Math1431 Section 1.3

Section 1.3: The Definition of a Limit The Limit of a Function If we let the limit of a function be equal to L and c be the fixed value that x approaches, then we can say lim→

if and only if for each 0, there exists a 0 such that if | | , then | | .

Example 1: Show that lim

→5 2 8 using the definition of a limit.

Identify the following pieces c = L =

Math1431 Section 1.3

Example 2: Give the largest that works with = 0.1 for the limit, lim→

1 2 3

Arithmetic Rules for Limits If lim

→ and lim

→ then:

1. lim

→ 2. lim

→∙

3. lim→

∙ 4. lim→

, 0

Example 3: Let lim

→3, lim

→2 and lim

→5 . Evaluate the following

lim→

2 ∙

Example 4: Evaluate the following limits:

a. lim→

3 22 1

Math1431 Section 1.3

b. lim

→2 3 4 12

c. lim→

3 22 3 1

d. lim→

2 21

So if a simple plug in didn’t give us an actual answer, which means we have to approach the limit a different way.

1. Factoring 2. Distribution, common denominators 3. Using the conjugate

Example 5: Back to 4d

lim→

2 21

Math1431 Section 1.3

Example 6: Evaluate: 0

7lim 4x

xx

Example 7: Evaluate: 3

3

2 54lim

3x

x

x

Example 8: Evaluate 25

25lim

5x

x

x

Math1431 Section 1.3

Example 9: Evaluate 1

1lim

1x

x

x

Example 10: Evaluate 0

1 14 4lim

h

hh

Limit of Piecewise Functions

Example 11: Let 2

5 , 0

( ) , 0 2

4 16, 2

x x

f x x x

x x

, find 0

lim ( )x

f x

Math1431 Section 1.3

Example 12:

2 16, 4

( ) 49, 4

xx

f x xx

, find 4

lim ( )x

f x

Limits at Infinity If a limit at infinity exists and it’s equal to a single real number L then they are written as lim ( )

xf x L

or

lim ( )x

f x L

. These limits deal with what is happening to the y-values to the far left or right side of the graph

(function). Limits at infinity problems often involve rational expressions (fractions). The technique we can use to evaluate limits at infinity is to recall some rules from Algebra used to find horizontal asymptotes. These rules came from “limits at infinity” so they’ll surely work for us here. The highest power of the variable in a polynomial is called the degree of the polynomial. We can compare the degree of the numerator with the degree of the denominator and come up with some generalizations.

If the degree of the numerator is smaller than the degree of the denominator, then ( )

lim 0( )x

f x

g x

If the degree of the numerator is the same as the degree of the denominator, then you can find ( )

lim( )x

f x

g x

by making a fraction from the leading coefficients of the numerator and denominator and then reducing to lowest terms.

If the degree of the numerator is larger than the degree of the denominator, then the limit does not exist.

Math1431 Section 1.3

Let’s see how these generalizations came to be. Example 13: Evaluate the following limits

1limx x

2

1limx x

1lim

nx x

Example 14: Evaluate 4 2

2 7lim

5 6x

x

x x

Math1431 Section 1.3

Example 15: Evaluate 2

2

3l

4im

2 1x

x

x

x

x

 

 

 

 

   

Example 15: Evaluate 2

4 33li

4m

1x

xx

x x

Math1431 Section 1.4

Section 1.4: Continuity A function is a continuous at a point if its graph has no gaps, holes, breaks or jumps at that point. If a function is not continuous at a point, then we say it is discontinuous at that point. The function 1 2 graphed below is continuous everywhere.

The function below is NOT continuous everywhere, it is discontinuous at 3 and 1

We’ll focus on the classifying them in a moment. The following types of functions are continuous over their domain.

Polynomials, Rational Functions, Root Functions, Trigonometric Functions, Inverse Trigonometric Functions, Exponential Functions, Logarithmic Functions

Math1431 Section 1.4

Theorem: If f and g are continuous at c, a is a real number then each are also continuous at c.

I. II.

III.

IV. f

gprovided 0

Theorem: If g is continuous at c and f is continuous at , then ∘ is continuous at c. Example 1: Discuss the continuity for: a. 7( ) 4g x x x Discontinuous Continuous

b. 2

3( )

6

xf x

x x

Discontinuous Continuous

c. 2

( )1 cos

xh x

x

Discontinuous Continuous

Types of Discontinuity Removable Discontinuity occurs when:

lim ( ) ( )x c

f x f c

(or the limit exist, but is undefined.

Jump Discontinuity occurs when:

lim ( )x c

f x

and lim ( )x c

f x

exists, but are not equal.

Infinite Discontinuity occurs when:

lim ( )x c

f x

on at least one side of c. Infinite discontinuities are generally associated with a vertical

asymptote

Math1431 Section 1.4

Example 2: Identify and state the discontinuity.

One-Sided Continuity A function f is called continuous from the left at c if lim ( ) ( )

x cf x f c

and continuous from the right at c if

lim ( ) ( )x c

f x f c

In the example 2, we have continuity from the right at x = 0 and continuity on the left at x = 2 Continuity Stated a Bit More Formally A function f is said to be continuous at the point x = c if the following three conditions are met: 1. ( )f c is defined. 2. lim ( )

x cf x

exists. 3. lim ( ) ( )

x cf x f c

To check if a function is continuous at a point, we’ll use the three steps above. This process is called the three step method.

Math1431 Section 1.4

Example 3: Let 3 1

1

x xf x

x x

is the function continuous at 1?

So let’s go through the process 1. Is 1 defined? 2. Check to see if

1lim ( )x

f x

exist.

So check:

1lim ( )x

f x

and 1

lim ( )x

f x

Does

1lim ( )x

f x

exist?

3.

1lim ( ) (1)x

f x f

?

If at least one of the three steps fails, identify the type of discontinuity

Math1431 Section 1.4

Example 4: Let

2

3

2 9 3

2 3

3

x x

f x x

x x

is the function continuous at 3?

So let’s go through the process 1. Is 3 defined? 2. Check to see if

3lim ( )x

f x

exist.

So check:

3lim ( )x

f x

and 3

lim ( )x

f x

Does

3lim ( )x

f x

exist?

3.

3lim ( ) (3)x

f x f

?

Example 5: Let 2

2 - 3 2

- 2

x xf x

cx x x

is the function continuous at 2?

So let’s go through the process 1. Is 2 defined? 2.

2lim ( )x

f x

must exist, so we need to make sure 2 2

lim ( ) lim ( )x x

f x f x

.

3. Set

2lim ( ) (2)x

f x f

to find c.

Math1431 Section 1.4

Example 6: Find A and B so that

22 1 2

2

3 2

x x

f x A x

Bx x

is continuous.

1. Find 2 . 2.

2lim ( )x

f x

must exist, so we need to make sure 2 2

lim ( ) lim ( )x x

f x f x

.

3. Since

2lim ( ) ( 2)x

f x f

then

Example 7: The function 25

5

xf x

x

is defined everywhere except at 25. If possible define at 25

so that it becomes continuous at 25.

Math1431 Section 1.4

Example 8: Given , find the points where the function is discontinuous and classify these points.

12

12 4

4 6

4 6

6

xx

xx

f x x x

x

x x