LESSON 1 REVIEW OF SOLVING NONLINEAR INEQUALITIES



LESSON 3 THE LIMIT OF A FUNCTION

Example Consider the function [pic].

The domain of f is the set of all real numbers except for 2. What is the value of f when x is close to 2?

|x |f(x) |

|2.5 |13.75 |

|2.1 |8.23 |

|2.01 |7.1203 |

|2.001 |7.012003 |

|2.0001 |7.00120003 |

|2.00001 |7.0001200003 |

|2.000001 |7.000012000003 |

|2.0000001 |7.00000120000003 |

|2.00000001 |7.0000001200000003 |

|2.000000001 |7.000000012000000003 |

|2.0000000001 |7.00000000120000000003 |

|2.00000000001 |7.0000000001200000000003 |

|2.000000000001 |7.000000000012000000000003 |

NOTE: [pic] =

[pic] =

[pic] = [pic] = 7.012003

It appears that as x is getting closer to 2, [pic] is getting closer to 7. We let x get closer to 2 through values that are larger than 2. What about values of x which are smaller than 2?

|x |f(x) |

|1.5 |1.75 |

|1.9 |5.83 |

|1.99 |6.8803 |

|1.999 |6.988003 |

|1.9999 |6.99880003 |

|1.99999 |6.9998800003 |

|1.999999 |6.999988000003 |

|1.9999999 |6.99999880000003 |

|1.99999999 |6.9999998800000003 |

|1.999999999 |6.999999988000000003 |

|1.9999999999 |6.99999999880000000003 |

|1.99999999999 |6.9999999998800000000003 |

|1.999999999999 |6.999999999988000000000003 |

NOTE: [pic] =

[pic] = [pic]

[pic] = 6.8803

Again, it appears that as x is getting closer to 2, [pic] is getting closer to 7.

Does the rational expression [pic] simplify? It will simplify if [pic] is a factor of [pic]. You might try factoring

[pic] by grouping together the first two terms, which have a [pic] in common, and the last two terms, which have a 5 in common.

[pic] = [pic] = [pic]

NOTE: We obtained the final factorization (product) by factoring out the common [pic]factor in the expression [pic] and in the expression [pic].

Thus, [pic] = [pic] = [pic] for all x such that [pic].

What does the graph of [pic] look like? It is the graph of [pic] with the point [pic] removed.

This graph was created using Maple.

Notation: The mathematical expression [pic] means “x approaches a”.

Thus, [pic] means “[pic] approaches L”.

For the example above, we may write the statement: As [pic], [pic]. We are going to define this statement as a limit using the following definitions.

Informal Definition of a Limit: To say that the limit of [pic] is L as x approaches a , denoted by [pic], means that [pic] gets closer and closer to L (that is, [pic] approaches L) as x gets closer and closer to a (that is, x approaches a).

That is, [pic] means “As [pic], [pic].”

Since as [pic], [pic] in the example above, then we may write [pic] or [pic].

Formal Definition of a Limit: To say that the limit of [pic] is L as x approaches a , denoted by [pic], means given any [pic], there exists a [pic] such that [pic] whenever [pic].

NOTE: That condition that [pic] in the definition above means that [pic].

NOTE: The expression [pic] is the distance between [pic] and L on the real number line and the expression [pic] is the distance between x and a on the real number line. Thus, the condition that [pic] whenever

[pic] is saying that whenever the distance between x and a is less than [pic] and [pic], then the distance between [pic] and L is less than [pic].

NOTE: The inequality [pic] is equivalent to the inequality

[pic] and this inequality is equivalent to the inequality

[pic]. This last inequality says that [pic] is in the interval

[pic].

( ( (

[pic] L [pic]

Similarly, the inequality [pic] is equivalent to the inequality [pic]. This inequality says that x is in the interval [pic]. Thus, the inequality [pic] is equivalent to saying that

[pic] and [pic]. This says that x is in the set

[pic]

( ( (

[pic] a [pic]

Sometimes, [pic].

Examples Find the following limits if they exist.

1. [pic]

[pic]

Since we are saying that [pic], this means that as x is getting closer to 1 ([pic]), [pic] is getting closer to 8 ([pic]). Let’s see if this is true. Let [pic].

[pic]

[pic]

[pic]

[pic]

[pic]

[pic]

[pic]

[pic]

To show that [pic] using the formal definition of limit, we need to show that for any [pic], there exists a [pic] such that [pic] whenever [pic]. Since [pic] = [pic] = [pic],

then [pic] whenever [pic], which implies that

[pic] whenever [pic]. Thus, given any [pic],

choose [pic].

Answer: 8

2. [pic]

[pic]

Since we are saying that [pic], this means that as x is getting closer to [pic] ([pic]), [pic] is getting closer to 38

([pic]). Check it out to see if this is true.

To show that [pic] using the formal definition of limit, we need to show that for any [pic], there exists a [pic] such that [pic] whenever [pic].

NOTE: [pic] = [pic] = [pic]

Since we are approaching [pic] in a limiting sense, we could restrict ourselves to the interval described by the inequality [pic]. This inequality is equivalent to the inequality [pic]. Thus, if [pic], then [pic] and [pic]. Since [pic], then [pic]. Since [pic] = [pic] from above, then [pic] = [pic] =

[pic] < [pic]. Thus, [pic] whenever

[pic], which implies that [pic] whenever [pic]. Thus, given any [pic], choose [pic] min[pic]

Answer: 38

3. [pic]

[pic]

Since we are saying that [pic], this means that as [pic], [pic]

Answer: [pic]

4. [pic]

[pic]

NOTE: The limit of a constant is that constant.

Answer: 8

5. [pic]

[pic]

NOTE: [pic] is called an indeterminate form because you can not determine any information from it.

We saw in the example at the beginning of this lesson that

[pic] can be factored by grouping and we obtained that

[pic] = [pic]. Thus, we have that

[pic] = [pic] = [pic] = 7

Answer: 7

6. [pic]

[pic] Indeterminate Form

[pic] = [pic] = [pic]

Answer: 10

7. [pic]

[pic] Indeterminate Form

[pic] = [pic] = [pic]

NOTE: Since [pic], then you can cancel the [pic] and [pic] as long as you provide a negative one in either the numerator or the denominator. For this problem, it easier to supply [pic] in the numerator.

Answer: [pic]

8. [pic]

[pic] Indeterminate Form

We can make use of a theorem from algebra called the Factor Theorem.

Factor Theorem If p is a polynomial in variable x and [pic], then [pic] is a factor of the polynomial p.

Thus, [pic] is a factor of [pic] and [pic].

Thus, [pic] = [pic] and

[pic] = [pic]

Thus, [pic] = [pic] = [pic]

Answer: [pic]

9. [pic]

[pic] Indeterminate Form

By the Factor Theorem given above, [pic] is a factor of

[pic] and [pic]. Thus,

[pic] = [pic] = [pic]

Answer: [pic]

10. [pic]

[pic] Indeterminate Form

By the Factor Theorem given above, [pic] is a factor of [pic]

and [pic]. This is equivalent to [pic] being a factor of

[pic] and [pic]. Thus,

[pic] = [pic] = [pic] =

[pic] = [pic] = [pic]

Answer: 2

11. [pic]

In order to evaluate [pic] for [pic], it is easier to use the factorization for [pic] since [pic] = [pic]. Since the factor [pic] is equal to zero when [pic], then [pic] = 0 when [pic]. Thus,

[pic] Indeterminate Form

[pic] = [pic]

We can cancel the factor of [pic] in the numerator with one on the two factors of [pic] in the denominator as long as we supply a [pic] in either the numerator or the denominator. Thus, we have that

[pic] = [pic] = [pic]

Now, we need to make use of the following theorem, which will be proved at the end of this lesson.

Theorem If [pic] and [pic], then [pic] does not exist (DNE).

Using this theorem, we have that

[pic] = [pic] = [pic] = [pic] DNE

Answer: DNE

12. [pic]

[pic]

The only available algebra step to try to simplify the fraction is to rationalize the numerator of the fraction. Thus, we have that

[pic] = [pic] =

[pic] = [pic] =

[pic] = [pic] = [pic] =

[pic] Answer: [pic]

13. [pic]

Just like in the previous example, we will need to rationalize the numerator. Thus,

[pic] = [pic] =

[pic] = [pic] =

[pic] = [pic] =

[pic] = [pic] = [pic] = [pic]

Answer: [pic]

14. [pic]

[pic] = [pic] = [pic]

We need to rationalize both the numerator and the denominator in the fraction. Thus,

[pic] =

[pic]

[pic] =

[pic] =

[pic] =

[pic] =

[pic] = [pic] = [pic] =

[pic] = [pic] = [pic] = [pic] Answer: [pic]

15. [pic]

COMMENT: We will need to calculate this limit in order to find the derivative of the function [pic] using the definition of derivative.

[pic] = [pic] = [pic]

[pic] =

[pic] =

[pic] =

[pic] =

[pic] = [pic] =

[pic] = [pic] Answer: [pic]

16. [pic]

COMMENT: We will need to calculate this limit in order to find the derivative of the function [pic] using the definition of derivative.

We will use the Binomial Expansion theorem in order to find the four terms in the expansion of [pic].

Binomial Expansion Theorem Let n be a positive integer. Let a and b be real numbers, then [pic], where [pic].

Thus, by the Binomial Expansion theorem, we have that

[pic]

[pic] = [pic] =

[pic] = [pic] =

[pic] = [pic] Answer: [pic]

17. [pic]

[pic]

By the Factor Theorem, [pic] is a factor of [pic]. This factorization can be found using the factorization formula for the sum of cubes, which is

[pic]

Since [pic] = [pic], then replace all the a’s in the formula by x and replace the b’s in the formula by 4. Thus, we have that

[pic] = [pic] = [pic] = [pic]

The factorization of [pic] can also be found using synthetic division. Since [pic] = [pic], then we write the following in order to carry out the synthetic division

[pic] [pic]

Thus, [pic] = [pic]. Thus, evaluating the given limit, we have that

[pic] = [pic] = [pic] =

[pic]

Answer: 48

18. [pic]

[pic]

By the Factor Theorem, [pic] is a factor of [pic]. This factorization can be found using the factorization formula for the difference of cubes, which is

[pic]

Since [pic] = [pic], then replace all the a’s in the formula by 2t and replace the b’s in the formula by 3. Thus, we have that

[pic] = [pic] = [pic] =

[pic]

The factorization of [pic] can also be found using synthetic division. Since [pic] = [pic], then we write the following in order to carry out the synthetic division

[pic] [pic]

Thus, [pic] = [pic]. Factoring out the 2 that is common to all terms in the second factor of [pic], we obtain that

[pic] = [pic] = [pic]. Now, distributing the 2 through the first factor of [pic], we obtain that [pic] = [pic]. Thus, evaluating the given limit, we have that

[pic] = [pic] = [pic] =

[pic]

Answer: 27

Theorem (Properties of Limits) If [pic] and [pic], and c is any real number, which will called a constant, then

1. [pic]

2. [pic] = [pic]

3. [pic] = [pic] + [pic]

4. [pic] = [pic]

5. [pic] = [pic]

6. [pic] = [pic], provided that [pic]

7. [pic] provided that [pic] is defined

Proof: These properties are proved using the formal definition of limit.

Example Find [pic] using properties of limits.

[pic] = [pic] = [pic] =

[pic] = [pic] = [pic] = [pic]

Theorem If [pic] and [pic], then [pic] does not exist (DNE).

Proof: Suppose that [pic] does exist, say that [pic]. Then

[pic] = [pic] = [pic] = [pic]. This is a contradiction to the fact that [pic]. Thus, [pic] does not exist.

NOTE: This type of proof in mathematics is called a proof by contradiction. In a proof by contradiction, an assumption is made and is used along with other information to obtain a conclusion that contradicts known information. The assumption, which is made, is usually the opposite of what you want to prove.

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