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# Finding Limits: Numerical and Graphical Approaches

Module by: First Last. E-mail the author

Summary: In this section, you will:

• Understand limit notation.
• Find a limit using a graph.
• Find a limit using a table.

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Intuitively, we know what a limit is. A car can go only so fast and no faster. A trash can might hold 33 gallons and no more. It is natural for measured amounts to have limits. What, for instance, is the limit to the height of a woman? The tallest woman on record was Jinlian Zeng from China, who was 8 ft 1 in.1 Is this the limit of the height to which women can grow? Perhaps not, but there is likely a limit that we might describe in inches if we were able to determine what it was.

To put it mathematically, the function whose input is a woman and whose output is a measured height in inches has a limit. In this section, we will examine numerical and graphical approaches to identifying limits.

## Understanding Limit Notation

We have seen how a sequence can have a limit, a value that the sequence of terms moves toward as the nu mber of terms increases. For example, the terms of the sequence

1, 1 2 , 1 4 , 1 8 ... 1, 1 2 , 1 4 , 1 8 ...

gets closer and closer to 0. A sequence is one type of function, but functions that are not sequences can also have limits. We can describe the behavior of the function as the input values get close to a specific value. If the limit of a function f(x)=L, f(x)=L, then as the input x xgets closer and closer to a, a, the output y-coordinate gets closer and closer to L. L. We say that the output “approaches” L. L.

Figure 1 provides a visual representation of the mathematical concept of limit. As the input value x xapproaches a, a, the output value f( x ) f( x ) approaches L. L.

We write the equation of a limit as

lim xa f(x)=L. lim xa f(x)=L.

This notation indicates that as x xapproaches a aboth from the left of x=a x=a and the right of x=a, x=a, the output value approaches L. L.

Consider the function

f(x)= x 2 6x7 x7 . f(x)= x 2 6x7 x7 .

We can factor the function as shown.

f(x)= (x7) (x+1) x7   Cancel like factors in numerator and denominator. f(x)=x+1,x7 Simplify. f(x)= (x7) (x+1) x7   Cancel like factors in numerator and denominator. f(x)=x+1,x7 Simplify.

Notice that x xcannot be 7, or we would be dividing by 0, so 7 is not in the domain of the original function. In order to avoid changing the function when we simplify, we set the same condition, x7, x7, for the simplified function. We can represent the function graphically as shown in Figure 2.

What happens at x=7 x=7 is completely different from what happens at points close to x=7 x=7 on either side. The notation

lim x7 f(x)=8 lim x7 f(x)=8

indicates that as the input x xapproaches 7 from either the left or the right, the output approaches 8. The output can get as close to 8 as we like if the input is sufficiently near 7.

What happens at x=7? x=7? When x=7, x=7, there is no corresponding output. We write this as

f(7) does not exist. f(7) does not exist.

This notation indicates that 7 is not in the domain of the function. We had already indicated this when we wrote the function as

f(x)=x+1,  x7. f(x)=x+1,  x7.

Notice that the limit of a function can exist even when f(x) f(x) is not defined at x=a. x=a. Much of our subsequent work will be determining limits of functions as x xnears a, a, even though the output at x=a x=a does not exist.

### A General Note: The Limit of a Function:

A quantity L Lis the limit of a function f( x ) f( x ) as x xapproaches a aif, as the input values of x xapproach a a(but do not equal a), a), the corresponding output values of f( x ) f( x ) get closer to L. L. Note that the value of the limit is not affected by the output value of f( x ) f( x ) at a. a. Both a aand L Lmust be real numbers. We write it as

lim xa f(x)=L lim xa f(x)=L

### Example 1

#### Problem 1

##### Understanding the Limit of a Function

For the following limit, define a,f(x), a,f(x), and L. L.

lim x2 ( 3x+5 )=11 lim x2 ( 3x+5 )=11
##### Solution

First, we recognize the notation of a limit. If the limit exists, as x xapproaches a, a, we write

lim xa f(x)=L. lim xa f(x)=L.

We are given

lim x2 ( 3x+5 )=11. lim x2 ( 3x+5 )=11.

This means that a=2,f(x)=3x+5, and L=11. a=2,f(x)=3x+5, and L=11.

##### Analysis

Recall that y=3x+5 y=3x+5 is a line with no breaks. As the input values approach 2, the output values will get close to 11. This may be phrased with the equation lim x2 (3x+5)=11 , lim x2 (3x+5)=11 , which means that as x xnears 2 (but is not exactly 2), the output of the function f(x)=3x+5 f(x)=3x+5 gets as close as we want to 3(2)+5, 3(2)+5, or 11, which is the limit L, L, as we take values of x xsufficiently near 2 but not at x=2. x=2.

### Try It:

#### Exercise 1

For the following limit, define a,f(x), a,f(x), and L. L.

lim x5 ( 2 x 2 4 )=46 lim x5 ( 2 x 2 4 )=46
##### Solution

a=5, a=5, f( x )=2 x 2 4, f( x )=2 x 2 4, and L=46. L=46.

### Understanding Left-Hand Limits and Right-Hand Limits

We can approach the input of a function from either side of a value—from the left or the right. Figure 3 shows the values of

f(x)=x+1,x7 f(x)=x+1,x7

as described earlier and depicted in Figure 2.

Values described as “from the left” are less than the input value 7 and would therefore appear to the left of the value on a number line. The input values that approach 7 from the left in Figure 3 are 6.9, 6.9, 6.99, 6.99, and 6.999. 6.999. The corresponding outputs are 7.9,7.99, 7.9,7.99, and 7.999. 7.999. These values are getting closer to 8. The limit of values of f( x ) f( x ) as x xapproaches from the left is known as the left-hand limit. For this function, 8 is the left-hand limit of the function f(x)=x+1,x7 f(x)=x+1,x7 as x xapproaches 7.

Values described as “from the right” are greater than the input value 7 and would therefore appear to the right of the value on a number line. The input values that approach 7 from the right in Figure 3 are 7.1, 7.1, 7.01, 7.01, and 7.001. 7.001. The corresponding outputs are 8.1, 8.1, 8.01, 8.01, and 8.001. 8.001. These values are getting closer to 8. The limit of values of f( x ) f( x ) as x xapproaches from the right is known as the right-hand limit. For this function, 8 is also the right-hand limit of the function f(x)=x+1,x7 f(x)=x+1,x7 as x xapproaches 7.

Figure 3 shows that we can get the output of the function within a distance of 0.1 from 8 by using an input within a distance of 0.1 from 7. In other words, we need an input x xwithin the interval 6.9<x<7.1 6.9<x<7.1 to produce an output value of f( x ) f( x ) within the interval 7.9<f(x)<8.1. 7.9<f(x)<8.1.

We also see that we can get output values of f(x) f(x) successively closer to 8 by selecting input values closer to 7. In fact, we can obtain output values within any specified interval if we choose appropriate input values.

Figure 4 provides a visual representation of the left- and right-hand limits of the function. From the graph of f(x), f(x), we observe the output can get infinitesimally close to L=8 L=8 as x xapproaches 7 from the left and as x xapproaches 7 from the right.

To indicate the left-hand limit, we write

lim x 7 f(x)=8. lim x 7 f(x)=8.

To indicate the right-hand limit, we write

lim x 7 + f(x)=8. lim x 7 + f(x)=8.

#### A General Note: Left- and Right-Hand Limits:

The left-hand limit of a function f(x) f(x) as x xapproaches a afrom the left is equal to L, L, denoted by

lim x a f(x)=L. lim x a f(x)=L.

The values of f(x) f(x) can get as close to the limit L Las we like by taking values of x xsufficiently close to a asuch that x<a x<a and xa. xa.

The right-hand limit of a function f(x), f(x), as x xapproaches a afrom the right, is equal to L, L, denoted by

lim x a + f(x)=L. lim x a + f(x)=L.

The values of f(x) f(x) can get as close to the limit L Las we like by taking values of x xsufficiently close to a abut greater than a. a. Both a aand L Lare real numbers.

### Understanding Two-Sided Limits

In the previous example, the left-hand limit and right-hand limit as x xapproaches a aare equal. If the left- and right-hand limits are equal, we say that the function f(x) f(x) has a two-sided limit as x xapproaches a. a. More commonly, we simply refer to a two-sided limit as a limit. If the left-hand limit does not equal the right-hand limit, or if one of them does not exist, we say the limit does not exist.

#### A General Note: The Two-Sided Limit of Function as x Approaches a :

The limit of a function f(x), f(x), as x xapproaches a, a, is equal to L, L, that is,

lim xa f(x)=L lim xa f(x)=L

if and only if

lim x a f(x)= lim x a + f(x). lim x a f(x)= lim x a + f(x).

In other words, the left-hand limit of a function f(x) f(x) as x xapproaches a ais equal to the right-hand limit of the same function as x xapproaches a. a. If such a limit exists, we refer to the limit as a two-sided limit. Otherwise we say the limit does not exist.

## Finding a Limit Using a Graph

To visually determine if a limit exists as x xapproaches a, a, we observe the graph of the function when x xis very near to x=a. x=a. In Figure 5 we observe the behavior of the graph on both sides of a. a.

To determine if a left-hand limit exists, we observe the branch of the graph to the left of x=a, x=a, but near x=a. x=a. This is where x<a. x<a. We see that the outputs are getting close to some real number L Lso there is a left-hand limit.

To determine if a right-hand limit exists, observe the branch of the graph to the right of x=a, x=a, but near x=a. x=a. This is where x>a. x>a. We see that the outputs are getting close to some real number L, L, so there is a right-hand limit.

If the left-hand limit and the right-hand limit are the same, as they are in Figure 5, then we know that the function has a two-sided limit. Normally, when we refer to a “limit,” we mean a two-sided limit, unless we call it a one-sided limit.

Finally, we can look for an output value for the function f( x ) f( x ) when the input value x xis equal to a. a. The coordinate pair of the point would be ( a,f( a ) ). ( a,f( a ) ). If such a point exists, then f( a ) f( a ) has a value. If the point does not exist, as in Figure 5, then we say that f( a ) f( a ) does not exist.

### How To:

Given a function f( x ), f( x ), use a graph to find the limits and a function value as x xapproaches a. a.

1. Examine the graph to determine whether a left-hand limit exists.
2. Examine the graph to determine whether a right-hand limit exists.
3. If the two one-sided limits exist and are equal, then there is a two-sided limit—what we normally call a “limit.”
4. If there is a point at x=a, x=a, then f( a ) f( a ) is the corresponding function value.

### Example 2

#### Problem 1

##### Finding a Limit Using a Graph
1. Determine the following limits and function value for the function f fshown in Figure 6.
1. lim x 2 f(x) lim x 2 f(x)
2. lim x 2 + f(x) lim x 2 + f(x)
3. lim x2 f(x) lim x2 f(x)
4. f(2) f(2)
2. Determine the following limits and function value for the function f fshown in Figure 7.
1. lim x 2 f(x) lim x 2 f(x)
2. lim x 2 + f(x) lim x 2 + f(x)
3. lim x2 f(x) lim x2 f(x)
4. f(2) f(2)
##### Solution
1. Looking at Figure 6:
1. lim x 2 f(x)=8; lim x 2 f(x)=8; when x<2, x<2, but infinitesimally close to 2, the output values get close to y=8. y=8.
2. lim x2 + f(x)=3; lim x2 + f(x)=3; when x>2, x>2, but infinitesimally close to 2, the output values approach y=3. y=3.
3. lim x2 f(x) lim x2 f(x) does not exist because lim x2 f(x) lim x2 + f(x); lim x2 f(x) lim x2 + f(x); the left and right-hand limits are not equal.
4. f( 2 )=3 f( 2 )=3 because the graph of the function f fpasses through the point ( 2,f( 2 ) ) ( 2,f( 2 ) ) or ( 2,3 ). ( 2,3 ).
2. Looking at Figure 7:
1. lim x2 f(x)=8; lim x2 f(x)=8; when x<2 x<2 but infinitesimally close to 2, the output values approach y=8. y=8.
2. lim x2 + f(x)=8; lim x2 + f(x)=8; when x>2 x>2 but infinitesimally close to 2, the output values approach y=8. y=8.
3. lim x2 f(x)=8 lim x2 f(x)=8 because lim x2 f(x)= lim x2 + f(x)=8; lim x2 f(x)= lim x2 + f(x)=8; the left and right-hand limits are equal.
4. f( 2 )=4 f( 2 )=4 because the graph of the function f fpasses through the point ( 2,f( 2 ) ) ( 2,f( 2 ) ) or ( 2,4 ). ( 2,4 ).

### Try It:

#### Exercise 2

Using the graph of the function y=f( x ) y=f( x ) shown in Figure 8, estimate the following limits.

##### Solution

a. 0; b. 2; c. does not exist; d. 2; 2; e. 0; f. does not exist; g. 4; h. 4; i. 4

## Finding a Limit Using a Table

Creating a table is a way to determine limits using numeric information. We create a table of values in which the input values of x xapproach a afrom both sides. Then we determine if the output values get closer and closer to some real value, the limit L. L.

Let’s consider an example using the following function:

lim x5 ( x 3 125 x5 ) lim x5 ( x 3 125 x5 )

To create the table, we evaluate the function at values close to x=5. x=5. We use some input values less than 5 and some values greater than 5 as in Figure 9. The table values show that when x>5 x>5 but nearing 5, the corresponding output gets close to 75. When x>5 x>5 but nearing 5, the corresponding output also gets close to 75.

Because

lim x 5 f(x)=75= lim x 5 + f(x), lim x 5 f(x)=75= lim x 5 + f(x),

then

lim x5 f(x)=75. lim x5 f(x)=75.

Remember that f( 5 ) f( 5 ) does not exist.

### How To:

Given a function f, f, use a table to find the limit as x xapproaches a aand the value of f(a), f(a), if it exists.

1. Choose several input values that approach a afrom both the left and right. Record them in a table.
2. Evaluate the function at each input value. Record them in the table.
3. Determine if the table values indicate a left-hand limit and a right-hand limit.
4. If the left-hand and right-hand limits exist and are equal, there is a two-sided limit.
5. Replace x xwith a ato find the value of f( a ). f( a ).

### Example 3

#### Problem 1

##### Finding a Limit Using a Table

Numerically estimate the limit of the following expression by setting up a table of values on both sides of the limit.

lim x0 ( 5sin(x) 3x ) lim x0 ( 5sin(x) 3x )
##### Solution

We can estimate the value of a limit, if it exists, by evaluating the function at values near x=0. x=0. We cannot find a function value for x=0 x=0 directly because the result would have a denominator equal to 0, and thus would be undefined.

f(x)= 5sin(x) 3x f(x)= 5sin(x) 3x

We create Figure 10 by choosing several input values close to x=0, x=0, with half of them less than x=0 x=0 and half of them greater than x=0. x=0. Note that we need to be sure we are using radian mode. We evaluate the function at each input value to complete the table.

The table values indicate that when x<0 x<0 but approaching 0, the corresponding output nears 5 3 . 5 3 .

When x>0 x>0 but approaching 0, the corresponding output also nears 5 3 . 5 3 .

Because

lim x 0 f(x)= 5 3 = lim x 0 + f(x), lim x 0 f(x)= 5 3 = lim x 0 + f(x),

then

lim x0 f(x)= 5 3 . lim x0 f(x)= 5 3 .

### Q&A:

Is it possible to check our answer using a graphing utility?

Yes. We previously used a table to find a limit of 75 for the function f(x)= x 3 125 x5 f(x)= x 3 125 x5 as x xapproaches 5. To check, we graph the function on a viewing window as shown in Figure 11. A graphical check shows both branches of the graph of the function get close to the output 75 as x xnears 5. Furthermore, we can use the ‘trace’ feature of a graphing calculator. By appraoching x=5 x=5 we may numerically observe the corresponding outputs getting close to 75. 75.

### Try It:

#### Exercise 3

Numerically estimate the limit of the following function by making a table:

lim x0 ( 20sin(x) 4x ) lim x0 ( 20sin(x) 4x )
##### Solution

lim x0 ( 20sin(x) 4x )=5 lim x0 ( 20sin(x) 4x )=5

### Q&A:

Is one method for determining a limit better than the other?

No. Both methods have advantages. Graphing allows for quick inspection. Tables can be used when graphical utilities aren’t available, and they can be calculated to a higher precision than could be seen with an unaided eye inspecting a graph.

### Example 4

#### Problem 1

##### Using a Graphing Utility to Determine a Limit

With the use of a graphing utility, if possible, determine the left- and right-hand limits of the following function as x xapproaches 0. If the function has a limit as x xapproaches 0, state it. If not, discuss why there is no limit.

f(x)=3sin( π x ) f(x)=3sin( π x )
##### Solution

We can use a graphing utility to investigate the behavior of the graph close to x=0. x=0. Centering around x=0, x=0, we choose two viewing windows such that the second one is zoomed in closer to x=0 x=0 than the first one. The result would resemble Figure 12 for [2,2] [2,2] by [3,3]. [3,3].

The result would resemble Figure 13 for [−0.1,0.1] [−0.1,0.1] by [−3,3]. [−3,3].

The closer we get to 0, the greater the swings in the output values are. That is not the behavior of a function with either a left-hand limit or a right-hand limit. And if there is no left-hand limit or right-hand limit, there certainly is no limit to the function f( x ) f( x ) as x xapproaches 0.

We write

lim x 0 ( 3sin( π x ) ) does not exist. lim x 0 ( 3sin( π x ) ) does not exist.
lim x 0 + ( 3sin( π x ) ) does not exist. lim x 0 + ( 3sin( π x ) ) does not exist.
lim x0 ( 3sin( π x ) ) does not exist. lim x0 ( 3sin( π x ) ) does not exist.

### Try It:

#### Exercise 4

Numerically estimate the following limit: lim x0 ( sin( 2 x ) ). lim x0 ( sin( 2 x ) ).

does not exist

### Media:

Access these online resources for additional instruction and practice with finding limits.

## Key Concepts

• A function has a limit if the output values approach some value L Las the input values approach some quantity a. a. See Example 1.
• A shorthand notation is used to describe the limit of a function according to the form lim xa f(x)=L, lim xa f(x)=L, which indicates that as x xapproaches a, a, both from the left of x=a x=a and the right of x=a, x=a, the output value gets close to L. L.
• A function has a left-hand limit if f( x ) f( x ) approaches L Las x xapproaches a awhere x<a. x<a. A function has a right-hand limit if f( x ) f( x ) approaches L Las x xapproaches a awhere x>a. x>a.
• A two-sided limit exists if the left-hand limit and the right-hand limit of a function are the same. A function is said to have a limit if it has a two-sided limit.
• A graph provides a visual method of determining the limit of a function.
• If the function has a limit as x xapproaches a, a, the branches of the graph will approach the same y- y- coordinate near x=a x=a from the left and the right. See Example 2.
• A table can be used to determine if a function has a limit. The table should show input values that approach a afrom both directions so that the resulting output values can be evaluated. If the output values approach some number, the function has a limit. See Example 3.
• A graphing utility can also be used to find a limit. See Example 4.

## Section Exercises

### Verbal

#### Exercise 5

Explain the difference between a value at x=a x=a and the limit as x xapproaches a. a.

##### Solution

The value of the function, the output, at x=a x=a is f( a ). f( a ). When the lim xa f( x ) lim xa f( x ) is taken, the values of x xget infinitely close to a abut never equal a. a. As the values of x xapproach a afrom the left and right, the limit is the value that the function is approaching.

#### Exercise 6

Explain why we say a function does not have a limit as x xapproaches a aif, as x xapproaches a, a, the left-hand limit is not equal to the right-hand limit.

### Graphical

For the following exercises, estimate the functional values and the limits from the graph of the function f fprovided in Figure 14.

#### Exercise 7

lim x 2 f(x) lim x 2 f(x)

–4

#### Exercise 8

lim x 2 + f(x) lim x 2 + f(x)

#### Exercise 9

lim x2 f(x) lim x2 f(x)

–4

f(−2) f(−2)

#### Exercise 11

lim x 1 f(x) lim x 1 f(x)

2

#### Exercise 12

lim x 1 + f(x) lim x 1 + f(x)

#### Exercise 13

lim x1 f(x) lim x1 f(x)

does not exist

f(1) f(1)

#### Exercise 15

lim x 4 f(x) lim x 4 f(x)

4

#### Exercise 16

lim x 4 + f(x) lim x 4 + f(x)

#### Exercise 17

lim x4 f(x) lim x4 f(x)

does not exist

#### Exercise 18

f(4) f(4)

For the following exercises, draw the graph of a function from the functional values and limits provided.

#### Exercise 19

lim x 0 f(x)=2, lim x 0 + f(x)=3, lim x2 f(x)=2,f(0)=4,f(2)=1,f(3) does not exist. lim x 0 f(x)=2, lim x 0 + f(x)=3, lim x2 f(x)=2,f(0)=4,f(2)=1,f(3) does not exist.

#### Exercise 20

lim x 2 f(x)=0, lim x 2 + =2, lim x0 f(x)=3,f(2)=5,f(0) lim x 2 f(x)=0, lim x 2 + =2, lim x0 f(x)=3,f(2)=5,f(0)

#### Exercise 21

lim x 2 f(x)=2, lim x 2 + f(x)=3, lim x0 f(x)=5,f(0)=1,f(1)=0 lim x 2 f(x)=2, lim x 2 + f(x)=3, lim x0 f(x)=5,f(0)=1,f(1)=0

#### Exercise 22

lim x 3 f(x)=0, lim x 3 + f(x)=5, lim x5 f(x)=0,f(5)=4,f(3) does not exist. lim x 3 f(x)=0, lim x 3 + f(x)=5, lim x5 f(x)=0,f(5)=4,f(3) does not exist.

#### Exercise 23

lim x4 f(x)=6, lim x 6 + f(x)=1, lim x0 f(x)=5,f(4)=6,f(2)=6 lim x4 f(x)=6, lim x 6 + f(x)=1, lim x0 f(x)=5,f(4)=6,f(2)=6

#### Exercise 24

lim x3 f(x)=2, lim x 1 + f(x)=2, lim x3 f(x)=4,f(3)=0,f(0)=0 lim x3 f(x)=2, lim x 1 + f(x)=2, lim x3 f(x)=4,f(3)=0,f(0)=0

#### Exercise 25

lim xπ f(x)= π 2 , lim xπ f(x)= π 2 , lim x 1 f(x)=0,f(π)= 2 ,f(0) does not exist. lim xπ f(x)= π 2 , lim xπ f(x)= π 2 , lim x 1 f(x)=0,f(π)= 2 ,f(0) does not exist.

##### Solution

For the following exercises, use a graphing calculator to determine the limit to 5 decimal places as x xapproaches 0.

#### Exercise 26

f(x)= ( 1+x ) 1 x f(x)= ( 1+x ) 1 x

#### Exercise 27

g(x)= ( 1+x ) 2 x g(x)= ( 1+x ) 2 x

7.38906

#### Exercise 28

h(x)= ( 1+x ) 3 x h(x)= ( 1+x ) 3 x

#### Exercise 29

i(x)= ( 1+x ) 4 x i(x)= ( 1+x ) 4 x

54.59815

#### Exercise 30

j(x)= ( 1+x ) 5 x j(x)= ( 1+x ) 5 x

#### Exercise 31

Based on the pattern you observed in the exercises above, make a conjecture as to the limit of f(x)= ( 1+x ) 6 x , f(x)= ( 1+x ) 6 x , g(x)= ( 1+x ) 7 x , g(x)= ( 1+x ) 7 x , and h(x)= ( 1+x ) n x . and h(x)= ( 1+x ) n x .

##### Solution

e 6 403.428794, e 6 403.428794, e 7 1096.633158, e 7 1096.633158, e n e n

For the following exercises, use a graphing utility to find graphical evidence to determine the left- and right-hand limits of the function given as x xapproaches a. a. If the function has a limit as x xapproaches a, a, state it. If not, discuss why there is no limit.

#### Exercise 32

(x)={ | x |1, if x1 x 3 , if x=1  a=1 (x)={ | x |1, if x1 x 3 , if x=1  a=1

#### Exercise 33

(x)={ 1 x+1 , if x=2 (x+1) 2 , if x2  a=2 (x)={ 1 x+1 , if x=2 (x+1) 2 , if x2  a=2

##### Solution

lim x2 f(x)=1 lim x2 f(x)=1

### Numeric

For the following exercises, use numerical evidence to determine whether the limit exists at x=a. x=a. If not, describe the behavior of the graph of the function near x=a. x=a. Round answers to two decimal places.

#### Exercise 34

f(x)= x 2 4x 16 x 2 ;a=4 f(x)= x 2 4x 16 x 2 ;a=4

#### Exercise 35

f(x)= x 2 x6 x 2 9 ;a=3 f(x)= x 2 x6 x 2 9 ;a=3

##### Solution

lim x3 ( x 2 x6 x 2 9 )= 5 6 0.83 lim x3 ( x 2 x6 x 2 9 )= 5 6 0.83

#### Exercise 36

f(x)= x 2 6x7 x 2  7x ;a=7 f(x)= x 2 6x7 x 2  7x ;a=7

#### Exercise 37

f(x)= x 2 1 x 2 3x+2 ;a=1 f(x)= x 2 1 x 2 3x+2 ;a=1

##### Solution

lim x1 ( x 2 1 x 2 3x+2 )=2.00 lim x1 ( x 2 1 x 2 3x+2 )=2.00

#### Exercise 38

f(x)= 1 x 2 x 2 3x+2 ;a=1 f(x)= 1 x 2 x 2 3x+2 ;a=1

#### Exercise 39

f(x)= 1010 x 2 x 2 3x+2 ;a=1 f(x)= 1010 x 2 x 2 3x+2 ;a=1

##### Solution

lim x1 ( 1010 x 2 x 2 3x+2 )=20.00 lim x1 ( 1010 x 2 x 2 3x+2 )=20.00

#### Exercise 40

f(x)= x 6 x 2 5x6 ;a= 3 2 f(x)= x 6 x 2 5x6 ;a= 3 2

#### Exercise 41

f(x)= x 4 x 2 +4x+1 ;a= 1 2 f(x)= x 4 x 2 +4x+1 ;a= 1 2

##### Solution

lim x 1 2 ( x 4 x 2 +4x+1 ) lim x 1 2 ( x 4 x 2 +4x+1 ) does not exist. Function values decrease without bound as x xapproaches –0.5 from either left or right.

#### Exercise 42

f(x)= 2 x4 ; a=4 f(x)= 2 x4 ; a=4

For the following exercises, use a calculator to estimate the limit by preparing a table of values. If there is no limit, describe the behavior of the function as x xapproaches the given value.

#### Exercise 43

lim x0 7tanx 3x lim x0 7tanx 3x

##### Solution

lim x0 7tanx 3x = 7 3 lim x0 7tanx 3x = 7 3

#### Exercise 44

lim x4 x 2 x4 lim x4 x 2 x4

#### Exercise 45

lim x0 2sinx 4tanx lim x0 2sinx 4tanx

##### Solution

lim x0 2sinx 4tanx = 1 2 lim x0 2sinx 4tanx = 1 2

For the following exercises, use a graphing utility to find numerical or graphical evidence to determine the left and right-hand limits of the function given as x xapproaches a. a. If the function has a limit as x xapproaches a, a, state it. If not, discuss why there is no limit.

#### Exercise 46

lim x0 e e 1 x lim x0 e e 1 x

#### Exercise 47

lim x0 e e   1 x 2 lim x0 e e   1 x 2

##### Solution

lim x0 e e   1 x 2 =1.0 lim x0 e e   1 x 2 =1.0

#### Exercise 48

lim x0 | x | x lim x0 | x | x

#### Exercise 49

lim x1 | x+1 | x+1 lim x1 | x+1 | x+1

##### Solution

lim x 1 | x+1 | x+1 = (x+1) (x+1) =1 lim x 1 | x+1 | x+1 = (x+1) (x+1) =1 and lim x 1 + | x+1 | x+1 = (x+1) (x+1) =1; lim x 1 + | x+1 | x+1 = (x+1) (x+1) =1; since the right-hand limit does not equal the left-hand limit, lim x1 | x+1 | x+1 lim x1 | x+1 | x+1 does not exist.

#### Exercise 50

lim x5 | x5 | 5x lim x5 | x5 | 5x

#### Exercise 51

lim x1 1 ( x+1 ) 2 lim x1 1 ( x+1 ) 2

##### Solution

lim x1 1 ( x+1 ) 2 lim x1 1 ( x+1 ) 2 does not exist. The function increases without bound as x xapproaches 1 1 from either side.

#### Exercise 52

lim x1 1 ( x1 ) 3 lim x1 1 ( x1 ) 3

#### Exercise 53

lim x0 5 1 e 2 x lim x0 5 1 e 2 x

##### Solution

lim x0 5 1 e 2 x lim x0 5 1 e 2 x does not exist. Function values approach 5 from the left and approach 0 from the right.

#### Exercise 54

Use numerical and graphical evidence to compare and contrast the limits of two functions whose formulas appear similar: f(x)=| 1x x | f(x)=| 1x x | and g(x)=| 1+x x | g(x)=| 1+x x | as x xapproaches 0. Use a graphing utility, if possible, to determine the left- and right-hand limits of the functions f( x ) f( x ) and g( x ) g( x ) as x xapproaches 0. If the functions have a limit as x xapproaches 0, state it. If not, discuss why there is no limit.

### Extensions

#### Exercise 55

According to the Theory of Relativity, the mass m mof a particle depends on its velocity v v. That is

m= m o 1( v 2 / c 2 ) m= m o 1( v 2 / c 2 )

where m o m o is the mass when the particle is at rest and c cis the speed of light. Find the limit of the mass, m, m, as v vapproaches c . c .

##### Solution

Through examination of the postulates and an understanding of relativistic physics, as vc, vc, m. m. Take this one step further to the solution,

lim v c m= lim v c m o 1( v 2 / c 2 ) = lim v c m= lim v c m o 1( v 2 / c 2 ) =
(33)

#### Exercise 56

Allow the speed of light, c, c, to be equal to 1.0. If the mass, m, m, is 1, what occurs to m mas vc? vc? Using the values listed in Table 1, make a conjecture as to what the mass is as v vapproaches 1.00.

 vv mm 0.5 1.15 0.9 2.29 0.95 3.20 0.99 7.09 0.999 22.36 0.99999 223.61

## Footnotes

1. https://en.wikipedia.org/wiki/Human_height and http://en.wikipedia.org/wiki/List_of_tallest_people

## Glossary

left-hand limit:
the limit of values of f( x ) f( x ) as x xapproaches from a athe left, denoted lim x a f(x)=L. lim x a f(x)=L. The values of f(x) f(x) can get as close to the limit L Las we like by taking values of x xsufficiently close to a asuch that x<a x<a and xa. xa. Both a aand L Lare real numbers.
limit:
when it exists, the value, L, L, that the output of a function f( x ) f( x ) approaches as the input x xgets closer and closer to a abut does not equal a. a. The value of the output, f(x), f(x), can get as close to L Las we choose to make it by using input values of x xsufficiently near to x=a, x=a, but not necessarily at x=a. x=a. Both a aand L Lare real numbers, and L Lis denoted lim xa f(x)=L. lim xa f(x)=L.
right-hand limit:
the limit of values of f( x ) f( x ) as x xapproaches a afrom the right, denoted lim x a + f(x)=L. lim x a + f(x)=L. The values of f(x) f(x) can get as close to the limit L Las we like by taking values of x xsufficiently close to a awhere x>a, x>a, and xa. xa. Both a aand L Lare real numbers.
two-sided limit:
the limit of a function f(x), f(x), as x xapproaches a, a, is equal to L, L, that is, lim xa f(x)=L lim xa f(x)=L if and only if lim x a f(x)= lim x a + f(x). lim x a f(x)= lim x a + f(x).

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