 13
Move everything to the side where x is positive and then put things in alphabetical order
and equal to 0.
3x + 4y – 13 = 0
Answer:
To draw, you can go from the equation in slope/point form. Plot the y-intercept first and
then do the slope.
Ex. For 3x + 2y – 6 = 0, if x increases by 3, how does y change?
First find the slope either by algebra (putting it in y = mx + b form) or the fast way.
Fast way:
Slope:
𝑐𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡 𝑜𝑓 𝑥
𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 𝑠𝑖𝑔𝑛 𝑐𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡 𝑜𝑓 𝑦
3
−2
3
∆𝑦
=
−2
3
Cross multiply to get −2∆𝑦 = 9
9
∆𝑦 = − 2
9
Therefore, y decreases by − 2
The Tangent Problem
“tangens”: Latin for touching
For simple curves, the tangent intersects once. But for more complicated functions, that
isn’t true.
Given a complex curve problem, how can we find the slope of a tangent line?
i.e. How can we find the instantaneous rate of change in y with respect to x?
Q
P
The slope of the secant line PQ gives the average rate of change in y as x changes
from 𝑥1 to 𝑥2 . The slope of the tangent line at P gives the instantaneous rate of change
in y at x. So, to find the slope at P, we have to find the limiting value of mPQ as Q
approaches P.
Therefore, the slope of the tangent at P = lim 𝑚𝑃𝑄
𝑄→𝑃
(Q is the variable point and P is the fixed point)
Ex. Find the equation of the line tangent to 𝑦 = 𝑥 2 at the point (1, 1).
The problem here is that we only have one point and we need two points to make a
slope. BUT we can choose a point close to the one we have and approximate.
Choose x ≠ 1 so that Q ≠ P.
𝑦 −𝑦
So, since slope = 𝑥2−𝑥1 our slope for this line would
2
1
be:
(x, y)
(1, 1)
𝑦−1
slope = 𝑥−1
Since Q is on parabola, 𝑦 = 𝑥 2 , you can replace y with 𝑥 2 .
Therefore,
𝑦−1
=
𝑥−1
𝑥 2 −1
𝑥−1
=
(𝑥−1)(𝑥+1)
𝑥−1
=x+1
Now, a point close to (1, 1) would be (1.1, 1.21). The 1.21 comes from (1.1)2. Plug that
into the formula:
1.21−1
1.1−1
=
0.21
0.1
= 2.1
OR, plug the 1.1 into the x + 1 formula: 1.1 + 1 = 2.1
Now look at the values approaching from the right and left.
From the right x > 1.
x
2
1.5
1.1
1.01
1.001
From the left x < 1
mPQ
3
2.5
2.1
2.01
2.001
x
0
0.5
0.9
0.99
0.999
mPQ
1
1.5
1.9
1.99
1.999
The closer the point Q is to P, the closer x is to 1 and the slope is to 2. That suggests
that our slope should be 2!
Therefore, lim 𝑚𝑃𝑄 = 2 and lim
𝑄→𝑃
𝑥 2 −1
𝑥→1 𝑥−1
=2
Then the equation of a line with slope 2 and going through (1, 1) is:
y = mx + b
1 = 2(1) + b
-1 = b
Answer:
y = 2x – 1
Homework: p. 9 #1 – 7, 11
Lesson 4: The Limit of a Function (1.2)
Limits also arise when we find velocity or other rates of change. Limits are basic to all
calculus.
lim 𝑓(𝑥) = 𝐿 means:
𝑥→𝑎
The limit of f(x) as x approaches a, is L. i.e. as x approaches a, f(x) (or in other words,
y) approaches L.
Important to note: when finding the limit, you need to find the approach from below and
from above.
Look at p. 11 example in the textbook. If you look at the table of values, when x < 3, the
limit approaches 0.5. When x > 3, the limit approaches 0.5 also. Notice that it is not
equal to x because it is not defined at x (vertical asymptote).
If you look at the graph, there is an open circle at (3, 0.5) showing that it is not actually
equal. So we would say:
𝑥−3
= 0.5
𝑥→3 𝑥 2 − 4𝑥 + 3
lim
So it gets closer and closer to 0.5 but never actually gets there because x ≠ a.
Check out the following example:
Examples: Find the limit using the graph of f ( x) provided.
1. lim f ( x)
x 1
2. lim f ( x)
x 3
3. lim f ( x)
x 3
Answers: 1. 4
2. 1
3. DNE (does not exist)
We can find the limits using the lovely properties on page 13. You do not have to
memorize these (many of them are intuitive) but than can be helpful.
**read through these with the students**
Basic limits:
lim 𝑥 = 𝑎
𝑥→𝑎
lim 𝑐 = 𝑐
c = constant
𝑥→𝑎
lim 𝑥 𝑛 = 𝑎𝑛
𝑥→𝑎
𝑛
𝑛
𝑛
lim √𝑥 = √𝑎
if √𝑎 exists
𝑥→𝑎
Find the following limits:
(general rule: substitute in the number. If it works, great! If it doesn’t, we need to use
some type of algebra)
1. lim 𝑥 3 − 3𝑥 2 = (2)3 − 3(2)2 = −4
𝑥→2
2. lim
𝑥→2
𝑥 3 −3𝑥 2
𝑥−3
=
(2)3 −3(2)2
2−3
−4
= −1 = 4
3. lim √3𝑥 2 − 𝑥 3 −= √3(2)2 − (2)3 = √4 = 2
𝑥→2
4. lim
𝑥→3
𝑥 3 −3𝑥 2
𝑥−3
=
𝑥 2 (𝑥−3)
𝑥−3
= 𝑥 2 = (3)2 = 9
**for this one you need to factor first, otherwise the denominator would be 0, thus
being undefined.
A function, f(x), is continuous (no gaps) at x = a if and only if lim 𝑓(𝑥) = 𝑓(𝑎).
𝑥→𝑎
Any polynomial function P(x) is continuous everywhere; so lim 𝑃( 𝑥) = 𝑃(𝑎).
𝑥→𝑎
𝑃(𝑥)
Any rational function 𝑓(𝑥) = 𝑄(𝑥), where P(x) and Q(x) are polynomials, is continuous at
every number a such that Q(a) ≠ 0.
𝑃(𝑥)
𝑃(𝑎)
lim 𝑄(𝑥) = 𝑄(𝑎), Q(a) ≠ 0.
𝑥→𝑎
Sometimes we will have to factor first and reduce like we did in Example 4 above.
Ex. lim
𝑥 2 −2𝑥−24
𝑥+4
𝑥→−4
=
(𝑥−6)(𝑥+4)
𝑥+4
=𝑥−6
Once we have simplified , then we can substitute in x = -4. Therefore, lim (- 4 – 6 = -10)
𝑥→−4
𝑃(𝑎)
0
Tip: If 𝑄(𝑎) = 0, then x – a is probably a factor….so try simplifying.
Ex. lim
𝑥→−3
𝑥 2 −9
𝑥 3 +2𝑥 2 −5𝑥−6
0
= 0 when we plug in -3.
So, we need to factor. The numerator is easy because it is a difference of squares.
The bottom is a cubic function. We are going to use our tip that x + 3 might be a factor.
Use long or synthetic division.
-3
1
1
Answer:
2
-5
-6
-3
-1
3
-2
6
0
𝑥 2 − 𝑥 − 2 = (x – 2)(x + 1)
So therefore our limit would look like:
𝑥2 − 9
(𝑥 − 3)(𝑥 + 3)
𝑥−3
=
=
𝑥→−3 𝑥 3 + 2𝑥 2 − 5𝑥 − 6
(𝑥 + 3)(𝑥 − 2)(𝑥 + 1) (𝑥 − 2)(𝑥 + 1)
lim
Now plug in x = -3.
−3 − 3
−6
−6 −3
=
=
=
(−3 − 2)(−3 + 1) (−5)(−2) 10
5
These will be represented by a hole (open circle) in the graph. (Point of discontinuity
from Math 30-1)
Sometimes you need to multiply out and then cancel/factor to solve.
Ex.
(2 + ℎ)2 − 4 (2 + ℎ)(2 + ℎ) − 4 4 + 4ℎ + ℎ2 − 4 4ℎ + ℎ2 ℎ(4 + ℎ)
=
=
=
=
=4+ℎ
ℎ→0
ℎ
ℎ
ℎ
ℎ
ℎ
lim
So substitute in h = 0. Therefore, 4 + 0 = 4.
Sometimes you need to rationalize either the numerator or denominator to solve.
Ex.
16 + ℎ − 16
ℎ
√16 + ℎ − 4 √16 + ℎ − 4 (√16 + ℎ + 4)
=
×
=
=
ℎ→0
ℎ
ℎ
(√16 + ℎ + 4) ℎ(√16 + ℎ + 4) ℎ(√16 + ℎ + 4)
lim
=
1
(√16 + ℎ + 4)
=
1
(√16 + 0 + 4)
=
1
1
=
(4 + 4) 8
Note: don’t multiply out the denominator unless absolutely necessary….
Sometimes you need to multiply by the common denominator to solve
Ex.
1
1
−2
2
+
ℎ
lim
ℎ→0
ℎ
*Multiply by 2(2 + h)
=
2(2 + ℎ)(
1
1
) − (2)(2)(2 + ℎ) 2 − (2 + ℎ)
−ℎ
−1
2+ℎ
=
=
=
ℎ(2)(2 + ℎ)
2ℎ(2 + ℎ)
2ℎ(2 + ℎ) 2(2 + ℎ)
−1
−1
=
2(2 + 0)
4
Go through example 8 on p. 18 together showing that limits may not exist.
Homework: p. 18 #1 (all), 3 – 4 (odd letters), 5 – 6 (c – e)
Lesson 5: One-Sided Limits (1.3)
There are more complex limits. The ones we’ve done so far have been simple
𝑓(𝑥) = {𝑥 2
𝑖𝑓 𝑥 ≤ 1
3 − 𝑥 𝑖𝑓 𝑥 > 1
Ex.
Remember that a function is a rule: you need to follow what it says.
Go through the example on p. 21 together (ignore the bottom explanation).
The ordinary two-sided limit doesn’t exist because the function approaches different
values from left to right. (check out the picture at the top of p. 22 to help understand
visually).

A left-hand limit [ lim− 𝑓(𝑥)] is the limit as x approaches a from the left.

A right-hand limit [ lim+ 𝑓(𝑥)] is the limit as x approaches a from the right.

lim 𝑓(𝑥) exists if lim− 𝑓(𝑥) = lim+ 𝑓(𝑥)
𝑥→𝑎
𝑥→𝑎
𝑥→𝑎
𝑥→𝑎
𝑥→𝑎
The function is continuous at x = a if and only if lim− 𝑓(𝑥) = lim+ 𝑓(𝑥).
𝑥→𝑎
𝑥→𝑎
Ex. Show that lim|𝑥| = 0.
𝑥→0
Remember that |𝑥| = x if x ≥ 0 and –x if x < 0.
lim |𝑥| = lim−(−𝑥) = 0 and lim+|𝑥| = lim+ 𝑥 = 0
𝑥→0−
𝑥→0
𝑥→0
𝑥→0
Since the LHS = RHS, the limit is 0.
Ex. f(x) =
{ -x – 2
x
𝑥 2 − 2𝑥
x ≤ -1
-1 < x < 1
x≥1
First try the two lower equations for the lower limit at x = -1:
f(x) = -x – 2 = -(-1) – 2 = -1 (this represents the approach from the left)
f(x) = x = -1 (this represents the approach from the right)
Since both limits are the same, the limit is -1
Then try the two upper equations for the upper limit at x = 1.
f(x) = x = 1 (this represents the approach from the left)
f(x) = 𝑥 2 − 2𝑥 = (1)2 − 2(1) = −1 (this represents the approach from the right)
Since the limits are different, the limit does not exist at this point. This is illustrated by
the graph on p. 24. (check it out)
A function that is described by different formulas for different parts of the domain is
called a piecewise function. Below is a graph of a piecewise function.
Examples: Use the graph of f ( x) to find the indicated limits
1.
lim f ( x)
2.
3. lim f ( x)
4.
5.
lim f ( x)
6. lim f ( x)
lim f ( x)
8.
x 6
x 6
7.
x 4
x 1
10. lim f ( x)
11. lim f ( x)
13. lim f ( x)
14 lim f ( x)
15. lim f ( x)
x2
x 5
Answers:
2. -2
7. -1
12. DNE
lim f ( x)
x 1
12. lim f ( x)
x2
x2
x 5
x 5
1. -2
6. DNE
11. +∞
lim f ( x)
x 4
x 4
x 1
9. lim f ( x)
lim f ( x)
x 6
3. -2
8. -2
13. +∞
4. 1
9. DNE
14. +∞
5. -1
10. −∞
15. DNE
Also, determine where the function is continuous or discontinuous (discontinuous
means there’s a break in the graph)
1. At x = -6?
Answers:
1. Dis
2. At x = -4?
2. Dis
Ex. Find the lim 𝑓(𝑥) where f(x) =
𝑥→−1
LHS: 3(-1) + 6 = 3
3. At x = -1?
3. Dis
{3𝑥 + 6
{𝑥 2 + 2
RHS: (−1)2 + 2 = 3
4. At x = 5? 5. At x = 6?
4. Dis
𝑥 < −1
𝑥 ≥ −1
5. Con
Since LHS = RHS lim 𝑓(𝑥) = 3
𝑥→−1
Ex. Find the lim 𝑔(𝑥) where = g(x) = {𝑥 2 − 4𝑥 + 2
𝑥→5
{√10𝑥
LHS: (5)2 − 4(5) + 2 = 7
𝑥< 5
𝑥≥5
RHS: √10(5) = 7.07
Since LHS ≠ RHS, lim 𝑔(𝑥) 𝐷𝑁𝐸
𝑥→5
Ex. State where the functions are discontinuous and why.
a) 𝑓(𝑥) =|𝑥|
**Remember that earlier we found that the LHS = RHS for this function. This function is
continuous.
b) g(x) = {𝑥 2 + 2𝑥 + 1
{−𝑥 − 1
{√𝑥 − 1 − 1
𝑥 ≤ −1
−1<𝑥 <1
𝑥 ≥1
Check the lower number (x = -1) first in the two lower equations.
LHS: (−1)2 + 2(−1) + 1 = 0
RHS: −(−1) − 1 = 0
Therefore, it is continuous at x = -1
Check the higher number (x = 1) in the two upper equations.
LHS: −(1) − 1 = −2
RHS: √1 − 1 − 1 = −1
Therefore, since LHS ≠ RHS, the function is discontinuous at x = 1
If, for any number ε > 0, there exists another number 𝛿 such that |𝑓(𝑥) − 𝐿| < ε,
whenever 0 < |𝑥 − 𝑎| < 𝛿, then lim 𝑓(𝑥) = L.
𝑥→𝑎
In plain English:
1. Choose any number ε….how close you want the function to get to the limit.
2. Find 𝛿 that will guarantee you will get close to the limit.
3. If #1 and #2 can be done, no matter how small ε is, then the limit exists.
Check out the visual for further explanation:
L+ε
L
L-ε
a-𝛿
a
Ex. If f(x) = 5 – 3x, show that:
a+𝛿
|𝑓(𝑥) − 2| < 0.03 if |𝑥 − 1| < 0.01
Replace f(x) with the function:
|5 − 3𝑥 − 2| < 0.03
|3 − 3𝑥| < 0.03
|−3(𝑥 − 1)| < 0.03
|−3||𝑥 − 1| < 0.03
3|𝑥 − 1| < 0.03
|𝑥 − 1| < 0.01
Homework: p. 20 #10, 11 and p. 27 #2 – 7, 9, 10, 14
Lesson 6: Using Limits to Find Tangents (1.4)
In general, if a curve C has equation y = f(x) and we want to find a tangent to C at the
point P (a, f(a)), we consider a nearby point Q(x, f(x)) where x ≠ a and find the slope of
the secant line.
∆𝑦
mPQ = ∆𝑥 =
𝑓(𝑥)−𝑓(𝑎)
𝑥−𝑎
This should look familiar from section 1.1 where we found the slope by plugging in lots
of points and estimating. However, we are now going to use limits to help us find an
approximation of the slope so that we don’t have to guess and check.
Therefore, we let Q approach P as x approaches a. As mPQ approaches a number, the
tangent is the line through P with slope m.
In limits notation:
m = lim
∆𝑦
∆𝑥→0 ∆𝑥
= lim
𝑓(𝑥)−𝑓(𝑎)
𝑥→𝑎
𝑥−𝑎
Q
f(x) – f(a)
P
x-a
What this means is that the tangent line is the limiting position of the sequence of
secant lines (check out the picture on p. 31)
We will use the lovely formula to find slope.
Ex. Find the slope and the equation of the tangent line to the curve 𝑦 = 2𝑥 2 + 4𝑥 − 1 at
point (2, 15) and then graph.
We start by using the limit formula to find slope:
lim
𝑥→2
=
𝑓(𝑥)−𝑓(2)
𝑥−2
=
2𝑥 2 +4𝑥−1−[2(2)2 +4(2)−1]
𝑥−2
2(𝑥 + 4)(𝑥 − 2)
= 2(𝑥 + 4)
𝑥−2
Now substitute in x = 2.
=
2𝑥 2 +4𝑥−1−15
𝑥−2
=
2𝑥 2 +4𝑥−16
𝑥−2
=
2(𝑥 2 +2𝑥−8)
𝑥−2
2(2 + 4) = 12. Our slope is 12.
To find the equation, use y = mx + b.
15 = 12(2) + b
15 = 24 + b
-9 = b
Therefore the equation is:
y = 12x – 9
Now, to graph, you either need to complete the square or make a table of values for the
quadratic.
𝑦 = 2𝑥 2 + 4𝑥 − 1
𝑦 = 2(𝑥 2 + 2𝑥
)−1
𝑦 = 2(𝑥 2 + 2𝑥 + 1) − 1 − 2
𝑦 = 2(𝑥 + 1)2 ) − 3
Therefore, it has a vertex of (-1, -3) and has a y-intercept of (0, -1) and passes through
(2, 15)
Another formula (expression) for slope is:
mPQ =
𝑓(𝑎+ℎ)−𝑓(𝑎)
ℎ
, where h represents the distance between x and a.
As x approaches a, h approaches 0.
𝑓(𝑎 + ℎ) − 𝑓(𝑎)
ℎ→0
ℎ
lim
Q (a + h, f(a + h)
f(a+ h) – f(a)
P
a
a+h
Ex. Find the equation of the tangent line to 𝑓(𝑥) = 𝑥 2 − 3 at a = -2
(Use the first formula)
𝑥 2 − 3 − ((−2)2 − 3) 𝑥 2 − 3 − 1 𝑥 2 − 4 (𝑥 − 2)(𝑥 + 2)
=
=
=
=𝑥−2
𝑥→−2
𝑥 − −2
𝑥+2
𝑥+2
𝑥+2
lim
lim 𝑥 − 2 = −2 − 2 = −4
𝑥→−2
Therefore the slope is -4. Now we need to find the y-coordinate of the point by
substituting in x = -2 into the original equation.
𝑦 = 𝑥 2 − 3 = (−2)2 − 3 = 1
Now substitute into y = mx + b.
1 = -4(-2) + b
1=8+b
-7 = b
The answer is: y = -4x – 7
Ex. Find the equation of the tangent line to 𝑓(𝑥) = √1 − 𝑥 at a = -3.
(Use the second formula)
𝑓(𝑎 + ℎ) − 𝑓(𝑎) 𝑓(−3 + ℎ) − 𝑓(−3) √1 − (−3 + ℎ) − √1 − −3
=
=
ℎ→0
ℎ
ℎ
ℎ
lim
=
√4 − ℎ − 2
ℎ
**now multiply by the conjugate of top.
=
4−ℎ−4
−ℎ
−1
√4 − ℎ − 2 (√4 − ℎ + 2)
×
=
=
=
ℎ
(√4 − ℎ + 2) ℎ(√4 − ℎ + 2) ℎ(√4 − ℎ + 2) (√4 − ℎ + 2)
Now substitute in h = 0.
−1
(√4 − 0 + 2)
=
−1
−1
=
(2 + 2)
4
Now substitute x = -3 into the original equation to find the point.
𝑦 = √1 − −3 = 2. Point is (-3, 2)
Now use y = mx + b to find the equation.
2=
−1
4
(−3) + 𝑏
8 = -1(-3) + 4b
8 = 3 + 4b
5 = 4b
5
=𝑏
4
Answer: y =
−1
4
5
𝑥+4
𝑥
2
Ex. Find the equation of the line tangent to 𝑓(𝑥) = 𝑥+2 at (4, 3).
4+ℎ
4
4+ℎ 2
𝑓(4 + ℎ) − 𝑓(4) 4 + ℎ + 2 − 4 + 2 6 + ℎ − 3
lim
=
=
ℎ→0
ℎ
ℎ
ℎ
Multiply by the common denominator of 3(6 + h).
4+ℎ 2
− × 3(6 + h) 3(4 + ℎ) − 2(6 + h) 12 + 3h − 12 − 2ℎ
6+ℎ 3
=
=
ℎ × 3(6 + h)
3h(6 + h)
3h(6 + h)
3(6 + h) ×
=
ℎ
1
1
1
=
=
=
3h(6 + h) 3(6 + h) 3(6 + 0) 18
Now substitute in to y = mx + b for the equation.
2
1
(4) + 𝑏
=
3 18
2 2
= +𝑏
3 9
6 = 2 + 9b
4 = 9b
4
=𝑏
9
1
4
Answer: y = 18 𝑥 + 9
Homework: p. 35 #1, 7v (formula #1), 8, 10, 11
Lesson 7: Velocity and Other rates of change (1.5)
It is easy to find average velocity. To find the average velocity of a car driving on a trip,
you divide the distance travelled by the time elapsed.
However, if you watch a speedometer, you see that the speed doesn’t stay constant.
The object has a definite velocity at every moment, but how do we define instantaneous
velocity?
We can do it in a similar way that we found slope for tangent lines….we considered
limits.
s = position
∆𝑠 = 𝑐ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝑝𝑜𝑠𝑖𝑡𝑖𝑜𝑛 (𝑑𝑖𝑠𝑝𝑙𝑎𝑐𝑒𝑚𝑒𝑛𝑡)
∆𝑠
Vavg = ∆𝑡
It is also rate of change: lim
𝑥2 →𝑥1
𝑓(𝑥2 )−𝑓(𝑥1 )
𝑥2 −𝑥1
Instantaneous velocity is defined to be the limiting value of the average velocities as h
approaches 0.
Therefore, the velocity at time t (the limit of these average velocities as h approaches 0)
must be equal to the slope of the tangent line at P (the limit of the slopes of the secant
lines).
∆𝑠
V(a) = lim ∆𝑡 =
𝑡→0
𝑓(𝑎+ℎ)−𝑓(𝑎)
ℎ
This is the instantaneous velocity.
Ex. An object moves in a way that its position in metres is given by 𝑠(𝑡) = 𝑡 2 − 5𝑡 + 4,
where t is measure in seconds. Find the average velocity over the following intervals:
a) 2 ≤ t ≤ 4
𝑠(4) − 𝑠(2) (4)2 − 5(4) + 4 − [(2)2 − 5(2) + 4]
=
= 1 𝑚/𝑠
4−2
2
b) 2 ≤ t ≤ 3
𝑠(3) − 𝑠(2) (3)2 − 5(3) + 4 − [(2)2 − 5(2) + 4]
=
= 0 𝑚/𝑠
3−2
1
c) 2 ≤ t ≤ 2.5
𝑠(2.5) − 𝑠(2) (2.5)2 − 5(2.5) + 4 − [(2)2 − 5(2) + 4]
=
= −0.5 𝑚/𝑠 (𝑑𝑜𝑤𝑛 𝑠𝑙𝑜𝑝𝑒)
2.5 − 2
0.5
Ex. For the above example, find the instantaneous velocity at t = 2.
𝑠(𝑡) − 𝑠(2) 𝑡 2 − 5(𝑡) + 4 − [(2)2 − 5(2) + 4] 𝑡 2 − 5(𝑡) + 6 (𝑡 − 2)(𝑡 − 3)
lim
=
=
=
𝑡→2
𝑡−2
𝑡−2
𝑡−2
𝑡−2
lim 𝑡 − 3 = 2 − 3 = −1 𝑚/𝑠
𝑡→2
Ex. For the same example, find the instantaneous velocity at t = a.
(use either formula)
1st formula:
𝑓(𝑥) − 𝑓(𝑎) 𝑡 2 − 5(𝑡) + 4 − [(𝑎)2 − 5(𝑎) + 4] 𝑡 2 − 5𝑡 + 4 − 𝑎2 + 5𝑎 − 4
=
=
𝑛→∞
𝑥−𝑎
𝑡−𝑎
𝑡−𝑎
lim
𝑡 2 − 𝑎2 − 5𝑡 + 5𝑎 (𝑡 − 𝑎)(𝑡 + 𝑎) − 5(𝑡 − 𝑎) (𝑡 − 𝑎)[𝑡 + 𝑎 − 5]
=
=
=𝑡+𝑎−5
𝑡−𝑎
𝑡−𝑎
𝑡−𝑎
= 𝑎 + 𝑎 −5 = 2a – 5
2nd formula:
𝑠(𝑎 + ℎ) − 𝑠(𝑎) (𝑎 + ℎ)2 − 5(𝑎 + ℎ) + 4 − [𝑎2 − 5(𝑎) + 4]
=
ℎ→0
ℎ
ℎ
lim
𝑎2 + 2𝑎ℎ + ℎ2 − 5𝑎 − 5ℎ + 4 − 𝑎2 + 5𝑎 − 4 2𝑎ℎ + ℎ2 − 5ℎ ℎ(ℎ + 2𝑎 − 5)
=
=
=
ℎ
ℎ
ℎ
lim ℎ + 2𝑎 − 5 = 2𝑎 − 5
ℎ→0
Ex. For the above example, find the instantaneous velocity:
a) t = 0 seconds 2(0) – 5 = -5 m/s
b) t = 1 second 2(1) – 5 = -3 m/s
c) t = 3 seconds 2(3) – 5 = 1 m/s
Always remember that rate of change is just another word for slope…..
Ex. A stone is dropped into a pool of water, creating a circular ripple. Find the rate of
change in the area of the circle with respect to the radius when the radius is 20 cm.
The area of a circle is 𝐴 = 𝜋𝑟 2 . Use the limit formula to find rate of change.
𝐴(𝑟) − 𝐴(20) 𝜋𝑟 2 − 𝜋(20)2 𝜋(𝑟 2 − 400) 𝜋(𝑟 − 20)(𝑟 + 20)
=
=
=
= 𝜋(𝑟 + 20)
𝑟→20
𝑟 − 20
𝑟 − 20
𝑟 − 20
𝑟 − 20
lim
lim 𝜋(𝑟 + 20) = 𝜋(20 + 20) = 40𝜋 𝑐𝑚2 /𝑐𝑚
𝑟→20
Homework: p. 43 #1 – 3, 6 – 8
Lesson 8: Infinite Sequences (1.6)
Sequence: list of numbers in a definite order
𝑡1 , 𝑡2 , 𝑡3 , 𝑡𝑛 , 𝑒𝑡𝑐.
𝑡𝑛 = general term: a rule that works for every term in the sequence.
We are considering infinite sequences.
Ex. A sequence has a general term 𝑡𝑛 =
𝑛
1−2𝑛
.
a) Find the first 6 terms.
1
𝑡1 =
1−2(1)
𝑡2 =
1−2(2)
𝑡3 =
1−2(3)
2
= −1
2
3
= −3
3
= −5
4
𝑡4 =
1−2(4)
𝑡5 =
1−2(5)
𝑡6 =
1−2(6)
5
6
4
= −7
5
= −9
6
= −11
b) Find the limit.
To find the limit, first consider the following:
1
What is the lim 𝑛? Consider successive terms as n approaches infinity.
𝑛→∞
1 1 1 1 1
, , , , , 𝑒𝑡𝑐. Notice that the fraction gets smaller and smaller until the answer
approaches 0. Therefore, we say the limit is 0.
1 2 3 4 5
To find the limit of our question, we divide the numerator and denominator by the
highest power of n in the denominator.
𝑛
𝑛
1
1
1
𝑛
lim
=
=
=
=
1 2𝑛 1
𝑛→∞ 1 − 2𝑛
0 − 2 −2
𝑛− 𝑛
𝑛−2
Rule: lim
1
𝑛→∞ 𝑛𝑟
= 0 if r > 0.
Ex. Find the limit of the following if it exists:
10𝑛5 + 𝑛2 + 9
𝑛→∞ 2𝑛5 + 7𝑛3 − 𝑛
lim
Divide each term by 𝑛5 .
1
9
+ 5 10 + 0 + 0
3
𝑛
𝑛 =
lim
=5
7
1
𝑛→∞
2
+
0
+
0
2+ 2− 4
𝑛
𝑛
10 +
Ex. Find the limit of the following if it exists:
10𝑛5 + 𝑛2 + 9
𝑛→∞ 𝑛4 + 7𝑛3 − 𝑛
lim
Divide each term by 𝑛4
1
9
+
𝑛2 𝑛4 = 10𝑛 + 0 + 0 = 10𝑛 = 𝐷𝑁𝐸
7 1
1+0+0
1+𝑛− 3
𝑛
10𝑛 +
lim
𝑛→∞
This limit does not exist since it approaches infinity.
Shortcut Law:
Instead of dividing every term in the numerator and denominator, just take the highest
degree term in the numerator and divide it by the highest degree term in the
denominator.
If we look back at our two examples:
1. lim
10𝑛5
𝑛→∞ 2𝑛5
2.
lim
10𝑛5
𝑛→∞ 𝑛4
=5
= 10𝑛
Ex. Find the limit of the following if it exists:
1. lim
𝑛→∞
2.
lim
𝑛−4𝑛3 −𝑛5
2𝑛+7𝑛3 +𝑛4
10𝑛5 +𝑛2 +9
𝑛→∞ 𝑛+5𝑛3 −𝑛6
=
=
−𝑛5
𝑛4
= −𝑛 = 𝐷𝑁𝐸
10𝑛5
−𝑛6
10
= −𝑛 = 0
3. lim (−2)𝑛
𝑛→∞
4. lim (−1)𝑛
𝑛→∞
1
5. lim (− 2)𝑛
𝑛→∞
1
1
(−2)1 = −2
(−1)1 = −1
(− 2)1 = − 2
(−2)2 = 4
(−1)2 = 1
(− 2)2 = 4
(−2)3 = −8
etc.
(−1)3 = −1
etc.
(− 2) = − 8
etc.
DNE
DNE
0
Rule: lim 𝑟 𝑛 = 0 if |𝑟| < 1.
𝑛→∞
Read through Zeno’s paradox on p. 49.
Homework: p. 50 #1 (all), 3 (odd letters)
1
1 3
1
1
Lesson 9: Infinite Series (1.7)
Does it make sense to talk about adding infinitely many numbers?
Remember back to your junior high days?
̅̅̅̅ = 0.444444 … …
0.4
4
4
4
4
This is actually 10 + 100 + 1000 + ⋯ = 9
Another limit
Read through another of Zeno’s paradoxes on p. 52.
Series: 𝑆𝑛 = 𝑡1 + 𝑡2 + 𝑡3 + ⋯ + 𝑡𝑛
A series is the sum of the first n terms in a sequence.
Convergent Series: a series that has a sum.
∞
lim 𝑆𝑛 = ∑ 𝑡𝑛 = 𝑡1 + 𝑡2 + 𝑡3 + ⋯ = 𝐿
𝑛→∞
𝑛=1
Divergent Series: a series that has no sum….the limit does not exist.
Ex. Find the sum of the geometric series:
𝑎 + 𝑎𝑟 + 𝑎𝑟 2 + 𝑎𝑟 3 + ⋯ + 𝑎𝑟 𝑛−1
Recall from Math 20-1:
The geometric series formula is: 𝑆𝑛 =
𝑎(1−𝑟 𝑛 )
1−𝑟
We will now look at 4 cases:
Case #1:
|𝑟| < 1
(which means r > - 1 or r < 1)
𝑎−𝑎𝑟 𝑛
lim 𝑆𝑛 =
𝑛→∞
1−𝑟
According to section 1.6, lim 𝑟 𝑛 = 0 if |𝑟| < 1. Therefore, 𝑎𝑟 𝑛 = 0
𝑛→∞
So our formula is actually:
𝑎
1−𝑟
So, for |𝑟| < 1, the series is convergent.
Case #2:
r=1
The series would be a + a + a + a + …..
This series does not have a sum and therefore is divergent.
Case #3:
r = -1
The series would be a – a + a – a + ….
Again, this series does not have a limit….it bounces between 0 and a. Therefore, it has
no sum and is divergent.
|𝑟| > 1
Case #4:
The ratio causes the terms to get progressively bigger and therefore, the limit does not
exist. Therefore, it has no sum and is divergent.
Conclusion:
𝑎
𝑆 = lim 𝑆𝑛 = 1−𝑟 for |𝑟| < 1. For all other values of r, the series is divergent.
𝑛→∞
Ex. Find the sum of the series or state divergent:
2
1
1
a) 4 – 3 + 9 –54 + …
1
a = 4 and r = − 6
Therefore the sum is:
𝑎
1−𝑟
=
4
1−−
1
6
=
4
7
6
=
24
7
.
b)
∞
2
∑ 5( )𝑛
3
𝑛=1
2
a = 5 and r = 3 (if you weren’t sure of this, you could find the first couple terms
just to double check)
10
3
2
1−
3
=
10
3
1
÷ 3 = 10.
̅̅̅̅̅ as a fraction.
Ex. Express 4.7328
= 4.7 + 0.0328 + 0.0000328 + …
If we ignore the 4.7 and just look at the repeating parts, it is a series with a = 0.0328 and
r = 0.001.
Therefore, it is a series with a convergent sum so we can use the formula:
𝑎
0.0328
0.0328
328
328
47 328
46953 + 328
=
=
=
= 4.7 +
=
+
=
1 − 𝑟 1 − 0.001
0.999
9990
9990 10 9990
9990
=
47281
9990
Reduce if possible.
Ex. For what value of x is the series convergent? Then find an expression for the sum:
1+
𝑥 + 3 (𝑥 + 3)2 (𝑥 + 3)3
+
+
+⋯
2
4
8
For it to be convergent, |𝑟| < 1. (r < 1 and r > -1)
𝑥+3
2
<1
𝑥+3
2
> −1
x+3<2
x + 3 > -2
x < -1
x > -5
Therefore, -5 < x < -1
Use the sum formula:
𝑎
1
2
2
2
=
=
=
𝑂𝑅 −
1 − 𝑟 1 − 𝑥 + 3 2 − (𝑥 + 3) −𝑥 − 1
𝑥+1
2
Homework p. 56 #1 – 4 (even letters)