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12.3: Finding Limits - Properties of Limits

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    Learning Objectives

    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.

    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 f is the same as the function g(x)=x+1? g(x)=x+1?

    The answer is no. Function f f does not have x=7 x=7 in its domain, but g g does. 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.

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

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

    Not necessarily. Remember, in determining a limit of a function as x x approaches 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 x equals a. a.

    Look again at Figure 1 and Figure 2. Notice that in both graphs, as x x approaches 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.

    Properties of Limits

    Let a,k,A, a,k,A, and B B represent real numbers, and f f and g g be 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

    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 xa f(x) ] n = A n , lim xa [f(x)] n = [ lim xa 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)
    Table 1

    Example 1

    Evaluating the Limit of a Function Algebraically

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

    Answer

    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 #1

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

    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 x approaches a a is 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

    Evaluating the Limit of a Function Algebraically

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

    Answer

    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 #2

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

    Example 3

    Evaluating the Limit of a Polynomial Algebraically

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

    Answer

    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 #3

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

    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

    Evaluating a Limit of a Power

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

    Answer

    We will take the limit of the function as x x approaches 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 #4

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

    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

    Evaluating the Limit of a Quotient by Factoring

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

    Answer

    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 #5

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

    Example 6

    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 ).

    Answer

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

    d8421804b5f226372bf131a17acb812f6a433f67

    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 #6

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

    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

    Evaluating a Limit Containing a Root Using a Conjugate

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

    Answer

    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 #7

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

    Example 8

    Evaluating the Limit of a Quotient of a Function by Factoring

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

    Answer

    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 #8

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

    Evaluating the Limit of a Quotient with Absolute Values

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

    Answer

    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 #9

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

    Media

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

    12.2 Section Exercises

    Verbal

    1.

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

    2.

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

    3.

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

    Algebraic

    For the following exercises, evaluate the limits algebraically.

    4.

    lim x0 ( 3 ) lim x0 ( 3 )

    5.

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

    6.

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

    7.

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

    8.

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

    9.

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

    10.

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

    11.

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

    12.

    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 )

    13.

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

    14.

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

    15.

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

    16.

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

    17.

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

    18.

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

    19.

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

    20.

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

    21.

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

    22.

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

    23.

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

    24.

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

    25.

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

    26.

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

    27.

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

    28.

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

    29.

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

    30.

    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 .

    31.

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

    32.

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

    33.

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

    For the following exercises, evaluate the following limits.

    34.

    lim x2 cos( πx ) lim x2 cos( πx )

    35.

    lim x2 sin( πx ) lim x2 sin( πx )

    36.

    lim x2 sin( π x ) lim x2 sin( π x )

    37.

    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)

    38.

    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)

    39.

    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)

    40.

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

    41.

    lim x 2 + (2x[x]) lim x 2 + (2x[x])

    42.

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

    43.

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

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

    44.

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

    45.

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

    46.

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

    47.

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

    48.

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

    49.

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

    50.

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

    51.

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

    52.

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

    53.

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

    Graphical

    54.

    Find an equation that could be represented by Figure 3.

    Graph of increasing function with a removable discontinuity at (2, 3).
    Figure 3
    55.

    Find an equation that could be represented by Figure 4.

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

    For the following exercises, refer to Figure 5.

    Graph of increasing function from zero to positive infinity.
    Figure 5
    56.

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

    57.

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

    Real-World Applications

    58.

    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 ] .

    59.

    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.

    60.

    The amount of money in an account after t t years 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.


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