12.4 Limits of Infinity

Limits of Infinity are basically asking about the horizontal asymtote. In order to find the solution to a limit as it aproaches infinity, we just need to find the horizonatal asymtote.

Finding the horizontal asymtote is basically finding where the graph will go when x becomes REALLY big.

When x becomes really big, only the first term is significant. Anything after that is so minor that it won't factor into changing the final answer, so it can just be disreguarded.

When the powers of x are the same, you are adding the same amount to both the numerator of the equation and the denominator so the ration of the x value will remain the same, so the horizontal asymtote is the ratio of the x's.

When the Power of x in the numerator is greater than the power of x in the denominator, the function will continue to increase forever, not approaching any horizontal asymtote, so therefore, the asymtote doesn't exist.

When the power of x in the numerator is less than the power of x in the denominator the horizontal asymtote is zero because the bottom will continue to become bigger and bigger, creating a smaller and smaller fraction, thus approaching zero.

To find solutions to limits as x approaches infinity, go to wolframalpha

Section 12-2 Techniques for Evaluating Limits

Previously, we had 3 ways to evaluate limits: Graphing, a Numeric Approach (using tables), and Direct Substitution.

Today, we added a few more techniques to our ever increasing bag of tricks. After all, a wise man often tells us, "Variety is the spice of life."

When given a problem such as...

You MUST factor to check for holes or asymptotes

The (x-3) terms cancel. Leaving you with...

From there, you use direct substitution and plug 3 in for x. Therefore, x=-1.

Next, we moved on to another type of situation. When given a problem such as...

You again have to factor this to check for holes/asymptotes. You may be scratching your heads and saying, "How do I factor a cubic?" Well my friends, I'll remind you of something we learned all the way back in Algebra 2. SYNTHETIC DIVISION!!!

To start, you plug the number that x is approaching into your box. In this specific case, we will use 2.

If it factors, there is a hole in the graph and it has a limit. However, if it does not factor, the graph has an asymptote and therefore a limit DOES NOT EXIST.

When you use synthetic division on the original problem,

You get a remainder of -2 meaning that it has an asymptote and therefore not a limit.

Rationalizing the Numerator

In order to solve a problem like this one,

You have to multiply it by the conjugate. Usually, you would do this to rationalize the denominator, but for limit problems you rationalize either the numerator or denominator to help you solve the problem.

This leaves you with...

From there, the x's cancel out and you are left with...

You then use direct substitution and plug 0 in for x. The problem then simplifies to 1/2.

Now for your viewing pleasure, here is a song about limits. Aspects of it hold true to my life. It may for some of you as well. Enjoy!!! :)

Here is the link (the video player wasn't working)

http://www.youtube.com/watch?v=BnHuZs7ar7I

Chapter 12.1 Part Two-Introduction to Limits

Properties of Limits

1.
lim b=b
^x→a

example
lim5=5
^x→2













2.
lim x=a
^x→a

example
lim x=2
^x→2











3.
lim xⁿ=a^n
x→a

example
lim x²=4
^x→2











4.

^x→c

example

<sup>x→8

Operations with Limits

Let f(x)=L
x→a
and
g(x)=M
x→a

Scalar Multiple
lim[c × f(x)]=c × L
x→a</sup>

Sum or Difference
lim[f(x)± g(x)]=L ± M
^x→a

Product
lim[f(x) × g(x)]=L × M
^x→a

Quotient
lim[f(x) ÷ g(x)]= L ÷ M M&ne0
^x→a

Power
lim[f(x)]ⁿ=L^n
x→a

Direct Substitution

is used when:
lim f(x)=f(a)
^x→a

• doesn't work with breaks, holes or asymptotes
  
• would be "false" on a quiz because it is not always true.
  

example
lim x^2
x→4

4²
16

Continuous

• no breaks, holes or asymptotes
  
• draw without lifting pencil
  
• lim f(x)=lim f(x)=f(a)
  ^{x→a+ x→a−}
• if the limit from the left and the limit from the right are both equal to f(a) then it is continuous

not continuous










Piecewise



lim f(x)
x→1−
2(1)-3
-1

lim f(x)
x→1+
-(1)²
-1

Since the limit coming from the right and coming from the left are equal(-1), the limit is -1

if the limits from both sides were different the limit does not exist.

For your viewing pleasure....

12.1 Intro to Limits (Part 1)

Definition of a 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.

lim f(x) = L

x--> c

Ex.

lim (3x- 2)

x--> 2
Let f(x) = 3x- 2. Then construct a table that shows values of f(x) when x is close to 2.

x           f(x)

1.9       3.7

1.99     3.97

1.999   3.997

2.0       ?

2.001   4.003

2.01     4.03

2.1       4.3

From the table, you can see that the closer x gets to 2, the closer f(x) gets to 4. So you can estimate the limit to be 4. For this equation you can substitute 2 for x to obtain the limit, so:

   lim (3x- 2) = 3(2) -2 = 4

    x--> 2

x--> 1+ (the + meaning that x approaches 1 from the right; right-hand limit)
x--> 1- (the - meaning that x approaches 1 from the left; left-hand limit)

Limits That Fail to Exist:

1.) When the one-sided limits are not equal

      ex.

   

 

 

2.) Unbounded Behavior

      ex.

When x > 100 or x<-100

3.) Oscillating Behavior

      ex.

WARNING: Use the above information with caution.

Section 9.5: The Binomial Theorem

A binomial is a polynomial that has two terms. This section will teach you how to raise a binomial to a power in a relatively quick way.

The Binomial Theorem:

nCr=

For example, you have the following toppings at a pizza store:

Pepperoni

Sausage

Onion

Ham

Bacon

You are allowed to put 2 toppings on each pizza. How many different combinations of toppings do you have?

n=5

r=2

There are 77,520 different combinations of pizza

Pascal's Triangle

Pascal's Triangle starts with row 0 and gradually gets larger.

Pascal's Triangle contains many different patterns, but we are going to focus on how the numbers in each row are similar to the coefficients when raising (x+y) to different powers.

(x+y)^0=1

(x+y)^1=1x+1y

(x+y)^2=1x^2+2xy+1y^2

(x+y)^3=1x^3+3x^2y+3xy^2+1y^3

(x+y)^4=1x^4+4x^3y+6x^2y^2+4xy^3+1y^4

(x+y)^5=1x^5+5x^4y+10x^3y^2+10x^2y^3+5xy^4+1y^5

The red numbers are the coefficients and make the shape and pattern of Pascal's Triangle.

Lets try an example:

1x^5+5x^4*4+10x^3*4^2+10x^2*4^3+5x*4^4+1*4^5

*Note: The binomial is raised to the 5th power. This tells you to look at row 5 of Pascal's Triangle and the numbers in this row will be the coefficients

Find the coefficient of the term:

17C10 * (2x^2)^10 * (-3i)^7

19448 * 1024x^20 * -823543i^7

The coefficient of the term is

Final Note:

=

nCr*a^r*b^(n-r)

Geometric Sequence

A geometric sequence is different from an arithmetric sequence by it is multipying instead of adding
The Common Ratio(r) is what the previous number is being multiplied by each time
example 3,6,12,24,48,96 r=3

Arithemetric Equation v. Geometric Equation

Arithmetic

a₁ =a₁
a₂ =a₁+d
a₃ =a₁+d+d
a₄ =a₁+d+d+d
.
.
.
a_n=a₁=d(n-1)

Geometric
a_a=a₁
a₂ =a₁× r
a₃ =a₁× r × r
a₄ =a₁× r × r × r
a_n=a₁× r^n-1 -equation to find nth term of a geometric sequence

Partial Sum

S_n=a₁+a₁r+a₁r²+....a₁r^n-2+a₁r^n-1
r × S_n=a₁r+a₁r²+....a₁r^n-2+a₁r^n-1+a₁rⁿ
S_n-r × S_n=a₁-a₁rⁿ S_n(1-r)=a₁(1-rⁿ) S_n=[a₁(1-rⁿ)]÷ (1-r)


example-
12r³=96
r³=8
r=2

a₃=a₁ × r^n-1
12=a₁ × r^3-1
12=a₁ × r^2
12⁄_r²=a₁
¹²⁄₄=a₁
S_n=3(1-2¹⁰)÷(1-2)
S_n=3069

Infinite Geometric Sequence

S=a₁(¹/_(1-r))
S=^a₁/_1-r

Section 9.2 Arithmetic Sequences and Partial Sums

A sequence whose consecutive terms have a common difference is called an arithmetic sequence





Definition of Arithmetic Sequence

A sequence is arithmetic if the differences between consecutive terms are the same. So, the sequence

is arithmetic if there is a number d such that

The number d is the common difference of the arithmetic sequence

The nth Term of an Arithmetic Sequence



The nth term of an arithmetic sequence has the form



a_n = dn + c




where d is the common difference between consecutive terms of the sequence and c = a_1 - d





The Sum of a Finite Arithmetic Sequence




The sum of a finite arithmetic sequence with n terms is



S_n = (n/2)(a_1 + a_n).



Example:



Find the sum of of the integers from 1 to 100



S_n = 1+2+3+4+5+6+...+99+100


= (n/2)(a_1 + a_n)


= (100/2)(1+100)


= 50(101)


=5050

Section 9.1: Sequences

A sequence is defined as a function whose domain consists of natural numbers.  

The function values

are the terms of the sequence. 

Finding the Terms of a Sequence

• To find the sequence's recursive formula, look at the first few terms. Then use previous terms to define all other terms

(next = now + 5)

• To find the sequence's explicit formula, look for an apparent pattern. 

Factorials
If n is a positive integer, n factorial is defined by

n! = 1 • 2 • 3 • 4 ... (n - 1) • n

Here are some values of n!:

0! : 1

1! : 1

2!: 1 • 2 • = 2

3!: 1 • 2 • 3 = 6

Common Sequences

Some sequences occur more frequently than others. Some common sequences include:

• Powers of 2: 2, 4, 8, 16...
• Powers of 3: 3, 9, 27, 81...
• Perfect Squares: 1, 4, 9, 16...
• Factorial Numbers: 1, 1, 2, 6, 24...

Pre Calc Section 7.2

This section contains information on Summation Notation.

A sequence is an ordered list of numbers

The domain of a sequence must be natural numbers, as opposed to a function, which has the domain of all real numbers.

Sigma Notation/Summation Notation

F(i)= is the explicit formula for a sequence you are taking the sum of.

n= the upper limit

i=1 is the lower limit

F(i)=F1+F2+F3+F4+..................F(N)

You take the sum, starting from the lower limit (1), until you hit the upper limit (N).

YOU DONT HAVE N # OF TERMS, YOU STOP SUMMATION WHEN YOU HIT THE NTH TERM!

To solve in graphing calculator

SUM(SEQ(explicit formula, variable, lower limit, upper limit))

Sigma Takes Commas out and turns them into pluses.

FORMS A SERIES

Partial Sums

Let N = 5

Series Starts at 1 and goes until N, in this case 5.

S15=15th Partial Sum

Section 7.1

This section is all about solving systems of equations and the different ways it can be done!

There are 3 main ways to solve a system of equations...
1) Substitution
2) Elimination
3) Graphing

Substitution
Example:
x + y = 4
2x + 1 = y

By plugging the second equation into the first we can conclude that:

x + (2x+1) = 4
3x + 1 = 4
3x = 3
x = 1

Some simple steps for substitution include...
1. Solve one equation for one variable in terms of the other
2. Substitute the expression found in step 1 into the other equation to obtain an equation in one variable
3. Solve the equation obtained in step 2
4. Back- substitute the solution in step 3 into the expression obtained in step 1 to find the value of the other variable
5. Check that the solution satisfies each of the original equations

Elimination
Example:
3x + 5y =10
2x - 5y = 5

Given this set of equations, we can elimate the 5y and -5y and solve the equation...

3x = 10
2x = 5
_______
5x = 15
x = 3

Remember a few easy steps when solving a system of equations using elimination:
1. Obtain coefficients for x (or y) that differ only in sign by multiplying all terms of one or both equations by suitably chosen constants
2. Add the equations to eliminate one variable; solve the resulting equation
3. Back- substitute the value obtained in Step 2 into either of the original equations and solve for the other variable
4. Check your solution in both of the original equations

Graphing
Graphing to find the solution of a system of equations comes in handy when the set is too challenging to solve algebraically. It should not be a first resort when solving the system though. When you graph the two equations, remember the solution is where the two graphs intersect!

Exponential and Logarithmic Wrap-Up

One to One:

Exponential:

2^x = 2^3x-1 original problem

x = 3x-1 because they are one to one, you can disregard their base

1/2 = x solve

Logarithmic:

log(x-4) - log(3x-10) = log(1/x) original problem

use the logarithmic laws

because they are one to one, you can disregard the logs

continue solving

Inverse (key to everything):

Exponential:

e^x-5 = 0

lne^x = ln 5

x = ln5

use calculator

Logarithmic:

take 10 to the power of each side (exponentiation)

continue solving

Solving Exponential and Logarithmic Equations

One-to-One Properties:

if and only if x=y

if and only if x=y

Inverse Properties:

=x


=x

Solving Exponential and Logarithmic Equations

- rewrite the equation using one to one properties

Ex.

x=3

- rewrite an exponential equation as a logarithmic equation and use the inverse properties

Ex.

since there is an e in the equation, we use natural log (ln) rather than log

the ln cancels out the e since an e is already included when we use ln

x= ln5

-Rewrite a logarithmic equation as an exponential equation and use the inverse properties

log x= 2

10^(log x) = 10^2

the 10 and the log cancel out

x= 10^2

x=100

Examples from class:

Ex. 1

2+3^(x-4) = 12

subtract the 2

3^(x-4) = 10

change to a logarithmic equation ( the exponent (x-4) will be multiplied by the logarithm)

(x-4) log 3= log 10

log 10 is equal to one (10^1 = 10)

(x-4) log 3= 1

divide by the logarithm

(x-4) = 1/(log 3)

add the 4

x= 1/(log 3) + 4

solve using a calculator

x= 6.10

Ex. 2

e^(2x) - 3e^x +2 = 0

let u = e^x

u^2 - 3u + 2 = 0

factor the equation

(u-1)(u-2) = 0

u=1 u=2

e^x=1 e^x=2

x=0 ln(e^x) =ln 2

x=.7