8.4: Polynomial Function Limits
You may recall that many of the rules you have learned to use when solving basic algebra equations often apply to the manipulation of complex equations also. You will find in this lesson that a similar concept applies to operations on limits.
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Khan Academy: Limit Examples (part 3)
Guidance
In this lesson, you will evaluate the limits of functions from an algebraic perspective. First you will require the proper tools for the job, the theorems necessary to calculate limits. These theorems are outlined in the two boxes below.
Box #1: Important Theorems of Limits Let a be a real number and suppose that and . Then: 1. , meaning the limit of the sum is the sum of the limits. 2. , meaning the limit of the difference is the difference of the limits. 3. , meaning the limit of the product is the product of the limits. 4. , meaning the limit of a quotient is the quotient of the limits (provided that the denominator does not equal zero.) 5. if n is even, meaning the limit of the n ^{ th } root is the n ^{ th } root of the limit. 

Other useful results follow from the above theorems:
Box #2 1. If a and k are real numbers, then . That is, if f ( x ) = k , a constant function, then the values of f ( x ) do not change as x is varied, thus the limit of f ( x ) is k . 2. If a is a real number then . That is, since f ( x ) = x is an identity function (its input equals its output), then as x → a , f ( x ) = x → a . 3. 4. 

Finally, we have one more theorem:
Theorem: The limit of a polynomial


Example A
Use the theorems above to find and justify each step.

 NOTE:

 This particular example was specifically chosen as "A", since you can verify your practice application of the theorems described above by simply substituting x = 1 directly into the polynomial:
Solution:
Using Equation (1) of Box #1 (the limit of the sum is the sum of the limits) we get
 From Equation (4) of Box #2, the first term becomes
 From Equations (2) and (3) of Box #2, the second term becomes
 Finally, from Equation (1) of Box #2, the third term becomes
Thus the limit of the above polynomial is
Example B
Find .
Solution:
According to the theorem above,
Example C
Find .
Solution:
Using Equation (4) of Box #1 (the limit of the quotient is the quotient of the limit),
 Making use of the limit of the polynomials theorem, we obtain,
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Guided Practice
1) Find the limit:
2) Find the limit:
3) Given:
Find:
4) Given:
Find:
5) Given:
Find:
Answers
1) Since the function has no specific applicable discontinuities, just solve for x = 256.
2) To find
 ..... First factor the numerator
 Since both numerator and denominator contain the function output is the same as the input for any number except where was undefined, 3.
 Since x = y for any number other than 3, as we get closer and closer to 3 on the input, we get closer and closer to 3 on the output (we can get as close as we like, we just can't hit it!).
3) This example uses rule #3 from above: the limit of the product is the product of the limits. Which means that we need to find the limit of each function, then multiply the limits.
Since we are given the limits, we simply multiply
4) Since these are both continuous functions, we can use
5) This example uses rule #4 from above: The limit of a quotient is the quotient of the limits (provided that the denominator does not equal zero).
 Since we are given the limits of and we just need to divide
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Find the limit:
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 Given: and , find:
 Given: and , find:
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Description
Learning Objectives
Here you will learn several important theorems involving limits, and you will explore the use of those theorems in finding the limits of polynomial functions.