MATH 1010E University Mathematics Lecture Notes (week 8) Martin Li

Transcription

1 MATH 1010E University Mathematics Lecture Notes (week 8) Martin Li 1 L Hospital s Rule Another useful application of mean value theorems is L Hospital s Rule. It helps us to evaluate its of indeterminate forms such as 0 0. Let s look at the following eample. Recall that we have proved in week 3 (using the sandwich theorem and a geometric argument) sin = 1. We say that the it above has indeterminate form 0 0 since both the numerator and denominator goes to 0 as 0. Roughly speaking, L Hospital s rule says that under such situation, we can differentiate the numerator and denominator first and then take the it. The result, if eists, should be equal to the original it. For eample, (sin ) () cos = = 1, 1 which is equal to the it before we differentiate! Theorem 1.1 (L Hospital s Rule) Let f, g : (a, b) R be differentiable functions in (a, b) and fi an 0 (a, b). Assume that (i) f( 0 ) = 0 = g( 0 ). (ii) 0 f () g () Then, we have Eample 1.2 Consider the it = L (i.e. the it eists and is finite). f() 0 g() = f () 0 g () = L. sin 2 1 cos, this is a it of indeterminate form 0 0. Therefore, we can apply L Hospital s Rule to obtain sin 2 1 cos = 1 (sin 2 ) (1 cos ),

2 if the it on the right hand side eists. Since the right hand side is the same as 2 sin cos = (2 cos ) = 2. sin Therefore, we conclude that sin 2 1 cos = 2. Eercise: Calculate the it in Eample 1.2 without using L Hospital s Rule (hint: sin 2 = 1 cos 2 ). Sometimes we have to apply L Hospital s Rule a few times before we can evaluate the it directly. This is illustrated by the following two eamples. Eample 1.3 Consider the it sin 3, this is of the form 0 0. Therefore, by L Hospital s rule sin 1 cos 3 = 3 2, if the right hand side eists. The right hand side is still in the form 0 0, therefore we can apply L Hospital s Rule again 1 cos sin 3 2 = 6, if the right hand side eists. But now the right hand side can be evaluated: sin 6 = 1 6 sin = 1 6. As a result, if we trace backwards, we conclude that the original it eists and sin 3 = 1 6. Eample 1.4 Consider the it e 1 1 cosh. Applying L Hospital s Rule twice, we can argue as in Eample 1.3 that e 1 1 cosh = e 1 sinh = 2 e cosh = 1 1 = 1.

3 After seeing these eamples, let us now go back to give a proof of L Hospital s Rule. Proof of L Hospital s Rule: Recall Cauchy s Mean Value Theorem which says that f(b) f(a) g(b) g(a) = f (ξ) g (ξ) for some ξ (a, b). Therefore, since f( 0 ) = g( 0 ) = 0, we have f() g() = f() f( 0) g() g( 0 ) = f (ξ) g (ξ) for some ξ between and 0. Notice that as 0, we must also have ξ 0. Therefore, we have This proves the L Hospital s Rule. f() 0 g() = f (ξ) ξ 0 g (ξ). 2 Other indeterminate forms When we evaluate its, there are other possible indeterminate forms, for eample 0 0, 0,, 00. (2.1) Note that these forms above are just formal epressions which does not have very precise mathematical meanings as is not a real number. Convention: We distinguish two infinities by writing := + and :=. Remark 2.1 Not all epressions involving 0 and would result in an indeterminate form. For eample, 0 = 0, =, + =, =. In this section, we will see that all the indeterminate forms in (2.1) can actually be rewritten into the standard form 0 0. Symbolically we have 1/0 =. Therefore, 0 = = 0 0, 3

4 = 1/0 1/0 = 0 0., 0 0 = ep(0 ln 0) = ep(0 ( )) = ep( 0 0 ). We should emphasize that the calculations above are just formal. They indicate the general idea of transforming the its rather than actual arithmetic of numbers. Using these ideas, we can actually handle all the determinate forms in (2.1) by the L Hospital s Rule. We have Theorem 2.2 (L Hospital s Rule) The same conclusion holds if we replace (i) by f() = ± = g(). 0 0 Remark 2.3 The theorem also holds in the case 0 = ± and for onesided its as well. We postpone the proof of Theorem 2.2 until the end of this section but we will first look at a few applications. Eample 2.4 Consider the one-side it ln. + This is of the form 0 ( ). However, we can rewrite it as ln = ln 1/, which is of the form as 0+. Therefore, we can apply Theorem 2.2 to conclude that ln + 1/ = 1/ + 1/ 2 = = 0. +( ) Therefore, we have + ln = 0. In words, this means that as 0 +, the linear function is going to 0 faster than the logarithm function ln going to. Eample 2.5 Sometimes we have to apply L Hospital s Rule a few times. For eample, 2 + e = 2 + e = 2 + e = 0. 4

5 Similarly, we can prove that + k = 0, for any k. e In other words, as +, the eponential function e is going to faster than any polynomial of. The following eample shows that L Hospital s Rule may not always work: sinh cosh = cosh sinh = sinh cosh, which gets back to the original it we want to evaluate! So L Hospital s Rule leads us nowhere in such situation. For this eample, we have to do some cancellations first, sinh cosh = e e e = + e 1 e 2 = e 2 3 Some tricky eamples of L Hospital s Rule Sometimes it is not very obvious how we should transform a it into a standard indeterminate form. Eample 3.1 Evaluate that it sin 1. We can choose to transform it to either sin 1 = sin(1/) 1/ or sin 1 = 1/ sin(1/). The first one has the form 0 0 and the second one has the form as. Therefore, we can apply L Hospital s Rule to both cases. For the first case, we have sin(1/) 1 cos 1 = 2 1/ 1 2 However, for the second case, we have 1/ sin(1/) = 5 = cos 1 = cos(1/) 2 sin 2 (1/),

6 which doesn t seem to simplify after L Hospital s Rule. Therefore, sometimes we have to choose a good way to transform the it before we apply the L Hospital s Rule. A general rule of thumb here is that the epression should get simpler after taking the derivatives. Eample 3.2 Evaluate the it ( 1 sin 1 This it has the indeterminate form, which we haven t mentioned. There is in fact no general way to evaluate its of such forms. But for this particular eample, we can transform it as ). 1 sin 1 = sin sin, which has the standard indeterminate form 0 0. Therefore, we can apply L Hospital s Rule a few times to get sin sin = Eample 3.3 Evaluate the it 1 cos sin + cos = 1. sin 2 cos sin = 0. Recall that if a > 0, b are real numbers, we define a b := ep(b ln a). Therefore, ( ) ( ) ( ) 1 1 = ep ln ln 1/ = ep = ep = e 0 = 1. 1 Note that we can move the it into the function ep since the eponential function ep is continuous. We end this section with a proof of Theorem 2.2. Proof of Theorem 2.2 : The idea is that if f( 0 ) = ± = g( 0 ), then we have 1 f( 0 ) = 0 = 1 g( 0 ). Therefore, we can apply L Hospital s Rule to conclude that f() 0 g() = 1/g() 0 1/f() = g ()/g() 2 0 f ()/f() 2. 6

7 The right hand side is ( g ) () 0 f () f()2 g() 2 Therefore, we have = 0 g () f () 0 f() 0 g() = g () 0 f () ( ) f() 2 g ( ) f() 2 = g() 0 f. () 0 g() ( ) f() 2. 0 g() Canceling and moving terms around, we obtain ( f() 0 g() = g ) () 1 f () 0 f = () 0 g (). This proves the theorem. Question: Spot the gaps in the above proof. Can you fi them? 4 Indefinite Integral Now, we know how to differentiate a function f() to get a new function f (). We want to ask whether we can reverse the process. Question: Given a function f : (a, b) R (say differentiable), can we find another function F : (a, b) R such that F () = f() for all (a, b)? Let s try to understand the above question by some simple eamples. Eample 4.1 Suppose f() = e. Can we solve for F () such that F () = f() = e? Well we know that F () = e is a solution since (e ) = e. Are there any other solutions? Yes, for eample, F () = e + 1 is another solution. In fact, for any constant C R, F () = e + C is a solution. Are these all the solutions then? The answer is indeed YES! 7

8 Question: Show that if there are two solutions F 1 () and F 2 () such that F 1 () = f() = F 2 () then F 1 () = F 2 () + C for some constant C. We make the following definition. Definition 4.2 If F () is a differentiable function such that F () = f(), we say that F () is a primitive function of f() and f() d := F () + C is said to be the indefinite integral of f(). Here, C R is an arbitrary constant called the integration constant. For eample, since we know that (e ) = e and = 1, e d = e + C, 1 d = + C. Proposition 4.3 We can evaluate some of the elementary indefinite integrals. 1. cos d = sin + C sin d = cos + C. n d = n+1 + C for any real number n 1. n d = ln + C. The following property helps us evaluate a much larger class of indefinite integrals. Proposition 4.4 (Linearity) Indefinite integrals are linear: 1. [f() ± g()] d = f() d ± g() d. 8

9 2. kf() d = k f() d for any constant k. Using just Proposition 4.3 and 4.4, we can evaluate a lot of indefinite integrals. First, we know how to compute the indefinite integrals of any polynomials. For eample, ( ) d = 4 d 3 d d = ( ) ( ) C C 2 + 7( + C 3 ) = (C 1 3C 2 + 7C 3 ) = C where C is ANY constant. Note that in the end, we can simply group all the constants together to form a single constant C since all these constants are arbitrary. We can also evaluate the indefinite integrals of some rational functions. For eample, ( + 2) d = d ( = ) d 1 = d + 4 d + 4 d = ln + C. The powers in the rational function do not need to be integers in some cases d = ( ) d = 5 5/2 5/ /2 1/2 + C = C. Proposition 4.5 We can evaluate the following indefinite integrals of trigonometric functions. 9

10 sec 2 d = tan + C. csc 2 d = cot + C. sec tan d = sec + C. csc cot d = csc + C. sec d = ln sec + tan + C. csc d = ln csc + cot + C. The first four integrals are easy. The last two will be proved after we learn t-substitution in the net class. 10