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Finding Limits: Properties of Limits

Module by: First Last. E-mail the author

Summary: In this section, you will:

  • Find the limit of a sum, a difference, and a product.
  • Find the limit of a polynomial.
  • Find the limit of a power or a root.
  • Find the limit of a quotient.

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Consider the rational function

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

The function can be factored as follows:

f(x)= ( x7 ) ( x+1 ) x7 , which gives us f(x)=x+1,x7. f(x)= ( x7 ) ( x+1 ) x7 , which gives us f(x)=x+1,x7.

Does this mean the function f fis the same as the function g(x)=x+1? g(x)=x+1?

The answer is no. Function f fdoes not have x=7 x=7 in its domain, but g gdoes. Graphically, we observe there is a hole in the graph of f( x ) f( x ) at x=7, x=7, as shown in Figure 1 and no such hole in the graph of g( x ), g( x ), as shown in Figure 2.

Figure 1: The graph of function f fcontains a break at x=7 x=7 and is therefore not continuous at x=7. x=7.
Graph of an increasing function where f(x) = (x^2-6x-7)\(x-7) with a discontinuity at (7, 8)
Figure 2: The graph of function g gis continuous.
Graph of an increasing function where g(x) = x+1

So, do these two different functions also have different limits as x xapproaches 7?

Not necessarily. Remember, in determining a limit of a function as x xapproaches a, a, what matters is whether the output approaches a real number as we get close to x=a. x=a. The existence of a limit does not depend on what happens when x xequals a. a.

Look again at Figure 1 and Figure 2. Notice that in both graphs, as x xapproaches 7, the output values approach 8. This means

lim x7 f(x)= lim x7 g(x). lim x7 f(x)= lim x7 g(x).

Remember that when determining a limit, the concern is what occurs near x=a, x=a, not at x=a. x=a. In this section, we will use a variety of methods, such as rewriting functions by factoring, to evaluate the limit. These methods will give us formal verification for what we formerly accomplished by intuition.

Finding the Limit of a Sum, a Difference, and a Product

Graphing a function or exploring a table of values to determine a limit can be cumbersome and time-consuming. When possible, it is more efficient to use the properties of limits, which is a collection of theorems for finding limits.

Knowing the properties of limits allows us to compute limits directly. We can add, subtract, multiply, and divide the limits of functions as if we were performing the operations on the functions themselves to find the limit of the result. Similarly, we can find the limit of a function raised to a power by raising the limit to that power. We can also find the limit of the root of a function by taking the root of the limit. Using these operations on limits, we can find the limits of more complex functions by finding the limits of their simpler component functions.

A General Note: Properties of Limits:

Let a,k,A, a,k,A, and B Brepresent real numbers, and f fand g gbe functions, such that lim xa f(x)=A lim xa f(x)=A and lim xa g(x)=B. lim xa g(x)=B. For limits that exist and are finite, the properties of limits are summarized in Table 1

Table 1
Constant, k lim xa k=k lim xa k=k
Constant times a function lim xa [ kf(x) ]=k lim xa f(x)=kA lim xa [ kf(x) ]=k lim xa f(x)=kA
Sum of functions lim xa [ f(x)+g(x) ]= lim xa f(x)+ lim xa g(x)=A+B lim xa [ f(x)+g(x) ]= lim xa f(x)+ lim xa g(x)=A+B
Difference of functions lim xa [ f(x)g(x) ]= lim xa f(x) lim xa g(x)=AB lim xa [ f(x)g(x) ]= lim xa f(x) lim xa g(x)=AB
Product of functions lim xa [ f(x)g(x) ]= lim xa f(x) lim xa g(x)=AB lim xa [ f(x)g(x) ]= lim xa f(x) lim xa g(x)=AB
Quotient of functions lim xa f(x) g(x) = lim xa f(x) lim xa g(x) = A B ,B0 lim xa f(x) g(x) = lim xa f(x) lim xa g(x) = A B ,B0
Function raised to an exponent lim xa [f(x)] n = [ lim x f(x) ] n = A n , lim xa [f(x)] n = [ lim x f(x) ] n = A n , where n n is a positive integer
nth root of a function, where n is a positive integer lim xa f(x) n = lim xa [ f(x) ] n = A n lim xa f(x) n = lim xa [ f(x) ] n = A n
Polynomial function lim xa p(x)=p(a) lim xa p(x)=p(a)

Example 1

Problem 1

Evaluating the Limit of a Function Algebraically

Evaluate lim x3 ( 2x+5 ). lim x3 ( 2x+5 ).

Solution
lim x3 (2x+5)= lim x3 (2x)+ lim x3 (5) Sum of functions property                      = 2lim x3 (x)+ lim x3 (5) Constant times a function property                      =2(3)+5  Evaluate                      =11 lim x3 (2x+5)= lim x3 (2x)+ lim x3 (5) Sum of functions property                      = 2lim x3 (x)+ lim x3 (5) Constant times a function property                      =2(3)+5  Evaluate                      =11

Try It:

Exercise 1

Evaluate the following limit: lim x12 ( 2x+2 ). lim x12 ( 2x+2 ).

Solution

26

Finding the Limit of a Polynomial

Not all functions or their limits involve simple addition, subtraction, or multiplication. Some may include polynomials. Recall that a polynomial is an expression consisting of the sum of two or more terms, each of which consists of a constant and a variable raised to a nonnegative integral power. To find the limit of a polynomial function, we can find the limits of the individual terms of the function, and then add them together. Also, the limit of a polynomial function as x xapproaches a ais equivalent to simply evaluating the function for a a.

How To:

Given a function containing a polynomial, find its limit.

  1. Use the properties of limits to break up the polynomial into individual terms.
  2. Find the limits of the individual terms.
  3. Add the limits together.
  4. Alternatively, evaluate the function for a a.

Example 2

Problem 1

Evaluating the Limit of a Function Algebraically

Evaluate lim x3 ( 5 x 2 ). lim x3 ( 5 x 2 ).

Solution
lim x3 (5 x 2 )=5 lim x3 ( x 2 ) Constant times a function property                 =5( 3 2 ) Function raised to an exponent property                 =45 lim x3 (5 x 2 )=5 lim x3 ( x 2 ) Constant times a function property                 =5( 3 2 ) Function raised to an exponent property                 =45

Try It:

Exercise 2

Evaluate lim x4 ( x 3 5). lim x4 ( x 3 5).

Solution

59

Example 3

Problem 1

Evaluating the Limit of a Polynomial Algebraically

Evaluate lim x5 ( 2 x 3 3x+1 ). lim x5 ( 2 x 3 3x+1 ).

Solution
lim x5 (2 x 3 3x+1)= lim x5 (2 x 3 ) lim x5 (3x)+ lim x5 (1) Sum of functions                                = 2lim x5 ( x 3 ) 3lim x5 (x)+ lim x5 (1) Constant times a function                                =2( 5 3 )3(5)+1 Function raised to an exponent                                =236 Evaluate lim x5 (2 x 3 3x+1)= lim x5 (2 x 3 ) lim x5 (3x)+ lim x5 (1) Sum of functions                                = 2lim x5 ( x 3 ) 3lim x5 (x)+ lim x5 (1) Constant times a function                                =2( 5 3 )3(5)+1 Function raised to an exponent                                =236 Evaluate

Try It:

Exercise 3

Evaluate the following limit: lim x1 ( x 4 4 x 3 +5 ). lim x1 ( x 4 4 x 3 +5 ).

Solution

10

Finding the Limit of a Power or a Root

When a limit includes a power or a root, we need another property to help us evaluate it. The square of the limit of a function equals the limit of the square of the function; the same goes for higher powers. Likewise, the square root of the limit of a function equals the limit of the square root of the function; the same holds true for higher roots.

Example 4

Problem 1

Evaluating a Limit of a Power

Evaluate lim x2 ( 3x+1 ) 5 . lim x2 ( 3x+1 ) 5 .

Solution

We will take the limit of the function as x xapproaches 2 and raise the result to the 5th power.

lim x2 (3x+1) 5 = ( lim x2 (3x+1)) 5                       = (3(2)+1) 5                       = 7 5                       =16,807 lim x2 (3x+1) 5 = ( lim x2 (3x+1)) 5                       = (3(2)+1) 5                       = 7 5                       =16,807

Try It:

Exercise 4

Evaluate the following limit: lim x4 ( 10x+36 ) 3 . lim x4 ( 10x+36 ) 3 .

Solution

64 64

Q&A:

If we can’t directly apply the properties of a limit, for example in lim x2 ( x 2 +6x+8 x2 ) lim x2 ( x 2 +6x+8 x2 ), can we still determine the limit of the function as x x approaches a a?

Yes. Some functions may be algebraically rearranged so that one can evaluate the limit of a simplified equivalent form of the function.

Finding the Limit of a Quotient

Finding the limit of a function expressed as a quotient can be more complicated. We often need to rewrite the function algebraically before applying the properties of a limit. If the denominator evaluates to 0 when we apply the properties of a limit directly, we must rewrite the quotient in a different form. One approach is to write the quotient in factored form and simplify.

How To:

Given the limit of a function in quotient form, use factoring to evaluate it.

  1. Factor the numerator and denominator completely.
  2. Simplify by dividing any factors common to the numerator and denominator.
  3. Evaluate the resulting limit, remembering to use the correct domain.

Example 5

Problem 1

Evaluating the Limit of a Quotient by Factoring

Evaluate lim x2 ( x 2 6x+8 x2 ). lim x2 ( x 2 6x+8 x2 ).

Solution

Factor where possible, and simplify.

lim x2 ( x 2 6x+8 x2 )= lim x2 ( (x2)(x4) x2 ) Factor the numerator.                               = lim x2 ( (x2) (x4) x2 ) Cancel the common factors.                               = lim x2 (x4) Evaluate.                               =24=2 lim x2 ( x 2 6x+8 x2 )= lim x2 ( (x2)(x4) x2 ) Factor the numerator.                               = lim x2 ( (x2) (x4) x2 ) Cancel the common factors.                               = lim x2 (x4) Evaluate.                               =24=2
Analysis

When the limit of a rational function cannot be evaluated directly, factored forms of the numerator and denominator may simplify to a result that can be evaluated.

Notice, the function

f(x)= x 2 6x+8 x2 f(x)= x 2 6x+8 x2

is equivalent to the function

f(x)=x4,x2. f(x)=x4,x2.

Notice that the limit exists even though the function is not defined at x = 2. x = 2.

Try It:

Exercise 5

Evaluate the following limit: lim x7 ( x 2 11x+28 7x ). lim x7 ( x 2 11x+28 7x ).

Solution

3 3

Example 6

Problem 1

Evaluating the Limit of a Quotient by Finding the LCD

Evaluate lim x5 ( 1 x 1 5 x5 ). lim x5 ( 1 x 1 5 x5 ).

Solution

Find the LCD for the denominators of the two terms in the numerator, and convert both fractions to have the LCD as their denominator.

CNX_Precalc_EQ_12_02_001.jpg

Analysis

When determining the limit of a rational function that has terms added or subtracted in either the numerator or denominator, the first step is to find the common denominator of the added or subtracted terms; then, convert both terms to have that denominator, or simplify the rational function by multiplying numerator and denominator by the least common denominator. Then check to see if the resulting numerator and denominator have any common factors.

Try It:

Exercise 6

Evaluate lim x5 ( 1 5 + 1 x 10+2x ). lim x5 ( 1 5 + 1 x 10+2x ).

Solution

1 50 1 50

How To:

Given a limit of a function containing a root, use a conjugate to evaluate.

  1. If the quotient as given is not in indeterminate ( 0 0 ) ( 0 0 ) form, evaluate directly.
  2. Otherwise, rewrite the sum (or difference) of two quotients as a single quotient, using the least common denominator (LCD).
  3. If the numerator includes a root, rationalize the numerator; multiply the numerator and denominator by the conjugate of the numerator. Recall that a± b a± b are conjugates.
  4. Simplify.
  5. Evaluate the resulting limit.

Example 7

Problem 1

Evaluating a Limit Containing a Root Using a Conjugate

Evaluate lim x0 ( 25x 5 x ). lim x0 ( 25x 5 x ).

Solution

lim x0 ( 25x 5 x )= lim x0 ( ( 25x 5 ) x ( 25x +5 ) ( 25x +5 ) ) Multiply numerator and denominator by the conjugate.                                = lim x0 ( ( 25x )25 x( 25x +5 ) ) Multiply: ( 25x 5 )( 25x +5 )=( 25x )25.                                = lim x0 ( x x( 25x +5 ) ) Combine like terms.                                = lim x0 ( x x ( 25x +5 ) ) Simplify  x x =1.                                = 1 250 +5 Evaluate.                                = 1 5+5 = 1 10 lim x0 ( 25x 5 x )= lim x0 ( ( 25x 5 ) x ( 25x +5 ) ( 25x +5 ) ) Multiply numerator and denominator by the conjugate.                                = lim x0 ( ( 25x )25 x( 25x +5 ) ) Multiply: ( 25x 5 )( 25x +5 )=( 25x )25.                                = lim x0 ( x x( 25x +5 ) ) Combine like terms.                                = lim x0 ( x x ( 25x +5 ) ) Simplify  x x =1.                                = 1 250 +5 Evaluate.                                = 1 5+5 = 1 10

Analysis

When determining a limit of a function with a root as one of two terms where we cannot evaluate directly, think about multiplying the numerator and denominator by the conjugate of the terms.

Try It:

Exercise 7

Evaluate the following limit: lim h0 ( 16h 4 h ). lim h0 ( 16h 4 h ).

Solution

1 8 1 8

Example 8

Problem 1

Evaluating the Limit of a Quotient of a Function by Factoring

Evaluate lim x4 ( 4x x 2 ). lim x4 ( 4x x 2 ).

Solution

lim x4 ( 4x x 2 )= lim x4 ( (2+ x )(2 x ) x 2 ) Factor.                       = lim x4 ( (2+ x ) (2 x ) (2 x ) ) Factor −1 out of the denominator. Simplify.                       = lim x4 (2+ x ) Evaluate.                       =(2+ 4 )                       =4 lim x4 ( 4x x 2 )= lim x4 ( (2+ x )(2 x ) x 2 ) Factor.                       = lim x4 ( (2+ x ) (2 x ) (2 x ) ) Factor −1 out of the denominator. Simplify.                       = lim x4 (2+ x ) Evaluate.                       =(2+ 4 )                       =4

Analysis

Multiplying by a conjugate would expand the numerator; look instead for factors in the numerator. Four is a perfect square so that the numerator is in the form

a 2 b 2 a 2 b 2

and may be factored as

( a+b )( ab ). ( a+b )( ab ).

Try It:

Exercise 8

Evaluate the following limit: lim x3 ( x3 x 3 ). lim x3 ( x3 x 3 ).

Solution

2 3 2 3

How To:

Given a quotient with absolute values, evaluate its limit.

  1. Try factoring or finding the LCD.
  2. If the limit cannot be found, choose several values close to and on either side of the input where the function is undefined.
  3. Use the numeric evidence to estimate the limits on both sides.

Example 9

Problem 1

Evaluating the Limit of a Quotient with Absolute Values

Evaluate lim x7 | x7 | x7 . lim x7 | x7 | x7 .

Solution

The function is undefined at x=7 , x=7 ,so we will try values close to 7 from the left and the right.

Left-hand limit: | 6.97 | 6.97 = | 6.997 | 6.997 = | 6.9997 | 6.9997 =1 | 6.97 | 6.97 = | 6.997 | 6.997 = | 6.9997 | 6.9997 =1

Right-hand limit: | 7.17 | 7.17 = | 7.017 | 7.017 = | 7.0017 | 7.0017 =1 | 7.17 | 7.17 = | 7.017 | 7.017 = | 7.0017 | 7.0017 =1

Since the left- and right-hand limits are not equal, there is no limit.

Try It:

Exercise 9

Evaluate lim x 6 + 6x | x6 | . lim x 6 + 6x | x6 | .

Solution

−1 −1

Media:

Access the following online resource for additional instruction and practice with properties of limits.

Key Concepts

  • The properties of limits can be used to perform operations on the limits of functions rather than the functions themselves. See Example 1.
  • The limit of a polynomial function can be found by finding the sum of the limits of the individual terms. See Example 2 and Example 3.
  • The limit of a function that has been raised to a power equals the same power of the limit of the function. Another method is direct substitution. See Example 4.
  • The limit of the root of a function equals the corresponding root of the limit of the function.
  • One way to find the limit of a function expressed as a quotient is to write the quotient in factored form and simplify. See Example 5.
  • Another method of finding the limit of a complex fraction is to find the LCD. See Example 6.
  • A limit containing a function containing a root may be evaluated using a conjugate. See Example 7.
  • The limits of some functions expressed as quotients can be found by factoring. See Example 8.
  • One way to evaluate the limit of a quotient containing absolute values is by using numeric evidence. Setting it up piecewise can also be useful. See Example 9.

Section Exercises

Verbal

Exercise 10

Give an example of a type of function f fwhose limit, as x xapproaches a, a, is f( a ). f( a ).

Solution

If f fis a polynomial function, the limit of a polynomial function as x xapproaches a awill always be f( a ). f( a ).

Exercise 11

When direct substitution is used to evaluate the limit of a rational function as x xapproaches a aand the result is f( a )= 0 0 , f( a )= 0 0 , does this mean that the limit of f fdoes not exist?

Exercise 12

What does it mean to say the limit of f( x ) , f( x ) , as x xapproaches c ,c, is undefined?

Solution

It could mean either (1) the values of the function increase or decrease without bound as x xapproaches c, c, or (2) the left and right-hand limits are not equal.

Algebraic

For the following exercises, evaluate the limits algebraically.

Exercise 13

lim x0 ( 3 ) lim x0 ( 3 )

Exercise 14

lim x2 ( 5x x 2 1 ) lim x2 ( 5x x 2 1 )

Solution

10 3 10 3

Exercise 15

lim x2 ( x 2 5x+6 x+2 ) lim x2 ( x 2 5x+6 x+2 )

Exercise 16

lim x3 ( x 2 9 x3 ) lim x3 ( x 2 9 x3 )

Solution

6

Exercise 17

lim x1 ( x 2 2x3 x+1 ) lim x1 ( x 2 2x3 x+1 )

Exercise 18

lim x 3 2 ( 6 x 2 17x+12 2x3 ) lim x 3 2 ( 6 x 2 17x+12 2x3 )

Solution

1 2 1 2

Exercise 19

lim x 7 2 ( 8 x 2 +18x35 2x+7 ) lim x 7 2 ( 8 x 2 +18x35 2x+7 )

Exercise 20

lim x3 ( x 2 9 x5x+6 ) lim x3 ( x 2 9 x5x+6 )

Solution

6

Exercise 21

lim x3 ( 7 x 4 21 x 3 12 x 4 +108 x 2 ) lim x3 ( 7 x 4 21 x 3 12 x 4 +108 x 2 )

Exercise 22

lim x3 ( x 2 +2x3 x3 ) lim x3 ( x 2 +2x3 x3 )

Solution

does not exist

Exercise 23

lim h0 ( ( 3+h ) 3 27 h ) lim h0 ( ( 3+h ) 3 27 h )

Exercise 24

lim h0 ( ( 2h ) 3 8 h ) lim h0 ( ( 2h ) 3 8 h )

Solution

12 12

Exercise 25

lim h0 ( ( h+3 ) 2 9 h ) lim h0 ( ( h+3 ) 2 9 h )

Exercise 26

lim h0 ( 5h 5 h ) lim h0 ( 5h 5 h )

Solution

5 10 5 10

Exercise 27

lim x0 ( 3x 3 x ) lim x0 ( 3x 3 x )

Exercise 28

lim x9 ( x 2 81 3 x ) lim x9 ( x 2 81 3 x )

Solution

108 108

Exercise 29

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

Exercise 30

lim x0 ( x 1+2x 1 ) lim x0 ( x 1+2x 1 )

Solution

1

Exercise 31

lim x 1 2 ( x 2 1 4 2x1 ) lim x 1 2 ( x 2 1 4 2x1 )

Exercise 32

lim x4 ( x 3 64 x 2 16 ) lim x4 ( x 3 64 x 2 16 )

Solution

6

Exercise 33

lim x 2 ( |x2| x2 ) lim x 2 ( |x2| x2 )

Exercise 34

lim x 2 + ( | x2 | x2 ) lim x 2 + ( | x2 | x2 )

Solution

1

Exercise 35

lim x2 ( | x2 | x2 ) lim x2 ( | x2 | x2 )

Exercise 36

lim x 4 ( | x4 | 4x ) lim x 4 ( | x4 | 4x )

Solution

1

Exercise 37

lim x 4 + ( | x4 | 4x ) lim x 4 + ( | x4 | 4x )

Exercise 38

lim x4 ( | x4 | 4x ) lim x4 ( | x4 | 4x )

Solution

does not exist

Exercise 39

lim x2 ( 8+6x x 2 x2 ) lim x2 ( 8+6x x 2 x2 )

For the following exercise, use the given information to evaluate the limits: lim xc f(x)=3, lim xc f(x)=3, lim xc g( x )=5 lim xc g( x )=5

Exercise 40

lim xc [ 2f(x)+ g(x) ] lim xc [ 2f(x)+ g(x) ]

Solution

6+ 5 6+ 5

Exercise 41

lim xc [ 3f(x)+ g(x) ] lim xc [ 3f(x)+ g(x) ]

Exercise 42

lim xc f(x) g(x) lim xc f(x) g(x)

Solution

3 5 3 5

For the following exercises, evaluate the following limits.

Exercise 43

lim x2 cos( πx ) lim x2 cos( πx )

Exercise 44

lim x2 sin( πx ) lim x2 sin( πx )

Solution

0

Exercise 45

lim x2 sin( π x ) lim x2 sin( π x )

Exercise 46

f(x)={ 2 x 2 +2x+1, x0 x3,  x>0 lim x 0 + f(x) f(x)={ 2 x 2 +2x+1, x0 x3,  x>0 lim x 0 + f(x)

Solution

3 3

Exercise 47

f(x)={ 2 x 2 +2x+1, x0 x3,  x>0 lim x 0 f(x) f(x)={ 2 x 2 +2x+1, x0 x3,  x>0 lim x 0 f(x)

Exercise 48

f(x)={ 2 x 2 +2x+1, x0 x3,  x>0 lim x0 f(x) f(x)={ 2 x 2 +2x+1, x0 x3,  x>0 lim x0 f(x)

Solution

does not exist; right-hand limit is not the same as the left-hand limit.

Exercise 49

lim x4 x+5 3 x4 lim x4 x+5 3 x4

Exercise 50

lim x 2 + (2x〚x〛) lim x 2 + (2x〚x〛)

Solution

2

Exercise 51

lim x2 x+7 3 x 2 x2 lim x2 x+7 3 x 2 x2

Exercise 52

lim x 3 + x 2 x 2 9 lim x 3 + x 2 x 2 9

Solution

Limit does not exist; limit approaches infinity.

For the following exercises, find the average rate of change f(x+h)f(x) h . f(x+h)f(x) h .

Exercise 53

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

Exercise 54

f(x)=2 x 2 1 f(x)=2 x 2 1

Solution

4x+2h 4x+2h

Exercise 55

f(x)= x 2 +3x+4 f(x)= x 2 +3x+4

Exercise 56

f(x)= x 2 +4x100 f(x)= x 2 +4x100

Solution

2x+h+4 2x+h+4

Exercise 57

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

Exercise 58

f(x)=cos(x) f(x)=cos(x)

Solution

cos(x+h)cos(x) h cos(x+h)cos(x) h

Exercise 59

f(x)=2 x 3 4x f(x)=2 x 3 4x

Exercise 60

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

Solution

1 x(x+h) 1 x(x+h)

Exercise 61

f(x)= 1 x 2 f(x)= 1 x 2

Exercise 62

f(x)= x f(x)= x

Solution

1 x+h + x 1 x+h + x

Graphical

Exercise 63

Find an equation that could be represented by Figure 3.

Figure 3
Graph of increasing function with a removable discontinuity at (2, 3).

Exercise 64

Find an equation that could be represented by Figure 4.

Figure 4
Graph of increasing function with a removable discontinuity at (-3, -1).
Solution

f( x )= x 2 +5x+6 x+3 f( x )= x 2 +5x+6 x+3

For the following exercises, refer to Figure 5.

Figure 5
Graph of increasing function from zero to positive infinity.

Exercise 65

What is the right-hand limit of the function as x xapproaches 0?

Exercise 66

What is the left-hand limit of the function as x xapproaches 0?

Solution

does not exist

Real-World Applications

Exercise 67

The position function s(t)=16 t 2 +144t s(t)=16 t 2 +144t gives the position of a projectile as a function of time. Find the average velocity (average rate of change) on the interval [ 1,2 ] [ 1,2 ] .

Exercise 68

The height of a projectile is given by s(t)=64 t 2 +192t s(t)=64 t 2 +192t Find the average rate of change of the height from t=1 t=1 second to t=1.5 t=1.5 seconds.

Solution

52

Exercise 69

The amount of money in an account after t tyears compounded continuously at 4.25% interest is given by the formula A= A 0 e 0.0425t , A= A 0 e 0.0425t ,where A 0 A 0 is the initial amount invested. Find the average rate of change of the balance of the account from t=1 t=1 year to t=2 t=2 years if the initial amount invested is $1,000.00.

Glossary

properties of limits:
a collection of theorems for finding limits of functions by performing mathematical operations on the limits

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