Precalculus 12.5b Area of a Plane Region

12.5b Area of a Plane Region

Area of a Region bounded by:

x = a, x = b, y = 0 and y = f (x) where f(x) is less than or equal to 0.

The boundary of the area is from x₀ = a to x_n = b.

What is the area under the curve?

George Riemanns - German who came up with technique to find the orange area.

1. Subdivide the interval from a to b into smaller intervals. This is called a partition of [a,b] specifically:

a = x₀ ≤ x₁ ≤ x₂ ≤ x₃ ... ≤ x_{n - 1} ≤ x_n = b

2. Choose a point w_i in interval [x_{i - 1}, x_i] , i = 1 ... n

so...

when i = 1, w₁ is contained in [x₀, x₁]

when i = 2, w₂ is contained in [x₁, x₂]

3. Let Δ x_i = x_i - x_i-1 for i = 1 ... n

i= 1; Δ x₁ = x₁ - x₀

i = 2; Δ x₂ = x₂ - x₁

Δ x_i = length of the i^th subinterval.

4. Form f(w_i)(Δx_i) for i = 1 ... n

5. The area of the region:

A = f (w₁)(Δ x₁) + f(w₂)(Δ x₂) + ... + f (w_n)(Δx_n)

This is called a Riemann's Sum.

So by increasing the number of rectangles that you make, you can obtain a closer and closer approximation

Therefore to find the Area of a Plane Region:

Let "f" be continuous function and nonnegative on the interval [a, b]. The Area A of the region bounded by the graph of "f", the x-axis (y = 0), and the vertical lines x = a and x = b is

Area of a rectangle = height times width

check out this website: http://tutorial.math.lamar.edu/classes/calcI/areaproblem.aspx

Example 1:
Find the Area of the Region: f(x) = 2 - x², -1 £ x £ 1

Let's let n = 4

so the width = (1 - ^-1)/4 = 1/2

height =
[-1, -1/2] = 1.75
[-1/2, 0] = 2
[0, 1/2] = 1.75
[1/2, 1] = 1

so the Area when f(x) = 2 - x² and is divided up into 4 rectangles =
(1/2)(1.75) + (1/2)(2) + (1/2)(1.75) + (1/2)(1) = 3.25

but to find the area with more rectangles, let's use the above formulas:

Therefore to find the area, use the summation formulas:

To check this on the graphing calculator:

put in y₁ = 2 - x²

Now on your homescreen, math arrow down to #9 fnInt (Y₁, x, -1, 1) enter

should give you 10/3.

In the calculator (function , variable, lower bound, upper bound) = area under the curve.

Precalculus 12.5 The Area Problem

12.5 The Area Problem

I. Limits of Summation

S = a₁ + a₁r + a₁r² + ... =

_¥
å (a₁(r^i-1)) =
^{i = 1}

a₁/(1-r) provided that the absolute value of r < 1

Using Limit Notation, this sum can be written as:

S =
¥
lim å (a₁ r^i-1)
^{n®¥ i = 1}

= lim a₁ (1 - rⁿ)/(1-r)
^n®¥

Recall that since r < 1 thus making it a fraction, when you take the limit or rⁿ as x approaches infinity, this would equal zero. So:

= (a₁ (1 - 0))/ (1 - r)

= a₁/(1 - r) which you have already seen in this coursework.

Summation Formulas and Properties:

Example 1: Evaluating a Summation:

What is the sum of
i² when i = 1 to i = 30?

=(30)(30 + 1)(2(30) + 1)/6 = (30)(31)(61)/6 = 56730/6 = 9455

Example 2: Evaluating a Summation:

What is the sum of:
(2i + 1) when i = 1 to i = 50?

You have to break this down into parts:

2( å i ) from i = 1 to i = 50 AND
å(1) from i = 1 to i = 50 is equal to

2(50(50+1)/2) + 1(50) = (50)(51) + 50 = 2550 + 50 = 2600

II. Finding the Limit of a Summation

_n
å (2i + 3)/(n²)
^{i = 1}

= (1/n²)(2(n(n+1))/2) + 3n) = (1/n²)(n(n+1) + 3n)
= (1/n)(n + 1 + 3) = (n + 4)/n

which in essence is infinity over infinity so
Therefore the Summation of
lim S_n =
^n®¥

lim (n + 4)/n = 1/1 = 1 so
^n®¥

lim S_n = 1
^n®¥

You can check this by using a table of values:

(n, S(n)) = {(1, 5), (10, 1.4), (100, 1.04), (1000, 1.004), (10000, 1.0004)...}
So you can see the values of S(n) approach 1.

Example 3: Finding the Limit of a Summation:

_n
å (3/n³)(1 + i²)=
^{i = 1}

= (3/n³)(n + (n(n + 1)(2n + 1)/6) =
(3/n² + (3(2n² + 3n + 1)(6n²)
= 3/n² + (6n² + 9n + 3)/(6n²)

lim 3/n² = 0 and
^{n ®¥}

lim (6n² + 9n + 3)/(6n²) = 6/6 = 1
^{n ®¥}

So:

lim S_n = 1
^{n ®¥}

Again, you can check this using a table of values:
(n, S(n)) = {(1, 6), (10, 1.185), (100, 1.0154), (1000, 1.0015), (10000, 1.0002)...}

Precalculus 12.4 Limits at Infinity and Limits of Sequences

12.4 Limits at Infinity and Limits of Sequences

Definition of Limits at Infinity
If "f" is a function and L₁ and L₂ are real numbers, the statements

lim f(x) = L₁
^x®-¥

AND

lim f(x) = L₁
^x®+¥

denote the limits at infinity. The first is read "the limit of f(x) as x approaches negative infinity is L₁," and the second is read "the limit of f(x) as x approaches infinity is L₂"

To help evaluate limits at infinity, you can use the following:

Limits at Infinity
If "r" is a positive real number, then

lim (1/(x^r)) = 0 Limit toward the right.
^x®+¥

Furthermore, if x^r is defined when x is less then 0, then

lim (1/(x^r) = 0 the limit toward the left.
^{x®- ¥}

Thought: -1/10 = -.1, -1/100 = -.01, -1/1000 = -.001, -1/10000 = -.0001
so as you can see, in the denominator as x approaches negative infinity, f(x) approaches zero.

Example 1: Evaluating a Limit at Infinity:

lim (5/(2x)) = 0
^x®¥

As you can see when you graph this function, as x gets larger, f(x) gets closer to zero so that is how we can conclude the the limit equals zero.

Example 2:
lim (1 - 3/(x²)) =
^x®¥

We can separate each part of the function to:

lim (1 )= 1
^x®¥

and

lim (3/(x²)) = 0
^x®¥

so
lim (1 - 3/(x²)) = 1 - 0 = 1
^x®¥

Compare the following limits:
Find the limit as x approaches infinity:

a). f(x) = (-5x + 2)/(4x² - 1)

b). g(x) = (4x² + 1)/(3x² - 1)

c). j(x) = (-3x⁴ + 1)/(2x³-1)


When you take the limit of each of these functions as x approaches infinity, you get:

1st, look in the denominator and find the largest exponent and use this term by dividing all the terms of the function by that variable raised to the largest exponent.

a).
lim (-5x + 2)/(4x² - 1)=
^x®¥

so divide each term by x²

lim [(-5x)/(x²) + 2/(x²)]/[(4x²)/(x²) - 1/(x²)]
^x®¥

= (-0 + 0)/(4 - 0) = 0

b).
lim (4x² + 1)/(3x² - 1)=
^x®¥

so divide each term by x²

lim [4 + 1/(x²)]/[3 - 1/(x²)]=
^x®¥

(4 + 0)/(3 - 0) = 4/3

c).
lim (-3x⁴ + 1)/(2x³-1)
^x®¥

and so divide each of the terms by x³

lim (-3x + 1/x³)/(2 - 1/x³)=
^x®¥

-3x/2 so by putting infinity in for x, the limit does not exist because the numerator decreases without bound as the denominator approaches 2.

Therefore we can conclude:

Limits at Infinity for Rational Functions

for the rational function f(x) = N(x)/D(x) when

N(x) = a_nxⁿ + ... + a₀

And

D(x) = b_mx^m + ... + b₀

the limit as x approaches positive or negative infinity is as follows:

lim f(x) = 0, when n is less than m
^x®¥

lim f(x) = a_n/b_m, when n = m
^x®¥

And if n is greater than m, the limit does not exist.

Limit of a Sequence:
Let "f" be a function of a real variable, such that
lim f(x) = L
^x®¥

If {a_n} is a sequence such that f(n) = a_n
for every positive integer "n", then the
lim a_n = L
^n®¥

Example:
Given: 1/3, 1/9, 1/27, 1/81, ...

As n increases without bound, the terms of the sequence
a_n = 1/3ⁿ is said to

CONVERGE to zero.

lim 1/3ⁿ = 0
^n®¥

___________________________________________

A sequence that does not converge is said to DIVERGE!

Example of this is: the sequence 1, -1, 1, -1, ... this diverges.

Find the limit of the Sequence:

Example 1:
lim (4x - 1)/(x + 3)=
^x®¥
{3/4, 7/5, 11/6, 15/7, ...} = 4

Example 2:
lim (3x + 2)/(2x² - 1)=
^x®¥
{5/1, 8/7, 11/17, 14/31, ...} = 0

Example 3:
lim (7x² - 1)/(8x²)=
^x®¥
{6/8, 27/32, 62/72, 111/128, ...} = 7/8

Use the site: http://www.calculus-help.com/funstuff/phobe.html

Chapter 1, lesson 4 to help with this lesson.

Precalculus 12.3 The Tangent Line Problem

12.3 The Tangent Line Problem

The Slope of the graph of a function can be used to analyze rates of change at particular points on the graph.

The Slope of a line indicates the rate at which a line rises or falls.
- For a straight line, this rate (or slope) is the same at every point on the line.
- For other graphs than lines, the rate at which the graph rises or falls changes from point to point.

Recall: slope formula = (the change in y's)/(the change in x's) = Δy/Δx

Review: With circles, you were taught about tangent lines. A tangent line is a line that intersects a circle in exactly one point, called the point of tangency. Using this process, to determine the rate at which a graph rises or falls at a single point, you can find the slope of the tangent line at that point.

The tangent line to a graph of a function "f" at a point P(x₁, y₁) is the line that best approximates the slope of the graph at that point.

Example1: f(x) = x² - 2x + 1

find the tangent line at the point (1, f(x))

f(1) = 1² -2(1) + 1 = 0
so the tangent line at (1,0) is a straight line and the equation of the tangent line at (1, f(x)) is y=0.

A more precise method then "eyeballing" the tangent line
is making use of the secant line through the point of tangency
and a second point on the line where (x, f(x)) is the first point
and (x + h, f(x + h)) is a second point on the graph of "f", the
slope of the secant line through these two points is:

M_secant = (f(x + h) - f(x))/h

Using the limit process, you can find the
exact slope of the tangent line at (x, f(x)).

Definition of the Slope of a Graph:

The slope m of the graph of "f" at the point (x, f(x)) is equal to the slope
of its tangent line at (x, f(x)) and is given by:

M =

lim m_sec =
^h®0

lim (f(x+h) - f(x))/h
^h®0

provided the limit exists!

Example 2: given f(x) = 2x + 5, find the slope of the tangent line at (-1, -3)

M_sec = (f(x+h) - f(x))/h

= (f(-1 + h) - f(-1))/h
= (2(-1 + h) + 5 - (2(-1) + 5))/h
= (-2 + 2h + 5 +2 - 5)/h
= (2h)/h
= 2
The graph has a slope of 2 at the point (-1, -3)
To find the equation of the tangent line:
y = mx + b
(-3) = 2(-1) + b
-1 = b

so the equation of the tangent line at point (-1, -3) is
y = 2x -1

Example 3: given f(x) = 10x - 2x² at point (3, 12)

M_sec = [10(x+h) - 2(x + h)² - (10x - 2x²)]/h

M_sec = [10x + 10h -2(x² +2xh + h²) -10x + 2x²]/h

M_sec = [10h - 2x² - 4xh - 2h² + 2x²]/h

M_sec = [10h - 4xh - 2h²]/h

M_sec = 10 - 4x - 2h

Slope of the tangent line = M =
lim M_sec =
^h®0

lim 10 - 4x - 2h = 10 - 4x so at point (3, 12), M = 10 - 4(3) = -2
^h®0

Therefore the slope of the tangent line at the point (3, 12) is -2.

To find the equation of the tangent line, use the point and the slope:

y = mx + b
12 = (-2)(3) + b
12 = -6 + b
18 = b

y = 2x + 18 is the equation of the tangent line at the point (3, 12)

Finding a Formula for the slope of a Graph:
Example4: g(x) = x³ at points (1, 1) and (-2, -8)

M_sec = [(x + h)³ - (x³)]/h

M_sec = [x³ + 3x²h + 3xh² + h³ - x³]/h

M_sec = [3x²h + 3xh² + h³]/h

M_sec = 3x² + 3xh + h² , where h¹0

Next take the limit of M_sec as h approaches 0.

The slope of the tangent line M =

lim 3x² + 3xh + h² = 3x^2
h®0

Now use this M = 3x² for the slope at (1,1)

M = 3(1)² = 3

now y = mx + b
1 = 3(1) + b
b = -2

so the equation of the tangent line at point (1,1) is y = 3x - 2

Again now at point (-2, -8)

Now use this M = 3x² for the slope at (-2,-8)

M = 3(-2)² = 3(4) = 12

now y = mx + b
-8 = 12(-2) + b
-8 = -24 + b
b = 16

so the equation of the tangent line at point (-2,-8) is y = 12x +16

The formula that you derived from the function f(x) = x³ and used the limit process to get M = 3x² represents the slope of the graph of "f" at the point (x, f(x)).

The Derived function is called the derivative of f at x. It is donoted by f ' (x), which is reas as "f prime of x".

Definition of the Derivative:

The Derivative of "f" at "x" is

f ' (x) = lim_h®0 (f(x+h) - f(x))/h

provided the limit exists!

Example 5: f(x) = x² - 3x + 4

f ' (x) = lim_h®0 [(x + h)² - 3(x + h) + 4 - (x² - 3x + 4)]/h

f ' (x) = lim_h®0 [x² + 2xh + h² - 3x -3h + 4 - x² + 3x -4]/h

f ' (x) = lim_h®0 [2xh + h² - 3h]/h

f ' (x) = lim_h®0 2x + h - 3 = 2x - 3

f '(x) = 2x - 3

So therefore the derivative of f(x) = x² - 3x + 4 is f ' (x) = 2x - 3

NOTE that in addition to f ' (x) , other notations can be used to denoted the derivatives of y = f(x). The most common are:

(dy)/(dx)

y'

(d/(dx))[f(x)]

and D_x[y]

Using the derivative of f(x) = x³ - x, find the slope of the tangent line at point (2,6).

f'(x) = lim_h®0 [(x + h)³ - (x + h ) - (x³ - x)]'h

f'(x) = lim_h®0 [x³ + 3x²h + 3xh² + h³ - x - h - x³ - x]/h

f'(x) = lim_h®0 [3x²h + 3xh² + h³ - h]/h

f'(x) = lim_h®0 (3x² + 3xh + h² - 1)

f ' (x) = 3x² - 1

So at the point (2, 6), the slope is
f ' (2) = 3(2)² - 1 = 3(4) - 1 = 12 - 1 = 11

So the equation of the tangent line at point (2,6) is
6 = 11(2) + b
6 = 22 + b
b = -16

so the equation of the tangent line at point (2,6) is y = 11x - 16

Example 6: Find the derivative of f. Use the derivative to determine any points on the graph of f where the tangent line is horizontal.

f(x) = x³ - 3x

f ' (x) = lim_h®0 [(x + h)³ - 3(x + h) - (x³ - 3x)]/h

f ' (x) = lim_h®0 [x³ + 3x²h + 3xh² + h³ - 3x -3h - x³ + 3x]/h

f ' (x) = lim_h®0 [3x²h+ 3xh² + h³ -3h]/h

f ' (x) = lim_h®0 [3x² + 3xh+ h² -3]

f ' (x) = 3x² - 3

Recall a horizontal line has a slope of zero so

f ' (x) = 0 so
0 = 3x² - 3
3 = 3x²
1 = x²
x = ±1

putting these values in for x:

So "f" has horizontal tangents at (1, 0) adn (-1, 0)


If you need extra help, watch the Chapter 2 lesson 1 tutorial at
http://www.calculus-help.com/funstuff/phobe.html

12.2b Precalculus - One-Sided Limits

12.2b One-Sided Limits

One-Sided Limits - when a limit approaches one value from the left side of "c" and a different value from the right side of "c"

lim f(x) = L
^x®c-

and

lim f(x) = K
^x®c+

where if L = K, then the limit exists but if L ¹K, then the limit does not exist for the value of x.

Example 1:

lim f(x)
^x®1

Given f(x) = 4 - x² when x£1 or f(x) = 3 - x when x is greater than 1

therefore:

lim f(x) = 4 - (1)² = 4 - 1 = 3
^x®1-

and

lim f(x) = 3 - 1 = 2
^x®1+

so since 3 ¹ 2, then
lim f(x) does not exist since the function approaches different values
^x®1
from the left and from the right.

I. Existence of a Limit:

If "f" is a function and "c" and "L" are real numbers, then

lim f(x) = L
^x®c

if and only if BOTH the left and the right limits exist and are equal to L.

Example 2:

f(x) = 3 - x when x is less than 1 or 3x - x² when x is greater than 1

lim f(x) = 2
^{x ®1-}

lim f(x) = 2
^{x ®1+}

Because the one-sided limits both exist and are equal to the same number 2, it follows that

lim f(x) = 2 so the limit does exist at x = 1.
^x®1

II. A limit from Calculus:

Evaluating a Limit from Calculus - the limit of a difference quotient:

lim (f(x+h) - f(x))/h where h¹0
^h®0

Direct substitution into the difference quotient ALWAYS produces the indeterminate
form 0/0.

Example 3: find the difference quotient of f(x) = 5 - 6x

lim [5 - 6(x + h) - (5 - 6x)]/h
^h®0

lim [5 - 6x - 6h - 5 + 6x]/h
^h®0

lim -6h/h = -6
^h®0

therefore:
lim [5 - 6(x + h) - (5 - 6x)]/h = -6
^h®0


Example 4: find the difference quotient of f(x)= 4 - 2x - x²

lim ([4 - 2(x + h) - (x+h)²] - [4 - 2x - x²])/h =
^h®0

lim ([4 - 2x -2h -(x² + 2xh + h²] - 4 + 2x + x²)/h =
^h®0

lim (4 - 2x -2h -x² -2xh - h² - 4 + 2x + x²)/h =
^h®0

lim (- 2h - 2xh - h²)/h =
^h®0

lim (- 2 -2x -h) = -2 -2x
^h®0

Precalculus - 12.2a Techniques for Evaluating Limits

12.2 Techniques for Evaluating Limits

I. Limits of Polynomial and Rational Functions

1. If p is a polynomial function and "c" is a real number, then

lim p(x) = p(c)
^x®c

2. If "r" is a rational function given by r(x) = p(x)/q(x) and "c" is a real number such that q(x) ¹0.

Example 1: Evaluating Limits by Direct Substitution

lim (.5x³ - 5x) = (.5)(.2)² - 5(2) = 6
^x®-2

Example 2:
lim (x² + 1)/(x) = (3² + 1)/ 3 = (9 + 1)/ 3 = 10/3
^x®3

Example 3:
lim (x² - 1)^1/3 = (3² - 1)^1/3 = (9 - 1)^1/3 = 8^1/3 = 2
^x®3

II. Factoring: Dividing out Technique

Example 4:
lim (x² - 3x)/x = (x)(x-3)/(x)
^x®0

lim x - 3 = -3
^x®0

Example 5:
lim (x² - 1)/(x + 1) =
^x®-1

lim(x + 1)(x - 1)/(x + 1) =
^x®-1

lim (x - 1) = -2
^x®-1

This technique should be applied only when direct substitution produces zero in both the numerator and the denominator. The resulting fraction 0/0, has no meaning as a real number.

It is called an indeterminate form because you cannot, from the form alone, determine the limit.

III. Rationalizing Technique

First Rationalize the numerator of the function.

Example 6:
lim (3 - Ö (x))/(x - 9)
^x®9

multiply the numerator and the denominator by the conjugate of the numerator:

lim[ (3 - Ö (x))(3 + Ö(x))] /[(x - 9) (3 + Ö(x))] =
^x®9

lim (9 - x) / [(x - 9 )(3 + Ö(x))] =
^x®9

As you can see in the numerator, you have (9 - x) and in the denominator you have (x - 9) so if we multiply the numerator by (-1) you would have:

lim (-1)(- 9 + x) / [(x - 9 )(3 + Ö(x))] =
^x®9

so now x - 9 divides out of the numerator and denominator

lim -1 / (3 + Ö(9)) = -1/ (3 + 3 ) = -1/6
^x®9

IV. Using Technology:

lim (1 + 2x)^1/x
x®0

Since you cannot use substitution, use a table:
(x, f(x)) = {(-.1, 9.313), (-.01, 7.540), (-.001, 7.404), (-.0001, 7.391), (-.00001, 7.3892), (-.000001, 7.38907), (.000001, 7.38904), (.00001, 7.3898), etc...}

so you can see that it is estimated about

lim (1 + 2x)^1/x »7.3890561 » e^2
x®0

Algebraically:
Choose 2 points that are equidistant from 0 and close to 0 and find their average:

(7.38907 + 7.38904)/2 »7.389055 » e²

Example 7:

lim (1/(2+x) - 1/2)/x =
^x®0

again using the calculator when x = -.0000000001, f(x) = -1/4
and when x = .0000000001, f(x) = -1/4

so

lim (1/(2+x) - 1/2)/x = -1/4
^x®0

Algebraically:

lim (1/(2+x) - 1/2)/x =
^x®0

simplifying the fraction:

[(2 - (2 + x))/(2 + x)(2)]/x =

(- x)/ [(2 + x)(2)(x)] =

-1/(4 + 2x) so:

lim (1/(2+x) - 1/2)/x = -1/(4 + 2(0)) = -1/4 same answer!!
^x®0

12.1 Introduction to Limits

12.1 Introduction to Limits

Watch the tutorials on lessons 1 - 3 on the following website.

http://www.calculus-help.com/funstuff/phobe.html

Lesson 1. What is a limit
Lesson 2. When does a limit exist.
Lesson 3. How do you evaluate a limit?

______________________________________________________

The Limit Concept - this notion of a limit is a fundamental concept of calculus.

Example 1: You are given 32 feet of fence and are asked to form a rectangular pen whose area is as large as possible. What is the largest area the pen would cover and what dimensions should the pen be?

If we let w be the width of the pen and l be the length of the pen.
Therefore we know that 2w + 2l = 32 for the perimeter.

Therefore if we solve for the length, l = 16 - w

then the area would be

A = lw
A = (16 - w)(w)
A = 16w - w²

Using this equation, we can experiment with different values of w to see how to obtain the maximum area.

(w, A) = {(1, 15), (2, 28), (3, 39), (4, 48), (5, 55), (6, 60), (7, 63), (8, 64), (9, 63), (10, 60), (11, 55), (12, 48), (13, 39) ...}

so as you can see it appears that (8, 64) is a maximum point so therefore when w = 8, the maximum area would be 64. The dimensions of the pen would be an 8 by 8 pen.

In limit terminology, you can say that "the limit of A as w approaches 8 is 64"
This is written as

lim A = 64
^w®8

Definition of Limit:

If f(x) becomes arbitrarily close to a unique number L as x approaches c from either side, the limit of f(x) as x approaches c is L. This is written as

lim f(x) = L.
^x®c

Example 2: Estimating a Limit Numerically

lim (4 - 3x)
^x®3

let's plug in some values and see what happens when x approaches 3.
(x,f(x)) = {(2.9, -4.7), (2.99, -4.97), (2.999, -4.997), (3, ?), (3.001, -5.003), (3.01, -5.03), (3.1, -5.3)}

So we can conclude that as x approaches 3, f(x) approaches -5.

So lim (4 - 3x) = -5
^x®3

this graph is continuous. For graphs that are not continuous, finding a limit can be more difficult.

Example 3: Estimating a Limit Numerically:

lim (x+1)/(x²-x-2)
^x®-1

Again, let's look at some values for x around -1.

(x, f(x)) = {(-1.1, -.3226), (-1.01, -.3322), (-1.001, -.3332), (-1, error),
(-.999, -.3334), (-.99, -.3344), (-0.9, -.3448)}

So as you can see in the table when x = -1, f(x) = error because if you place -1 in for x in the equation, f(x) = 0/0 which is impossible to have because you cannot divide by 0.

But if you place values close to -1, f(x) ® -1/3 so

lim (x+1)/(x²-x-2) = -1/3
^x®-1

Now try:

lim (x + 1)/(x²-x-2)
^x®2

again f(2) = 3/0 which you cannot do so:

when we look at the value of f(x) at values of x around 2, you have two different answers.

1. as x approaches 2 from the left side:

lim (x + 1)/(x²-x - 2)
^x®2-

you can see the limit approaches -¥

2. as x approaches 2 from the right side:

lim (x + 1)/(x²-x - 2)
^x®2+

you can see the limit approaches +¥

so since f(x) is not bound as x approaches 2, you can conclude that the limit does not exist at x = 2.

Example 4:

lim (3x²-12)/(x-2)
^x®2

Again plugging x = 2, f(2) = 0/0 which is impossible to have so

(x, f(x)) = {(0, 6), (1, 9), (2, error), (3, 15), (4, 18), (5, 21), ...} so you can see that it is suppose to be 12. we can do this if we use factoring:

(3x² - 12)/(x - 2) =

(3(x² - 4))/(x - 2) =

( 3 (x + 2)(x - 2))/ (x - 2) =

3 (x + 2) because the (x - 2) divide out so we can simplify the limit to look like:

lim 3(x + 2)
^x®2

= 3 ( 2 + 2) = 12

Example 5:

lim (cos (x) - 1)/ x
^x®0

again when x =0, f(0) = error but when we look at the graph, as x approaches 0 from the left side and the right side, f(x) = 0 so we can conclude

lim (cos (x) - 1)/ x = 0
^x®0

Example 6:

lim (cos (1/x))
^x®0

Try using different windows:
let x be between -2p and +2p and y between -1 and +2

Now zoom in alittle: let x be between -1 and +1 by .1

Now zoom in more: let x be between -.1 and +.1 by .01

Continue this, what do you notice?

The smaller the window, so the closer to x = 0 you get, the graph does not approach any particular number (it goes between +1 and -1).

Therefore this limit does not exist because no matter how close you are to zero, it is possible to choose values of x₁ and x₂ such that cos(1/x₁) and cos(1/x₂) that do not equal.

Example 7: Finding Limits by Direct Substitution:

lim (x + 4)/(x - 3)
^x®1

Do the numerator separately from the denominator:

lim (x + 4) = 5
^x®1

lim (x - 3) = -2 so
^x®1

lim (x + 4)/(x - 3) = -5/2
^x®1

____________________________________________________________

I. Conditions Under Which Limits Do not exist:

The Limit of f(x) as x ® c does not exist if any of the following conditions is true.

1. f(x) approaches a different number from the right side of "c" than from the left side of "c".

2. f(x) increases or decreases without bound as x approaches c.

3. f(x) oscillates between two fixed values as x approaches c.

II. Properties of Limits:

Let "b" and "c" be real numbers and let "n" be a positive integer.

1. lim b = b
^x®c

2. lim x = c
^x®c

3. lim xⁿ = c^n
x®c

4. lim ⁿ Ö(x) = ⁿÖ(c) for n even and c greater than zero.
^x®c

III. Operations with Limits:

check out: http://homepage.usask.ca/~mha040/Limit%20Formulas.pdf

Given: lim f(x) = L
^x®c
and

lim g(x) = K
^x®c

1. Scalar multiple:
lim [ bf(x)] = bL
^x®c

2a. Sum of limits:
lim [ f(x) + g(x)] = L + K
^x®c

2b. Difference of limits:
lim [ f(x) - g(x)] = L - K
^x®c

3. Product:
lim [ f(x) ´ g(x)] = L ´ K
^x®c

4. Quotient:
lim [ f(x) ¸ g(x)] = L ¸ K
^x®c

5. Power:
lim [ f(x)]ⁿ = L^n
x®c

Example 8:

lim f(x) = 3x
^x®c

and

lim g(x) = -2x + 1
^x®c

Find:
lim [ f(x) + g(x)] =3(3) + (-2)(3) + 1 = 9 - 6 + 1 = 4
^x®3

Find:
lim [ f(x) g(x)] = [(3)(4)][-2(4) + 1] = (12)(-8 + 1) = (12)(-7) = -84
^x®4

Example 9:
lim f(x) = 3
^x®c

and

lim g(x) = -2
^x®c

Find:
lim [ f(x) + g(x)]² = (3 + -2)² = 1
^x®c

Find:
lim [ 6f(x)g(x)] = 6(3)(-2) = -36
^x®c