12.2 Techniques for Evaluating Limits

Limits of Polynomial and Rational Functions

Lesson 12.2 discusses several processes by which limits may be evaluated:

Direct Substitution

Direct Substitution is a viable approach only when a function is continuous. For example, the function below is continuous, and by using direct substitution, the limit is revealed:

Substitute 6 for x, and 0 results. Therefore, the limit is 0.

Definition of a Continuous:

However, not all limits can be evaluated by direct substitution, as stated above. Another approach must be taken to evalute these limits. The first of these techniques is found below:

Dividing Out

Direct substitution is not an option for the function below as it is not continutous:

If direct substitution is used both the numerator and denominator will be zero. However, unlike in previous lessons, this does not mean that the limit is undefined. It means that the limit is Indeterminate, that is, a limit may or may not exist. Another method must be employed to find out.

Dividing out involves the canceling out of common terms in a function. Factoring is one method that may be used to accomplish this. When the above limit is factored, the following results, and (x - 1) may be canceled out. In addition, another way to view the discontinuity of the function is by noticing that a hole exists where x = 1:

Once (x - 1) has been canceled out, direct substitution may be employed:

7 is revealed to be the limit.

Also, in cases where factoring is not an option, other techniques can be used in the dividing out method. For example, synthetic division can be used to evaluate the function below:

In addition to the dividing out method, another method exists for evaluating limits of discontinuous functions:

Rationalizing

In this case, rationalization must be used as dividing out is not possible:

Attempting to use direct substitution reveals the function is discontinuous and the limit indeterminate.

So,

the function will be rationalized:

And what results:

Simplify and cross out like terms:

As with the dividing out method, direct substitution can now be employed to evaluate the limit:

The limit is one-half.

12.1 Limits

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.

Note: As x gets closer to c, the closer y gets to L

Example:

• One way to solve this is by putting 2 in for x.
• Another way is to make a table

x 1.9 1.99 1.999 2 2.001 F(x) 4.7 4.97 4.997 5 5.003

The closer x gets to 2 (from both sides) the closer f(x) gets to 5

Describing the limit

Limits that Do Not Exist (DNR):

1. when the left and right behaviors are different

the limit from the left would be -2, but the limit from the right would be 2.

The limit from the left side is described as

The - sign above the 2 shows it is from the left side

The limit from the right side is 2 (c) with a + sign

2. If a function increases without a bound, f(x) can be as large as you want

The asymptote of x=0 causes you to be able to choose a number as close to 0 and it keeps decreasing

3. When it Oscillates between two different numbers

the limit just switches between 1 and -1.

Continuity:

If a function is continuous at c then

• A trick for knowing if a graph is continuous is if you can draw it out without lifting up your pencil.
• Discontinuous graphs include: BREAKS, HOLES AND ASYMPTOTES

9.5 The Binomial Theorem

A binomial is a polynomial that has two terms.

Lets look at the expansion of

n = 0

n=1

n=2

n=3 ... etc.

Observations:

1. In each expansion, there are n+1 terms

2. In each expansion, x and y have symmetric roles. The powers of x decrease by one in successive terms, whereas the powers of y increase by one

3. The sum of the powers in each term in n. For instance, in the expansion of (x+y)^5, the sum of the powers in each term is 5.

4. The coefficients increase then decrease in a symmetric pattern.

The coefficients of a binomial expansion are called Binomial Coefficients. To find them, you can use the Binomial Theorem.

The Binomial Theorem:

In the expansion of

the coefficient of is:

Pascal's Triangle:

By arranging the coefficients in a triangular pattern, you obtain the following array, which is called Pascal's Triangle.

->The first and last number in each row of Pascal's Triangle is 1. Every other number in each row is formed by adding the two numbers immediately above the number. Pascal noticed that numbers in this triangle are precisely the same numbers as the coefficients of binomial expansions.

->The top row in Pascal's Triangle is called the zero row because it corresponds to the binomial expansion

->The nth row in Pascal's Triangle gives the coefficients of

Binomial Expansions:

When you write out the coefficients for a binomial that is raised to a power, you are expanding a binomial.

Geometric Sequences and Series

DEFINITION OF GEOMETRIC SEQUENCE

A sequence is a geometric sequence if the ratios of the consecutive terms are the same.

The number r is the common ratio of the sequence

THE NTH TERM OF A GEOMETRIC SEQUENCE

Explicit:

The nth term of a geometric sequence has the form

THE PARTIAL SUM OF A FINITE GEOMETRIC SEQUENCE

 THE SUM OF AN INFINITE GEOMETRIC SERIES

Arithmatic Sequences and Partial Sums

Definition of an Arithmetic Sequence:

A sequence is arithmetic, if the differences between consecutive terms are the same, so if you have  a1, a2, a3, a4, ...
Then,

a2 - a1 = a3 - a2 = a4 - a3 = ..... d

d is the common difference of the arithmetic sequence


An example of an arithmetic sequence would be 3, 6, 9, 12, 15...

since 6-3 = 3,   9-6=3,  12-9=3,  15-12=3 ... and so on.

In this case, d = 3; The common difference of this arithmetic sequence is 3.

Finding The nth Term of an Arithmetic Sequence:





where d is the common difference in the consecutive terms of the sequence and 


 An example of finding the nth term would be- in the sequence  3,8,13,18...  you need to simply find the common difference which would be 8-3=5, with the common difference, place it in front of n in your equation which is now . With this equation, plug in a number from the sequence lets try 3,
 ; solve for c; c=-2. so plug d and c back into the original equation of  and you get .

The Sum of a Finite Arithmetic Sequence:


The sum can be found with the equation where sn is the sum, and n is the number of terms.

An example of this equation would be- With the sum of 1+3+5+7+9, replace n with 5 (since there are five terms), a1 with 1, an with 9. The new equation is . Solve and you get , So the sum of these five terms = 25.

9.1 Sequences and Series

=An infinite sequence is a function whose domain is the set of positive integers. The function values

a1, a2, a3, a4, … , an

are the terms of the sequence. If the domain of the function consists of the first n positive integers only, the sequence is a finite sequence.

There are two types of formulas to display these sequences: Recursive and Explicit.

For this sequence: -5, -2, 1, 4, 7, 10 here are the recursive and explicit formulas.

Recursive:
a1 = -5
ak+1 = ak+3

k represents the next term for the sequence. It shows how to find out the next term.

Explicit:

an = 3n-8

n represents the current term for the sequence and how to find it. For example, the first term (1) is found by multiplying 1 by 3 and then subtracting by 8. You can do this for each term following the first term with 2, 3 ,4 and so on.

Here are some other examples of some basic sequences:

n² = 1, 4, 9, 16, 25, 36, …
n³ = 1, 8, 27, 64, 125, …
2^n = 2, 4, 8, 16, 32, …
3^n = 3, 9, 27, 81, …
n! = 1, 2, 6, 24, 120, ...
! represents factorial. If n is a positive integer, n! = 1 · 2 · 3 · 4 · ... · (n-1) · n.

0!=1
1!=1
2! = 1 · 2 = 2
3! = 1 · 2 · 3 = 6
4! = 1 · 2 · 3 · 4 = 24
5! = 1 · 2 · 3 · 4 · 5 = 120

Summation Notation

Summation (Sigma) Notation

The sum of the first n terms of a sequence, represented by

1 is the lower limit of the sum, which is the starting input of a sequence

The i is the index of summation

The n is the upper limit of the sum, which is the final input of the sequence

Use the limits and equation to find the sum

Examples:

= 2(3) + 2(4) + 2(5) + 2(6) = 36

= (2(5²) + 5(5) + 2) + . . .

When sums become very large, like the second example, using a calculator is useful to find the sum

To solve a sum with a calculator, plug this in:

sum(seq(explicit formula, variable, lower limit, upper limit))

Now the example from above:

sum(seq(2x² + 5x + 2, x, 3, 6)) = 270

Series

The sum of the first n terms of the seque nce is called a finite series

The sum of all terms of an ongoing sequence is called an infinite series

Can have a finite solution if a fraction is raised to an exponent

Partial Sums

The sum of the first n factors of a sum

infinite series, solve for the 5th partial sum

Substitute 5 for n, then find the sum of the first 5 numbers