# More Properties of Limits: Order of Operations

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1 math 30 day 5: calculating its 6 More Proerties of Limits: Order of Oerations THEOREM 45 (Order of Oerations, Continued) Assume that!a f () L and that m and n are ositive integers Then 5 (Power)!a [ f ()] n [!a f ()] n L n 6 (Fractional Power) Assume that m n is reduced Then h i n/m [ f!a ()]n/m f () L n/m,!a rovided that f () 0 for near a if m is even EXAMPLE 44 Determine!3 (4 ) 5 Indicate which it roerties were used at each ste SOLUTION (4!3 0)5 Powers [ 4 0] 5 Diff [ 4 0] 5!3!3!3 EXAMPLE 45 Determine! each ste Const Mult [4!3!3 0] 5 Thm 5,5 [4(3) 0] 5 () 5 3 Indicate which it roerties were used at SOLUTION Notice that ( ) / is a fractional ower function In n the language of Theorem 45, m is reduced and m is even Near, f () is ositive So Theorem 45 alies and we may calculate the it as! / Frac Pow Poly! (3) / 3 EXAMPLE 46 Determine! 3 ( + ) 4/3 Indicate which it roerties were used at each ste SOLUTION ( + ) 4/3 is a fractional ower function with m n 4 3 which is reduced and m 3 is odd Near, f () + is ositive So Theorem 45 alies and we may calculate the it as! 3 / Frac Pow Sum, Prod +! 3 + (8 3 + ) / 4 EXAMPLE 47 Determine! 3 ( 5) 3/4 SOLUTION ( 5) 3/4 is a fractional ower function In the language of Theorem 45, m n 3 4 is reduced and m 4 is even Near 3, 5 is negative Since ( 5) 3/4 is not even defined near 3, this it does not eist We now look at some secial cases of its with familiar functions THEOREM 46 (Secial Functions) Let n be a ositive integer and c be any constant 7 (Monomials)!a c n ca n 8 (Polynomials) If () c n n + c n n + + c + c 0 is a degree n olynomial, then () (a)!a

2 math 30 day 5: calculating its 7 9 (Rational Functions) If r() () is a rational function, then for any oint a in the q() domain of r() r() r(a)!a Theorem 46 says that the it of olynomial or rational function as! a is the same as the value of the function at a This is not true of all its For sin eamle, we saw that, yet we can t even ut 0 into this function!!0 Those secial or nice functions where!a f () f (a) are called continuous at a We will eamine them in deth in a few days For the moment we can say that olynomials are continuous everywhere and rational functions are continuous at every oint in their domains Proof Let s see how it roerties 7 through 9 follow from the revious roerties of its To rove the monomial roerty, use!a cn Const Mult c[ n ] Powers n Thm 5 c[ ] ca n!a!a To rove the olynomial roerty, since () c n n + c n n + + c + c 0 is a degree n olynomial, then () [c n n + c n n + + c + c 0 ]!a!a Sum!a c n n +!a c n n + +!a c +!a c 0 ] Monomial, Thm 5 c n a n + c n a n + + c a + c 0 (a) The rational function result is simler, still If r() () is a rational function, q() then () and q() are olynomials So for any oint a in the domain of r() (ie, q(a) 6 0), () Quotient r()!a () Polynomial (a)!a!a q()!a q() q(a) r(a) EXAMPLE 48 To see how these last results greatly simlify certain it calculations, 4 + let s determine! 3 + SOLUTION Since we have a rational function and the denominator is not 0 at, we see that 4 + Rational 4() + ()! 3 + 3()+ 7 That was easy! Several Cautions Most of the its we will encounter this term will not be so easy to determine While we will use the roerties we ve develoed and others below, most its will start off in the indeterminate form 0 Tyically we will need to 0 carry out some sort of algebraic maniulation to get the it in a form where the basic roerties aly For eamle, while!5 5 5

3 math 30 day 5: calculating its 8 is a rational function, roerty 8 above does not aly to the calculation of the it since 5 (the number is aroaching) is not in the domain of the function Consequently, some algebraic maniulation (in this case factoring) is required 5!5 5 ( 5)( + 5) + 5 Poly 0!5 5!5 There are two additional things to notice The first is mathematical grammar We continue to use the it symbol u until the actual numerical evaluation takes lace Writing something such as the following is simly wrong:!5 ((((((((((((((((((((h hhhhhhhhhhhhhhhhhhh 5 ( 5)( + 5) Among other things, the function + 5 is not the same as the constant 0 An even worse calculation to write is X XX X5 X XX! X X or even ((((((((((((((h!5 hhhhhhhhhhhhh Undefined The eression 0 0 is indeed not defined (and is certainly not equal to ) However, the it is indeterminate Near (but not equal to) 5, the fraction is not yet 0 0 You need to do more work to determine the it The work may involve factoring or other algebraic methods to simlify the eression so that we can more easily see what it is aroaching Another thing to notice is that 5 5 and + 5 are the same function as long as 6 5 where the first function is not defined but the second is However, we are interested in a it as! 5 so remember that this involves being close to, but 5 not equal to, 5 Consequently and + 5 are indeed the same!!5 5!5 43 One-sided Limits We have now stated a number of roerties for its All of these roerties also hold for one-sided its, as well, with a slight modification for fractional owers THEOREM 47 (One-sided Limit Proerties) Limit roerties through 9 (the constant multile, sum, difference, roduct, quotient, integer ower, olynomial, and rational function rules) continue to hold for one-sided its with the following modification for fractional owers Assume that m and n are ositive integers and that m n is reduced Then ale n/m (a)!a +[ f ()]n/m f () rovided that f () 0 for near a with > a if m!a + is even ale n/m (b) [ f!a ()]n/m f () rovided that f () 0 for near a with > a if m!a is even The net few eamles illustrate the use of it roerties with iecewise functions ( 3 +, if < EXAMPLE 49 Let f () Determine the following its if they if eist (a)! f () (b) f () (c) f () (d) f ()! +!!0

4 math 30 day 5: calculating its 9 SOLUTION We must be careful to use the correct definition of f for each it (a) As! from the left, is less than so f () 3 + there Thus! f () <! 3 + Poly 3( ) + 3 (b) As! from the right, is greater than so f () Thus f () > Root ! +! + (c) To determine f () we comare the one sided its Since f () 6!! + f (), we conclude that f () DNE!! (d) To determine f () we see that the values of near 0 are less than So!0 f () 3 + there So f () < 3 + Poly!0!0 We don t need to use the other definition for f since it does not aly to values of near 0 8 >< 3, if ale EXAMPLE 40 Let f () +, if < ale 5 Determine the following its if >: + if > 5 they eist (a) (d)!!5 f () (b) f ()! + (c)! f () (e) f ()!5 + (f )!5 SOLUTION We must be careful to use the correct definition of f for each it Note how we choose the function! (a) (b)! f () <! 3 Poly < ale 5 f ()! +! + Poly + (c) Since f ()! +! (d) (e)!5 f () < ale 5!5 + Poly 6 f () > 5!5 +!5 + + (f ) Since f () 6!5 +!5 f (), we conclude that! f () 5 6 f (), we conclude that!5 f () DNE 44 Most Limits Are Not Simle Let s return to the original motivation for calculating its We were interested in finding the sloe of a curve and this led to looking at its that have the form!a f () f (a) a Assuming that f is continuous, this it cannot be evaluated by any of the basic it roerties since the denominator is aroaching 0 More secifically, as! a, this difference quotient has the indeterminate form 0 0 To evaluate this it we must do more work Let s look at an EXAMPLE 4 Let f () Determine the sloe of this curve right at

5 math 30 day 5: calculating its 0 SOLUTION To find the sloe of a curve we must evaluate the difference quotient f () f (4) ( 3 + ) 5!4 4!4 4 Though this is a rational function, the it roerties do not aly since the denominator is 0 at 4, and so is the numerator (check it!) Instead, we must do more work f () f (4) 3 4 ( 4)( + ) + Poly 5!4 4!4 4!4 4!4 Only at the very last ste were we able to use a it roerty The Indeterminate Form 0 0 Many of the most imortant its we will see in the course have the indeterminate form 0 0 as in the revious eamle To evaluate such its, if they eist, requires more work tyically of the following tye factoring using conjugates simlifying making use of known its Let s look at some eamles of each Recall that if a > 0, then a + b and a b are called conjugates Notice that ( a + b)( a b) a b There is no middle term EXAMPLE 4 (Factoring) Factoring is one of the most critical tools in evaluating the sorts of its that arise in elementary calculus Evaluate! + 8 SOLUTION Notice that this it has the indeterminate form 0 0 Factoring is the key %0 ( )( )! + 8 &0! ( + 4)( ) ( ) Rational! Only at the very last ste were we able to use a it roerty EXAMPLE 43 (Factoring) Evaluate! 3 + SOLUTION This it has the indeterminate form 0 0 Factoring is the key %0! 3 + & 0 ( + 6)( + ) ( + 6)! ( + )! 4 4 Only at the very last ste were we able to use a it roerty EXAMPLE 44 (Conjugates) Evaluate!4 8) SOLUTION Notice that this it has the indeterminate form 0 0 Let s see how conjugates hel!4 % 0 ( 4) &0! !4 ( 8)( + )!4 4 (( 4)( + )!4 ( + ) EXAMPLE 45 (Conjugates) Here s another: Evaluate! Root

6 math 30 day 5: calculating its SOLUTION Notice that this it has the indeterminate form 0 0 Use conjugates again! % 0 &0! ( + 4) 6! ( )( )! ( )( )! Root 6 EXAMPLE 46 (Conjugates) Evaluate! + 3 SOLUTION This it has the indeterminate form 0 0! % &0! ) ( )(! ( + 3) 4 ( )( + )( )!! ( + )( ) Prod, Root ( + ) 8 EXAMPLE 47 (Simlification) Sometimes its, like this net one, involve comound + fractions One method of attack is to carefully simlify them Evaluate! SOLUTION Notice that this it has the indeterminate form 0 0 Use common denominators to simlify! + % 0 &0! EXAMPLE 48 (Simlification) Evaluate! ( ) (+)( ) 4 %0! ( + )( )( ) &0 Rational! ( + )( ) 3 + SOLUTION Notice that this it has the indeterminate form 0 0 Use common denominators to simlify! + % 0 &0! (+) (+ %0! ( + )( ) &0! ( + ) EXAMPLE 49 (Simlification) Evaluate h!0 +h h SOLUTION Notice that this it has the indeterminate form 0 0 h!0 +h h &0 % 0 h!0 (+h) (+h)() h h h!0 ( + h)()(h) h!0 ( + h)()

7 math 30 day 5: calculating its EXAMPLE 40 (Simlification) Evaluate! 4 +7 SOLUTION Notice that this it has the indeterminate form 0 0 Use common denominators to simlify! 4 +7! 8 ( +7) ( +7)! ( +7)! ( )( + ) ( + 7)( ) ( + )! ( + 7) Practice Problems EXAMPLE 4 (Simlification) Evaluate!4 4 SOLUTION Notice that this it has the indeterminate form 0 0 Use factoring to simlify this rational function ( 4)( ) Linear!4 4!4 4!4 + EXAMPLE 4 (Simlification) Evaluate! 4 SOLUTION Notice that this it has the indeterminate form 0 0 Use factoring to simlify this rational function! + 4! + ( )( + )! 3 0 EXAMPLE 43 (Simlification) Evaluate!5 5 4 SOLUTION Notice that this it has the indeterminate form 0 0 Use factoring to simlify this rational function 3 0 ( + )( 5)!5 5!5 ( 5)( + 5) + 7! EXAMPLE 44 (Simlification) Evaluate! SOLUTION Notice that this it has the indeterminate form 0 0 Use factoring to simlify this rational function! 3 5 6! EXAMPLE 45 (Simlification) Evaluate!0 ( ) ( + )( 6) ( )( + )! ( + )( 6)! ( ) 6 SOLUTION Notice that this it has the indeterminate form 0 0 Use common denominators to simlify! !0 3 3(+) !0 6!0 ( + ) 6 6! +

8 math 30 day 5: calculating its 3 EXAMPLE 46 (Simlification) Evaluate! 4 SOLUTION Notice that this it has the indeterminate form 0 0 Use common denominators to simlify! 4! 4 4! EXAMPLE 47 (Simlification) Evaluate! 4 (4 )( ) ( )( + ) +! (4 )( )! SOLUTION Notice that this it has the indeterminate form 0 0 Use common denominators to simlify! + ( +)!! ( +)! EXAMPLE 48 (Simlification) Evaluate! ( + )( ) ( )( + )! ( + )( ) ( + )! ( + ) 4 SOLUTION Notice that this it has the indeterminate form 0 0 Use conjugates to simlify!! + +! ( )( + )! 3 EXAMPLE 49 (Simlification) Evaluate!3 + + Root SOLUTION Notice that this it has the indeterminate form 0 0 Use conjugates to simlify! ! EXAMPLE 430 (Simlification) Evaluate!0 4 ( 3)( + + )! ) ( 3)(!3 3 Root + + 4!3 SOLUTION Notice that this it has the indeterminate form 0 0 Use conjugates to simlify (4 ) 4!0!0 4 +!0 ( )( 4 + )!0 ( )( 4 + )!0 ( )( 4 + )!0 ( )( 4 + ) Root, Prod 4 4

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