3.7 Indeterminate Forms - l Hôpital s Rule

Transcription

1 3.7. INDETERMINATE FORMS - L HÔPITAL S RULE Indeterminate Forms - l Hôpital s Rule 3.7. Introduction An indeterminate form is a form for which the answer is not predictable. From the chapter on lits, we know that sin!!! is an indeterminate form. Recall that is. But when we evaluated it, it turned out to be. Now, consider 3. This is again =. Finally, 2!! 2. But if we simplify and evaluate, we get! 3 = 2 is again. If we simplify and evaluate the it, we get = =. Here are three its which all were of the form!. They produced di erent answers. This is why is called an indeterminate form. We cannot predict what it will be. There are many other indeterminate forms. They include:,,,,,,,. To help see why they are indeterminate, it helps if we break them apart. We illustrate this with each indeterminate form.. A fraction of the form a is but a fraction of the form a is. So, what is? If the numerator converges to faster than the denominator, then the numerator will win and will behave like a and thus will be. L Hôpital s rule provides another way to handle such forms. It only works on indeterminates of the form and. For the other forms, there are techniques which can be applied to change them to or so that we can then use l Hôpital s rule on them. To understand how to handle them, it helps to classify them according to the operations involved in them. We will refer to indeterminate quotients when talking about or ;to indeterminate product when talking about and to indeterminate powers when talking about,,,. We rst state L Hôpital s rule and see how to apply it to indeterminates of the form or. We then see how to change the other forms so we can apply l Hôpital s rule to them Main Result - Indeterminates Quotients (of the Form or ) Before we state the theorem, it is important to understand that it only applies to indeterminates of the form or. For any other form, some preinary

2 42 CHAPTER 3. APPLICATIONS OF THE DERIVATIVE work will have to be done rst. In addition, there are certain conditions which have to be met before the theorem can be applied. Theorem 27 (l Hopital s rule) Suppose that f and g are di erentiable and f () g () 6= on an open interval that contains a (ecept possibly at a). If!a g () is of the form or then f ()!a g () = f ()!a g () provided the latter it eists or is is taken as!. or. The rule also applies if the it Not only is it important to know the conclusion of the theorem, that is how to compute the it. It is also equally important to know the conditions under which the theorem applies. Before the theorem can be applied, there are several conditions which have to be satis ed. If the theorem is applied when it should not be applied, whatever conclusion is reached by applying it is not valid. We use the theorem as follows:. Verify that all the conditions are met. 2. Compute!a f () g (). f () 3. If the above it eists or is, then it is equal to, the it!a g () we were trying to nd. If it does not eist, we cannot conclude and must nd another way. Let us now look at some eamples. sin Eample 28 Find! You will recognize one of the fundamental trigonometric its we dealt with earlier. This function satis es the conditions of the theorem because both sin and are di erentiable near ecept possibly at ). In addition, as!, sin is of the form. We begin by computing (sin ) cos! () =! = =

3 3.7. INDETERMINATE FORMS - L HÔPITAL S RULE 43 Since the it eists, by l Hôpital s rule, we have: sin! = This is much easier than the way we found this it earlier. Remark 29 It is important to understand the logic behind these computations. We cannot claim the two its are equal until we know that the second one eists or is. This is why we compute the it of the quotient of the derivatives separately. If we nd it eists or is, we can then conclude the two its are equal. e Eample 2 Find! 2 The theorem can also be used if we are nding a it as!. Once again, this is an indeterminate form of the type, the function satis es the conditions of the theorem (they are both di erentiable for large ), so we can apply l Hôpital s rule. We begin by nding the it of the quotient of the derivatives, that is: (e )! ( 2 ) = e! 2 This, is again an indeterminate form of the type, the function satis es the conditions of the theorem, so we can apply l Hôpital s rule again. As before, we begin by nding the it of the quotient of the derivatives, that is: (e )! (2) = e! 2 = Because the last it is in nity, by l Hôpital s rule we have e l Hôpital s rule again, we have! 2.! e =. By 2 Remark 2 In some instances, we may need to apply the rule more than once, as in the previous eample. Eample 22 Find! sin cos If we attempt to apply l Hôpital s rule without rst checking if we are in a situation in which we can and should apply it, we would compute! (sin ) ( cos ) =! = cos sin

4 44 CHAPTER 3. APPLICATIONS OF THE DERIVATIVE Thus, we would conclude that! sin cos rules as follows:! sin cos =. This would be wrong. is not an indeterminate form. We can compute it with our it! Which is the correct answer. sin cos = ( ) = 2 = Remark 23 The last eample illustrates the importance of making sure the theorem we are using can indeed be used. Sometimes, if we use a theorem when we are not supposed to, it will not work and we will see it. Other times though, we will get an answer. The only problem is that the answer we got is not valid, but we don t know it. So, please, whenever you use a theorem, always check that all the conditions under which the theorem can be used are satis ed Indeterminates not of the form or We try to change the epression we are nding the it of so that it becomes an indeterminate of the form or. We then apply l Hôpital s rule to it. The way this is done depends on the form we have. We look at several cases. Indeterminate Products (of the form ) This usually happens when computing!a f () g (), one of the function goes to while the other goes to. The idea here is to notice that fg = f g = g f So that if for eample f ()! and g ()!, we rewrite f () g () = g () f() g() Since f ()!, f()!, so is of the form and we can apply l Hôpital s f() rule to it. We illustrate this with eamples. Eample 24 sin! As!,! and so sin! sin () =. This is an indeterminate of

5 3.7. INDETERMINATE FORMS - L HÔPITAL S RULE 45 the form. We apply the technique described above, replacing by. So, we have: sin! sin =! sin As!, nding the it of the quotient of the derivatives.!!. We can apply l Hôpital s rule to it. We begin by sin =! =! cos = 2 cos 2 Since the it eists, by l Hôpital s rule, sin =! Indeterminate Di erences (of the form ) This usually happens when computing (f () g ()), both functions go to!a. If fractions are involved, try to rewrite the epressions as one fraction. It should then reduce to one of the forms discussed above. If radicals are involved, use the conjugate. Eample 25! + sin If!,!. Also, sin! so that sin!. So, we have an indeterminate of the form. We rewrite the epression as one fraction.! + sin = sin! + sin We see that as! +, sin sin is of the form. We can apply l Hôpital s rule to it. We begin by nding the it of the quotient of the derivatives.! + (sin ) ( sin ) =! + cos sin + cos

6 46 CHAPTER 3. APPLICATIONS OF THE DERIVATIVE This is again an indeterminate of the form. So, we apply l Hôpital s rule to it. As before, we compute the it of the quotient of the derivatives. (cos )! + (sin + cos ) = sin! + cos + cos sin = 2 = cos By l Hôpital s rule, it follows that! + sin + cos again, it follows that =.! + sin =. By l Hôpital s rule Indeterminate Powers (of the form,,, ) This usually happens when computing!a f () g(). We remember from other problems involving powers (logarithmic di erentiation) that the function ln transforms powers into products. If the it we want is an indeterminate that has the form of a power, we proceed as follows:. Apply the natural logarithmic function (ln) to the epression. Note that this is no longer the problem you were given. 2. Using the properties of ln, rewrite powers as products. 3. Take the it of the new epression using the techniques describes above. 4. Assuming the it we just computed eists, it is not the answer to our problem, since we applied ln to the original problem. The answer we want is the eponential of what we found above. In other words, if the it in step 3 is A, then the answer to the original problem is e A. We illustrate this with an eample. Eample 26! As!, is an indeterminate of the form. We follow the steps outlined above. First, we change the problem to! as apply the properties of ln. ln =!! ln ln =! This is an indeterminate of the form. We apply l Hôpital s rule. We begin by nding the it of the quotient of the derivatives. (ln )! () =! =

7 3.7. INDETERMINATE FORMS - L HÔPITAL S RULE 47 By l Hôpital s rule, this means that original problem is Things to know! ln = e! = =. Therefore the answer to our Know how to use l Hôpital s rule for the indeterminate forms and. Know how to transform the other indeterminate forms so that we can apply l Hôpital s rule. Remember the technique of applying ln to change powers into products. By now you should realize that this technique is used often, when dealing with eponents. Be able to do problems such as #, 3, 5, 7, 8, 9, 7, 2, 25, 27, 29, 3, 33, 35, 37 on pages 33, 34