APPLICATIONS OF DIFFERENTIATION Indeterminate Forms and L Hospital s Rule In this section, we will learn: How to evaluate functions whose values cannot be found at certain points.
INDETERMINATE FORM TYPE 0/0 In general, if we have a limit of the form f( ) lim a g ( ) where both f() 0 and g() 0 as a, then this limit may or may not eist. 0 It is called an indeterminate form of type. 0 We met some limits of this type in Chapter 2.
INDETERMINATE FORMS For rational functions, we can cancel common factors: 2 ( 1) lim lim 1 2 1 ( 1)( 1) 1 1 lim 1 1 2
INDETERMINATE FORMS We used a geometric argument to show that: sin lim 1 0
INDETERMINATE FORMS However, these methods do not work for limits such as Epression 1. Hence, in this section, we introduce a systematic method, known as l Hospital s Rule, for the evaluation of indeterminate forms.
INDETERMINATE FORMS Epression 2 Another situation in which a limit is not obvious occurs when we look for a horizontal asymptote of F and need to evaluate the limit ln lim 1
INDETERMINATE FORMS It isn t obvious how to evaluate this limit because both numerator and denominator become large as. There is a struggle between the two. If the numerator wins, the limit will be. If the denominator wins, the answer will be 0. Alternatively, there may be some compromise the answer may be some finite positive number.
INDETERMINATE FORM TYPE / In general, if we have a limit of the form f( ) lim a g ( ) where both f() (or - ) and g() (or - ), then the limit may or may not eist. It is called an indeterminate form of type /.
INDETERMINATE FORMS We saw in Section 4.4 that this type of limit can be evaluated for certain functions including rational functions by dividing the numerator and denominator by the highest power of that occurs in the denominator. For instance, lim 1 1 1 1 0 1 lim 1 2 0 2 2 2 2 2 2 2 1
INDETERMINATE FORMS This method, though, does not work for limits such as Epression 2. However, L Hospital s Rule also applies to this type of indeterminate form.
L HOSPITAL S RULE Suppose f and g are differentiable and g () 0 on an open interval I that contains a (ecept possibly at a). Suppose or that lim f ( ) 0 and lim g( ) 0 a a lim f ( ) and lim g( ) a a In other words, we have an indeterminate form 0 of type or /. 0
L HOSPITAL S RULE Then, lim f ( ) f '( ) lim g( ) g '( ) a a if the limit on the right eists (or is or - ).
NOTE 1 L Hospital s Rule says that the limit of a quotient of functions is equal to the limit of the quotient of their derivatives provided that the given conditions are satisfied. It is especially important to verify the conditions regarding the limits of f and g before using the rule.
NOTE 2 The rule is also valid for one-sided limits and for limits at infinity or negative infinity. That is, a can be replaced by any of the symbols a +, a -,, or -.
NOTE 3 For the special case in which f(a) = g(a) = 0, f and g are continuous, and g (a) 0, it is easy to see why the rule is true.
NOTE 3 In fact, using the alternative form of the definition of a derivative, we have: lim f ( ) f ( a) f ( ) f ( a) lim f '( ) f '( a) a a lim a g '( ) g '( a) g( ) g( a) g( ) g( a) lim a a a f ( ) f ( a) lim a g ( ) g ( a ) a a f( ) lim a g ( )
NOTE 3 It is more difficult to prove the general version of l Hospital s Rule.
L HOSPITAL S RULE Find ln lim 1 1 Eample 1 lim ln ln1 0 lim( 1) 0 and 1 1 Thus, we can apply l Hospital s Rule: d (ln ) ln 1/ 1 lim lim d lim lim 1 1 1 1 d 1 1 1 ( 1) d
L HOSPITAL S RULE Calculate lim e 2 Eample 2 We have lim e and lim 2 So, l Hospital s Rule gives: d ( e ) e e lim lim d lim 2 d 2 2 ( ) d
L HOSPITAL S RULE Eample 2 As e and 2 as, the limit on the right side is also indeterminate. However, a second application of l Hospital s Rule gives: e e e lim lim lim 2 2 2
L HOSPITAL S RULE Calculate lim ln 3 Eample 3 3 ln 1/ lim lim 3 As ln and as, l Hospital s Rule applies: Notice that the limit on the right side 0 is now indeterminate of type. 0 1 3 2/3
L HOSPITAL S RULE However, instead of applying the rule a second time as we did in Eample 2, we simplify the epression and see that a second application is unnecessary: ln 1/ 3 lim lim lim 0 3 1 2 / 3 3 3 Eample 3
L HOSPITAL S RULE tan lim 0 3 Find Eample 4 Noting that both tan 0 and 3 0 as 0, we use l Hospital s Rule: 2 tan sec 1 lim lim 3 0 3 0 2
L HOSPITAL S RULE Eample 4 As the limit on the right side is still 0 indeterminate of type, we apply the rule 0 again: 2 2 sec 1 2sec tan lim lim 0 3 2 0 6
L HOSPITAL S RULE Eample 4 2 Since limsec 1, we simplify the 0 calculation by writing: 2 2sec tan 1 2 tan lim lim sec lim 0 6 3 0 0 1 tan lim 3 0
L HOSPITAL S RULE Eample 4 We can evaluate this last limit either by using l Hospital s Rule a third time or by writing tan as (sin )/(cos ) and making use of our knowledge of trigonometric limits.
L HOSPITAL S RULE Eample 4 Putting together all the steps, we get: tan sec 1 lim lim 0 3 0 3 2 lim 0 2 2sec 2 tan 6 1 tan 1 sec 1 lim lim 3 0 3 0 1 3 2
L HOSPITAL S RULE Find lim 1 sin cos Eample 5 If we blindly attempted to use l-hospital s rule, we would get: lim sin cos lim 1 cos sin
L HOSPITAL S RULE Eample 5 This is wrong. Although the numerator sin 0 as π -, notice that the denominator (1 - cos ) does not approach 0. So, the rule can t be applied here.
L HOSPITAL S RULE Eample 5 The required limit is, in fact, easy to find because the function is continuous at π and the denominator is nonzero there: sin sin 0 lim 0 1 cos 1 cos 1 ( 1)
L HOSPITAL S RULE The eample shows what can go wrong if you use the rule without thinking. Other limits can be found using the rule, but are more easily found by other methods. See Eamples 3 and 5 in Section 2.3, Eample 3 in Section 4.4, and the discussion at the beginning of this section.
L HOSPITAL S RULE So, when evaluating any limit, you should consider other methods before using l Hospital s Rule.
INDETERMINATE PRODUCTS If and (or - ), lim f ( ) 0 lim g ( ) a a then it isn t clear what the value of lim f ( ) g ( ), if any, will be. a
INDETERMINATE PRODUCTS There is a struggle between f and g. If f wins, the answer will be 0. If g wins, the answer will be (or - ). Alternatively, there may be a compromise where the answer is a finite nonzero number.
INDETERMINATE FORM TYPE 0. This kind of limit is called an indeterminate form of type 0.. We can deal with it by writing the product fg as a quotient: fg f g or fg 1/ g 1/ f This converts the given limit into an indeterminate form of type or /, so that we can use l Hospital s Rule. 0 0
INDETERMINATE PRODUCTS Eample 6 Evaluate lim ln 0 The given limit is indeterminate because, as 0 +, the first factor () approaches 0, whereas the second factor (ln ) approaches -.
INDETERMINATE PRODUCTS Eample 6 Writing = 1/(1/), we have 1/ as 0 +. So, l Hospital s Rule gives: ln 1/ lim ln lim lim 2 1/ 1/ lim( ) 0 0 0 0 0
INDETERMINATE PRODUCTS Note In solving the eample, another possible option would have been to write: lim ln lim 0 0 1/ ln This gives an indeterminate form of the type 0/0. However, if we apply l Hospital s Rule, we get a more complicated epression than the one we started with.
INDETERMINATE PRODUCTS Note In general, when we rewrite an indeterminate product, we try to choose the option that leads to the simpler limit.
INDETERMINATE PRODUCTS Eample 7 Use l Hospital s Rule to sketch the graph of f() = e. Since both and e become large as, we have: lim e
INDETERMINATE PRODUCTS Eample 7 However, as -, e 0. So, we have an indeterminate product that requires the use of l Hospital s Rule: 1 lim e lim lim e e lim e 0 Thus, the -ais is a horizontal asymptote.
INDETERMINATE PRODUCTS Eample 7 We use the methods of Chapter 4 to gather other information concerning the graph. The derivative is: f () = e + e = ( + 1)e Since e is always positive, we see that f () > 0 when + 1 > 0, and f () < 0 when + 1 < 0. So, f is increasing on (-1, ) and decreasing on (-, -1).
INDETERMINATE PRODUCTS Eample 7 f (-1) = 0 and f changes from negative to positive at = -1. Hence, f(-1) = -e -1 is a local (and absolute) minimum.
INDETERMINATE PRODUCTS Eample 7 The second derivative is: f '' 1 e e 2 e Since f () > 0 if > -2 and f () < 0 if < -2, f is concave upward on (-2, ) and concave downward on (-, -2). The inflection point is (-2, -2e -2 ).
INDETERMINATE PRODUCTS Eample 7 We use this information to sketch the curve. Thomson Higher Education
INDETERMINATE FORM TYPE - If and, then lim f ( ) a the limit is called an indeterminate form of type -. lim g ( ) a lim[ f ( ) g ( )] a
INDETERMINATE DIFFERENCES Again, there is a contest between f and g. Will the answer be (f wins)? Will it be - (g wins)? Will they compromise on a finite number?
INDETERMINATE DIFFERENCES To find out, we try to convert the difference into a quotient (for instance, by using a common denominator, rationalization, or factoring out a common factor) so that we have an indeterminate form of type or /. 0 0
INDETERMINATE DIFFERENCES Eample 8 Compute lim (sec tan ) ( / 2) First, notice that sec and tan as (π/2) -. So, the limit is indeterminate.
INDETERMINATE DIFFERENCES Here, we use a common denominator: lim (sec tan ) lim ( / 2) ( / 2) cos cos ( / 2) ( / 2) 1 sin 1 sin lim cos cos lim 0 sin Note that the use of l Hospital s Rule is justified because 1 sin 0 and cos 0 as (π/2) -. Eample 8
INDETERMINATE POWERS Several indeterminate forms arise from the limit ( ) lim[ f ( )] g a 1. lim f ( ) 0 and lim g( ) 0 type0 a a 2. lim f ( ) and lim g( ) 0 type a a 3. lim f ( ) 1 and lim g( ) type1 a a 0 0
INDETERMINATE POWERS Each of these three cases can be treated in either of two ways. Taking the natural logarithm: g( ) Let y [ f ( )], then ln y g( )ln f ( ) Writing the function as an eponential: [ f ( )] g e g f ( ) ( )ln ( )
INDETERMINATE POWERS Recall that both these methods were used in differentiating such functions. In either method, we are led to the indeterminate product g() ln f(), which is of type 0..
INDETERMINATE POWERS Eample 9 Calculate lim(1 sin 4 ) 0 cot First, notice that, as 0 +, we have 1 + sin 4 1 and cot. So, the given limit is indeterminate.
INDETERMINATE POWERS Let y = (1 + sin 4) cot Eample 9 Then, ln y = ln[(1 + sin 4) cot ] = cot ln(1 + sin 4)
INDETERMINATE POWERS So, l Hospital s Rule gives: Eample 9 lim ln y lim 0 0 0 ln(1 sin 4 ) tan 4cos 4 1 sin 4 lim 4 sec 2
INDETERMINATE POWERS Eample 9 So far, we have computed the limit of ln y. However, what we want is the limit of y. To find this, we use the fact that y = e ln y : cot lim(1 sin 4 ) lim 0 0 lim 0 y e e ln y 4
INDETERMINATE POWERS Find lim 0 Eample 10 Notice that this limit is indeterminate since 0 = 0 for any > 0 but 0 = 1 for any 0.
INDETERMINATE POWERS Eample 10 We could proceed as in Eample 9 or by writing the function as an eponential: = (e ln ) = e ln In Eample 6, we used l Hospital s Rule to show that lim ln 0 0 Therefore, ln 0 lim lim e e 1 0 0
CAUCHY S MEAN VALUE THEOREM (MVT) To give the promised proof of l Hospital s Rule, we first need a generalization of the Mean Value Theorem. The following theorem is named after another French mathematician, Augustin-Louis Cauchy (1789 1857).
CAUCHY S MVT Theorem 3 Suppose that the functions f and g are continuous on [a, b] and differentiable on (a, b), and g () 0 for all in (a, b). Then, there is a number c in (a, b) such that: f ' c f b f a g ' c g b g a
CAUCHY S MVT Notice that: If we take the special case in which g() =, then g (c) = 1 and Theorem 3 is just the ordinary Mean Value Theorem.
CAUCHY S MVT Furthermore, Theorem 3 can be proved in a similar manner. You can verify that all we have to do is: First change the function given by Equation 4 in Section 4.2 to the function f b f a h f f a g g a g b g a Then apply Rolle s Theorem as before.
L HOSPITAL S RULE We are assuming that: Proof lim f 0 and lim g 0 a a Let: L lim a f ' g ' We must show that: lim f g L a
L HOSPITAL S RULE Define: Proof F f if a g if a G 0 if a 0 if a Then, F is continuous on I since f is continuous on I a and Likewise, G is continuous on I. limf limf 0 F a a a
L HOSPITAL S RULE Proof Let I and a. Then, F and G are continuous on [a, ] and differentiable on (a, ) and G 0 there (since F = f and G = g ).
L HOSPITAL S RULE Proof Therefore, by Cauchy s MVT, there is a number y such that a < y < and F ' y F F a F G' y G G a G Here, we have used the fact that, by definition, F(a) = 0 and G(a) = 0.
L HOSPITAL S RULE Proof Now, if we let a +, then y a + (since a < y < ). Thus, lim f g lim a a F G lim ya F' G ' y y lim ya f ' g ' y y L
L HOSPITAL S RULE Proof A similar argument shows that the left-hand limit is also L. Therefore, f lim a g L This proves l Hospital s Rule for the case where a is finite.
L HOSPITAL S RULE Proof If a is infinite, we let t = 1/. Then, t 0 + as. So, we have: f f 1 t lim lim t0 1 g g t 2 2 f ' 1 t 1 t by l'hospital's Rule lim t0 g ' 1 t 1 t for finite a f ' 1 t f ' = lim lim t0 g ' 1 t g '