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# SI Calculus II MATH 1220

Rosa Farrell
Weber State University
GPA 3.53

Michael Wills

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COURSE
PROF.
Michael Wills
TYPE
Class Notes
PAGES
5
WORDS
KARMA
25 ?

## Popular in Math

This 5 page Class Notes was uploaded by Rosa Farrell on Wednesday October 28, 2015. The Class Notes belongs to MATH 1220 at Weber State University taught by Michael Wills in Fall. Since its upload, it has received 16 views. For similar materials see /class/230803/math-1220-weber-state-university in Math at Weber State University.

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Date Created: 10/28/15
ADDITIONAL NOTES AND EXERCISES FOR 48 MIKE WILLS 1i INDETERMINATE FORMS De nition 11 The set of the extended real numbers is 1 1 RU00700 where 00 and 700 are symbols which satisfy the inequality 12 700 lt a lt 00 for every real number a An expression like is unde ned in the real numbers However in the extended real numbers the expression has two reasonable interpretations either 00 or 700 In other words in the extended real numbers is determined up to sign There are still issues of course but the point is that if f x A 0 as I A 0 then 2 If 0113 me form is an expression for which a whole range of values can be assigned with equal lack of validity For example does not make sense even in the context of the extended real numbers it matters how we approach zero in the numerator and the denominator Indeterminant forms arise most naturally in the evaluation of limits For example the observation exists it must be either 00 or 700 By contrast an indeterminant lim am 7 y we 13 aixhlll z lin z 1 illustrates that can in some sense take every possible real value en evaluating a limit where an indeterminant form is present we cannot blindly misapply limit laws and hope to get a reasonable result This is where l Hopital7s rule comes into play 2 L7HoP1TAL7s RULE AND THE INDETERMINATE FORMS AND L7Hopitalls rule gives us an excellent way of computing limits of ratios of func tions whose numerator and denominator both convergediverge either to zero or to 00 The text discusses how one can use llHopitan ru e The point of this section is to highlight the pitfalls that occur when the rule is blindly misappliedi Here are some things to bear in mind First in many cases it is more effective and ef cient to use algebra as was done in the example in the last section Second l Hopitalls rule is ineffective when we are not working with an indeter minate form 2 MIKE WILLS For example 21 0 lim y im CO 1H0 1 cos I EHO sinz The right hand side is unde ned in the extended real numbers a Hence in general the limit of the ratio of the original functions is not necessarily the limit of the ratio of the derivatives even when both limits existi Third and more subtly llHopital7s rule may be inapplicable for technical reasons Consider the following expression sin z 2 lim 7 moo I Here the ratio of the limits is the indeterminate form which means l Hopitalls rule is potentially applicable lndeed sin 1 cos 1 2 3 lim lim 1H0 z 1H0 1 li The problem is that to prove that the derivative of sinz is cosz we had to show sinx that lirn 1 Thus using llHopital7s rule to evaluate our limit is begging the 1 I question Fourth the limit may exist and we may have an appropriate indeterminate form but llHopital7s rule is inapplicable either because the functions under study are not differentiable or because the limit of the ratio of the derivatives does not ex1sti Example 21 Let 24gt gltzgt 0 I N z z E Q Then 25 lim 91 lim 1 0 moo 40 but l Hopitan rule is inapplicable because the denominator is not differentiable Nevertheless 2 x 2 6 gin 7 0 Fifth llHopitan rule may be applicable but not yield any useful information As a somewhat silly example try applying l Hopitalls rule to ex lim 7 mace ET A somewhat more substantive example is given in the exercises 3 THE INDETERMINATE FORM 00 7 00 Although llHopital7s rule is not directly applicable to indeterminant forms of this type sometimes one can still use the rule by converting the difference into a quotienti The idea is best illustrated with an example ADDITIONAL NOTES AND EXERCISES FOR 48 3 1 Example 31 Compute lim cotz 7 7 110 1 Note that as I i 0 both cot z and diverge to 00 and hence we have an inde terminant form of type 00 7 00 We start by trying to rewrite the expression as a quotient 31 t 1 cosz 1 zcosz 7 sinz co I77 77 1 sin I z 1 sin I As I approaches zero7 both the numerator and denominator approach 0 and so we attempt to apply l Hopital7s rule Anticipating in advance that llHopital7s rule will work7 we compute zcosz7sinz cosz7sinz cosz 2cosz7sinz 32 lim lim lim 00 0210 zsmz x10 zzcosz x10 z1cosz 4 THE INDETERMINANT FORM 0 00 Suppose that 7gt ioo and 91 7 0 as I 7gt a We can then attempt to evaluate limina by writing f z 41 fltzgtgltzgt 7 E 91 or m 42 I z E lt gt f M gt 91 The rst rewriting leads to an indeterminant form of type while the second leads to an indeterminant form of type In both cases7 llHopital7s rule may be applicable Example 41 We evaluate lim re 4700 0 43 lim reg lim 7 lim 7700 7700 671 7700 671 1 Example 42 We evaluate lim ztan 7 700 1 sec2 T l E hm sec2 7 1 700 I HJHMH 44 139 mi 139 H 32 5 n32 5 INDETERMINANT FORMS INVOLVING POWERS The following unde ned expressions are in calculus at least1 regarded as inde terminant forms 0 00 m In all three situations7 to apply l Hopital7s rule we rst convert the exponential expression to a product As an example7 suppose that gt 0 and g 7 0 as I 7gt a Then 51 3091 eg 1nf1 1The expression 00 is frequently taken to be 1 for example when using summation notation we make this convention whenever 00 appears in the sum 4 MIKE WILLS Since ln is continuous lnf 7gt 700 as I 7gt 0 Since exp is continuous 52 um em Wm 691132 1 f I 1711 Which gives the indeterminant form 0 00 Similarlyiffzgt0 f7gtooandg7gt0asz7athenlnf7gtooasz7gtaand so we again have the indeterminant form 0 00 If f 7gt 1 andg 7gt ioo as I 7gt a then lnf 7gt 0 as I 7gt a and the same calculation gives the indeterminant form 00 0 One example should illustrate the method Example 51 We compute li 114 as lnz l 14 lim 7 ilim 7 ilim 7 53 mgr li e 41 e 1 4 e 11042 e no 4 1 x x 6 EXERCISES ln exercises one through ten evaluate the limits if they exist If they do not exist explain Why Exercise 61 s1nhz li x70 sinz Exercise 62 1 7 3 1325 i I gt Exercise 63 1 man im 7 17W cos I 1 Exercise 64 lim Inc n E N 1700 Exercise 65 lim 1 1700 Exercise 66 z 7 l allgi z 7 1 Exercise 67 tanz lim 7r g 1 7 5 Exercise 68 hm lnlnlzl 17gt 1 Exercise 69 E 2 lim 0 cost dt x70 13 Exercise 610 zlnz 11330 312 7 1 ADDITIONAL NOTES AND EXERCISES FOR 48 5 Exercise 611 Let Using trigonometric identities compute 6 1 aggro Explain why l Hopital7s rule is inapplicable even though we have the indeterminant form and the functions are differentiab e Exercise 612 Let fx Compute 62 lim mace Explain why l Hopital7s rule is ineffective The next two exercises show that the unde ned expression 0 is not an indeter minant form Exercise 613 Suppose that gt 0 and that A 0 91 A 00 as I A a Compute 63 lim fz9 EH11 Exercise 614 Suppose that gt 0 and that A 0 91 A 700 as I A a ompute 64gt gig MW

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