2 Indeterminate Forms and L’Hospital’s Rule INVERSE FUNCTIONS7.8Indeterminate Forms and L’Hospital’s RuleIn this section, we will learn:How to evaluate functions whosevalues cannot be found at certain points.
3 Suppose we are trying to analyze the behavior of the function INDETERMINATE FORMSSuppose we are trying to analyze the behavior of the functionAlthough F is not defined when x = 1, we need to know how F behaves near 1.
4 In particular, we would like to know the value of the limit INDETERMINATE FORMSExpression 1In particular, we would like to know the value of the limit
5 INDETERMINATE FORMSIn computing this limit, we can’t apply Law 5 of limits (Section 2.3) because the limit of the denominator is 0.In fact, although the limit in Expression 1 exists, its value is not obvious because both numerator and denominator approach 0 and is not defined.
6 INDETERMINATE FORM —TYPE 0/0 In general, if we have a limit of the formwhere both f(x) → 0 and g(x) → 0 as x → a, then this limit may or may not exist.It is called an indeterminate form of type .We met some limits of this type in Chapter 2.
7 For rational functions, we can cancel common factors: INDETERMINATE FORMSFor rational functions, we can cancel common factors:
8 We used a geometric argument to show that: INDETERMINATE FORMSWe used a geometric argument to show that:
9 However, these methods do not work for limits such as Expression 1. INDETERMINATE FORMSHowever, these methods do not work for limits such as Expression 1.Hence, in this section, we introduce a systematic method, known as l’Hospital’s Rule, for the evaluation of indeterminate forms.
10 INDETERMINATE FORMSExpression 2Another situation in which a limit is not obvious occurs when we look for a horizontal asymptote of F and need to evaluate the limit
11 There is a struggle between the two. INDETERMINATE FORMSIt isn’t obvious how to evaluate this limit because both numerator and denominator become large as x → ∞.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.
12 INDETERMINATE FORM —TYPE ∞/∞ In general, if we have a limit of the formwhere both f(x) → ∞ (or -∞) and g(x) → ∞ (or -∞), then the limit may or may not exist.It is called an indeterminate form of type ∞/∞.
13 INDETERMINATE FORMSWe 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 x that occurs in the denominator.For instance,
14 This method, though, does not work for limits such as Expression 2. INDETERMINATE FORMSThis method, though, does not work for limits such as Expression 2.However, L’Hospital’s Rule also applies to this type of indeterminate form.
15 L’HOSPITAL’S RULESuppose f and g are differentiable and g’(x) ≠ 0 on an open interval I that contains a (except possibly at a).Supposeor thatIn other words, we have an indeterminate form of type or ∞/∞.
16 if the limit on the right exists (or is ∞ or - ∞). L’HOSPITAL’S RULEThen,if the limit on the right exists (or is ∞ or - ∞).
17 NOTE 1L’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.
18 NOTE 2The rule is also valid for one-sided limits and for limits at infinity or negative infinity.That is, ‘x → a’ can be replaced by any of the symbols x → a+, x → a-, x → ∞, or x → - ∞.
19 NOTE 3For 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.
20 NOTE 3In fact, using the alternative form of the definition of a derivative, we have:
21 NOTE 3It is more difficult to prove the general version of l’Hospital’s Rule.
22 Find and Thus, we can apply l’Hospital’s Rule: L’HOSPITAL’S RULE Example 1FindandThus, we can apply l’Hospital’s Rule:
23 Calculate We have and So, l’Hospital’s Rule gives: L’HOSPITAL’S RULE Example 2CalculateWe have andSo, l’Hospital’s Rule gives:
24 However, a second application of l’Hospital’s Rule gives: Example 2As ex → ∞ and 2x → ∞ as x → ∞, the limit on the right side is also indeterminate.However, a second application of l’Hospital’s Rule gives:
25 Calculate As ln x → ∞ and as x → ∞, l’Hospital’s Rule applies: Example 3CalculateAs ln x → ∞ and as x → ∞, l’Hospital’s Rule applies:Notice that the limit on the right side is now indeterminate of type .
26 L’HOSPITAL’S RULEExample 3However, instead of applying the rule a second time as we did in Example 2, we simplify the expression and see that a second application is unnecessary:
27 L’HOSPITAL’S RULEExample 4FindNoting that both tan x – x → 0 and x3 → 0 as x → 0, we use l’Hospital’s Rule:
28 L’HOSPITAL’S RULEExample 4As the limit on the right side is still indeterminate of type , we apply the rule again:
29 Since , we simplify the calculation by writing: L’HOSPITAL’S RULEExample 4Since , we simplify the calculation by writing:
30 L’HOSPITAL’S RULEExample 4We can evaluate this last limit either by using l’Hospital’s Rule a third time or by writing tan x as (sin x)/(cos x) and making use of our knowledge of trigonometric limits.
31 Putting together all the steps, we get: L’HOSPITAL’S RULEExample 4Putting together all the steps, we get:
32 Find If we blindly attempted to use l-Hospital’s rule, we would get: Example 5FindIf we blindly attempted to use l-Hospital’s rule, we would get:
33 L’HOSPITAL’S RULEExample 5This is wrong.Although the numerator sin x → 0 as x → π -, notice that the denominator (1 - cos x) does not approach 0.So, the rule can’t be applied here.
34 L’HOSPITAL’S RULEExample 5The required limit is, in fact, easy to find because the function is continuous at π and the denominator is nonzero there:
35 L’HOSPITAL’S RULEThe example 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 Examples 3 and 5 in Section 2.3, Example 3 in Section 4.4, and the discussion at the beginning of this section.
36 L’HOSPITAL’S RULESo, when evaluating any limit, you should consider other methods before using l’Hospital’s Rule.
37 INDETERMINATE PRODUCTS If and (or -∞), then it isn’t clear what the value of , if any, will be.
38 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.
39 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:This converts the given limit into an indeterminate form of type or ∞/∞, so that we can use l’Hospital’s Rule.
40 INDETERMINATE PRODUCTS Example 6EvaluateThe given limit is indeterminate because, as x → 0+, the first factor (x) approaches 0, whereas the second factor (ln x) approaches -∞.
41 INDETERMINATE PRODUCTS Example 6Writing x = 1/(1/x), we have 1/x → ∞ as x → 0+.So, l’Hospital’s Rule gives:
42 INDETERMINATE PRODUCTS NoteIn solving the example, another possible option would have been to write:This gives an indeterminate form of the type 0/0.However, if we apply l’Hospital’s Rule, we get a more complicated expression than the one we started with.
43 INDETERMINATE PRODUCTS NoteIn general, when we rewrite an indeterminate product, we try to choose the option that leads to the simpler limit.
44 INDETERMINATE PRODUCTS Example 7Use l’Hospital’s Rule to sketch the graph of f(x) = xex.Since both x and ex become large as x → ∞, we have:
45 INDETERMINATE PRODUCTS Example 7However, as x→ -∞, ex → 0.So, we have an indeterminate product that requires the use of l’Hospital’s Rule:Thus, the x-axis is a horizontal asymptote.
46 INDETERMINATE PRODUCTS Example 7We use the methods of Chapter 4 to gather other information concerning the graph.The derivative is:f’(x) = xex + ex = (x + 1)exSince ex is always positive, we see that f’(x) > 0 when x + 1 > 0, and f’(x) < 0 when x + 1 < 0.So, f is increasing on (-1, ∞) and decreasing on (-∞, -1).
47 INDETERMINATE PRODUCTS Example 7f’(-1) = 0 and f’ changes from negative to positive at x = -1.Hence, f(-1) = -e-1 is a local (and absolute) minimum.
48 INDETERMINATE PRODUCTS Example 7The second derivative is:Since f”(x) > 0 if x > -2 and f”(x) < 0 if x < -2, f is concave upward on (-2, ∞) and concave downward on (-∞, -2).The inflection point is (-2, -2e-2).
50 INDETERMINATE FORM—TYPE ∞ - ∞ If and , then the limitis called an indeterminate form of type ∞ - ∞.
51 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?
52 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 ∞/∞.
53 INDETERMINATE DIFFERENCES Example 8ComputeFirst, notice that sec x → ∞ and tan x → ∞ as x → (π/2)-.So, the limit is indeterminate.
54 INDETERMINATE DIFFERENCES Example 8Here, we use a common denominator:Note that the use of l’Hospital’s Rule is justified because 1 – sin x → 0 and cos x → 0 as x → (π/2)-.
55 Several indeterminate forms arise from the limit INDETERMINATE POWERSSeveral indeterminate forms arise from the limit
56 Each of these three cases can be treated in either of two ways. INDETERMINATE POWERSEach of these three cases can be treated in either of two ways.Taking the natural logarithm:Writing the function as an exponential:
57 INDETERMINATE POWERSRecall that both these methods were used in differentiating such functions.In either method, we are led to the indeterminate product g(x) ln f(x), which is of type 0 . ∞.
58 INDETERMINATE POWERSExample 9CalculateFirst, notice that, as x → 0+, we have 1 + sin 4x → 1 and cot x → ∞.So, the given limit is indeterminate.
59 Then, ln y = ln[(1 + sin 4x)cot x] = cot x ln(1 + sin 4x) INDETERMINATE POWERSExample 9Let y = (1 + sin 4x)cot xThen, ln y = ln[(1 + sin 4x)cot x] = cot x ln(1 + sin 4x)
61 So far, we have computed the limit of ln y. INDETERMINATE POWERSExample 9So 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 = eln y:
62 INDETERMINATE POWERSExample 10FindNotice that this limit is indeterminate since 0x = 0 for any x > 0 but x0 = 1 for any x ≠ 0.
63 INDETERMINATE POWERSExample 10We could proceed as in Example 9 or by writing the function as an exponential: xx = (eln x)x = ex ln xIn Example 6, we used l’Hospital’s Rule to show thatTherefore,
64 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).
65 CAUCHY’S MVTTheorem 3Suppose that the functions f and g are continuous on [a, b] and differentiable on (a, b), and g’(x) ≠ 0 for all x in (a, b).Then, there is a number c in (a, b) such that:
66 CAUCHY’S MVTNotice that:If we take the special case in which g(x) = x, then g’(c) = 1 and Theorem 3 is just the ordinary Mean Value Theorem.
67 Furthermore, Theorem 3 can be proved in a similar manner. CAUCHY’S MVTFurthermore, 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 functionThen apply Rolle’s Theorem as before.
68 L’HOSPITAL’S RULEProofWe are assuming that:Let:We must show that:
69 Define: Then, F is continuous on I since f is continuous on and L’HOSPITAL’S RULEProofDefine:Then, F is continuous on I since f is continuous on andLikewise, G is continuous on I.
70 L’HOSPITAL’S RULEProofLetThen, F and G are continuous on [a, x] and differentiable on (a, x) and G’ ≠ 0 there (since F’ = f’ and G’ = g’).
71 L’HOSPITAL’S RULEProofTherefore, by Cauchy’s MVT, there is a number y such that a < y < x andHere, we have used the fact that, by definition, F(a) = 0 and G(a) = 0.
72 Now, if we let x → a+, then y → a+ (since a < y < x). L’HOSPITAL’S RULEProofNow, if we let x → a+, then y → a+ (since a < y < x).Thus,
73 A similar argument shows that the left-hand limit is also L. L’HOSPITAL’S RULEProofA similar argument shows that the left-hand limit is also L.Therefore,This proves l’Hospital’s Rule for the case where a is finite.
74 If a is infinite, we let t = 1/x. L’HOSPITAL’S RULEProofIf a is infinite, we let t = 1/x.Then, t → 0+ as x → ∞.So, we have: