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# Complex Variables MATH 3810

Weber State University

GPA 3.53

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This 7 page Class Notes was uploaded by Rosa Farrell on Wednesday October 28, 2015. The Class Notes belongs to MATH 3810 at Weber State University taught by Staff in Fall. Since its upload, it has received 18 views. For similar materials see /class/230804/math-3810-weber-state-university in Math at Weber State University.

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Date Created: 10/28/15

ADDITIONAL EXERCISES AND NOTES FOR 25 1 MORE ON LIMITS When we gave the formal de nition of a limit we assumed that a and L were nite However there are occasions when this point of View is too limiting We state here some limit de nitions when either a or L or both is in nite De nition 1 Let I be an open interval containing a Let f be a realvalued function de ned on I except possibly at a We say that f approaches 00 as I approaches a if given N gt 0 there exists 6 gt 0 such that gt N whenever 0 lt lz 7 al lt 6 In this situation we write lim f 00 711 We say that f approaches 700 as I approaches a if given N gt 0 there exists 6 gt 0 such that lt 7N whenever 0 lt lz 7 al lt 6 In this situation we write lim f 700 711 With some minor modi cations this de nition can be modi ed to onesided limits ii Let I be an interval of the form 1200 Let f be a function de ned on I We say that f approaches L a nite number as I approaches 00 if given 6 gt 0 there exists M gt 0 such that 7 Ll lt 6 whenever I gt M In this situation we write lim L 1700 We say that f approaches 00 as I approaches 00 if given N gt 0 there exists M gt 0 such that gt N whenever I gt M In this situation we write lim L 1700 We say that f approaches 700 as I approaches 00 if given N gt 0 there exists M gt 0 such that lt 7N whenever I gt M In this situation we write lim f z L 1700 Limits as I approaches 700 are de ned analogously Example 2 Let fx Then lim fx 0 170 Proof Let N gt 0 When computing limits rigorously this is an excellent rst statement We need to nd 6 gt 0 such that if 0 lt lt 6 then gt N Now if lt 6 then i gt and hence g gt 671 By choosing 6 gt 0 so that W 1 lt1 7 N the desired condition on 6 will be satis ed Solving for 6 in equation 1 we obtain a i W Thus if 1 0 lt lt 7 W then gt N as required D 2 ADDITIONAL EXERCISES AND NOTES FOR 25 The above proof contained a lot of repetition As one acquires a facility with these kinds of proofs much of the repetition can be dispensed with However there is rarely any harm in overwriting a proof and you are encouraged to write a lot when asked to provide a proof In any case you will only occasionally be asked to give formal proofs since the focus of the calculus course here at Weber State is computational There is a time and place for serious proof writing and it is called analysis Math 3810 4210 and 4220 at Weber State Example 3 Let fr Si Then lim fr 0 a Proof Let 6 gt 0 We need to nd M gt 0 such that lt 6 whenever I gt M Now if z gt M gt 0 then 1 1 i 7 0 M gt z gt Since lsinzl S l for any real number I it follows that sinz l l 2 7 lt H i ltgt was I IltM By choosing M the calculation in 2 yields the desired result D There are several important limit laws that can aid with calculations Some of the laws are mentioned in the text We mention a few more here Theorem 4 Suppose that a is a nite number and f is a function de ned on an interval containing a except possibly at a Then lim ezists and only both xaa lim and lim exist and are equal daat xaa This result is frequently used to show that a limit does not exist For example lim dalzil does not exist because the two one sided limits are not equal Theorem 5 In this result a may be nite or ioo Suppose that lim 00 xaa lim 91 00 and lim hz c where f g and h are functions de ned near a daa 1 and c is a positive number Then the following hold 239 mom gltzgtgt co lim 00 fltzgthltzgt oo 239va fryz foo v gig fzhz foo Notice that there was no result for lim 7 This is because that limit may not exist and even when it doesftife value of the limit could be anywhere in 70000 that is any real number or ioo The expression mm 7 grim oo 7 oo is an example of an indeterminate form Another indeterminate form is 0 00 We will discuss such forms in the second semester of this course when we look at l Hospital s rule ADDITIONAL EXERCISES AND NOTES FOR 25 3 If we replace all the limits in the preceding theorem with onesided limits all approaching from the same side of a the results remain vali i Theorem 6 Comparison theorem Suppose that S 91 and that limfz L and limgz M Then L S M Theorem 7 The squeeze theorem Suppose that S hz S 91 and that limfz limgz L Then limhz L Welve implicitly used the squeeze theorem which follows from the comparison theorem in several examples mostly involving sin and cos Exercise 8 Compute the following limits without a formal proofi If the limit is unde ned explain why a b hm c lim 13 7 z Hint Factor I rst l 0 05213100 4 e l1m 7 xao z 2 2 f Elimocos t s1n ti Exercise 9 Prove that limi4 ooi xHOI 2 MORE ON CONTINUITY Continuity is a fundamental concept in advanced mathematicsi Most of the func tions that we discuss in calculus are continuous or at least piecewise continuousli This is a bit misleading continuity is a rather special condition and in some sense the probability of picking a function at random de ned on all of R 70000 that is continuous anywhere at all is zero There are two reasons that we usually restrict our attention to functions that are continuous First and this is not really that good of a reason they are relatively easy to handle second many physical are A A 39 t by quot nctionsi We recall the de nition of continuity from the text De nition 10 Let I cd be an open interval and let a be in I Let f z I A R be a function Then f is continuous at a is lim fx fai xaa If given any a in I f is continuous at a then f is continuous on L If c and d are nite numbers and f is de ned at c and d then f is continuous on ad if f is continuous on 1 mm M fc mac and E133 fltzgt fd 1Piecewise continuity is de ned below 4 ADDITIONAL EXERCISES AND NOTES FOR 25 If f is not continuous at a then f is discontinuous at a For technical reasons it is often convenient for us to de ne continuity on closed intervals rather than open Theorem 11 ff andg are continuous at a and c is a constant then fig fg and cf are all continuous at a Ifga 0 then g is also continuous at a The proof follows immediately from the corresponding results for limits Theorem 12 Ifg is continuous at a and f is continuous at 9a then f 09 is continuous at a Thus the composition of continuous functions is continuous Example 13 Let P Q be polynomials with Q not identically 0 Then P is con tinuous on all of R and g is continuous away from the roots of Trigonometric functions and root functions are also continuous on their domains In previous courses you may have seen exponential and logarithmic functions These functions are also continuous It is often worthwhile considering how a function can fail to be continuous De nition 14 Suppose that lim L and lim M with L and M Eaat EH07 both nite If L M but fa L possibly because u is unde ned then f has a removable discontinuity at z a If f is unde ned at z a and a is a removable discontinuity of f at z a a continuous extension of f to a is given y M w y a 3 91 11m 161 1 a xaa Example 15 Note that g is unde ned at z 0 However the disconti nuity is removable and 91 z is a continuous extension of f to z 0 De nition 16 Suppose that lirnJr L and lim M with L and M xaa xaa both nite If L y M then f has a jump discontinuity at a Example 17 The discontinuities of the greatest integer function are jump discon tinuities De nition 18 If f is discontinuous at a but the discontinuity is neither a jump discontinuity nor a removable discontinuity then the discontinuity is essential If at least one of the onesided limits is ioo then the discontinuity is in nite Example 19 The function sin has an essential discontinuity at z 0 Try to graph f on a graphing calculator or on a computer algebra system Example 20 If P and Q are nonzero polynomials and Pa 0 Qa then 3 has an in nite discontinuity at z a Exercise 21 Classify the discontinuities of each of the following functions I ltagtfltzgt 18 ADDITIONAL EXERCISES AND NOTES FOR 25 5 ltbgtfltzgtltj gtf om easing 1ng 4712 120 ltdgtfltzgt 13 M0 Exercise 22 Explain why in 120 is continuous on all of R De nition 23 Suppose that f is a function de ned on a closed bounded interval I If f has nitely many discontinuities none of which are essential7 then f is piecewise continuous on Example 24 Any continuous function on I is piecewise continuous The greatest integer function is piecewise continuous on any closed bounded interval The func tion 52174 fails to be piecewise continuous on any closed bounded interval containing 2 or 72 Piecewise continuity plays an important role when we think about integration We close with an important result relating differentiability and continuity Theorem 25 If f is di erentiable at a then f is continuous at a Proof We need to show that lim fa 5 ADDITIONAL EXERCISES AND NOTES FOR 25 Since f is differentiable at a we know that 1 7 1 04 f a De nition of derivative M 7 fa hm Ha I 7 a 17m Limit of a constant is a constant lim M 7 lim fa 0 17m 1 7 a 17m lim a 7 fagt 0 lim 17m lim 17m 17m Difference of limits is limit of difference mm 7agtltgiga ZZ i if 7 W 7 0 gig ltz 7agt f 5 7 z 7 antLa 7 o 4 Product of limits is limit of product gig 1W 7 W 7 z 7 antLa 7 o 53101 7 W 7 z 7 antLa 7 77113 I 7 anta Right hand side is just a fancy way of writing zeroi gig 1W 7 fa 7 z 7 agtf ltagt giggle 7 Loxa 7 0 gig f17 m 7 0 Sum of limits is limit of sum gig 1W 7 m 7 fa 7 fa mum 7 fa 53 M 7 fa Limit of a constant is a constant gig M 7 fa Sum of limits is limit of sum The last line is exactly what we wished to showi D This proof looks more complicated than it really is The main idea is to manipu late the various limits involved using the limit laws and being careful not to assume a limit exists before we have shown the existence of the limit in other words7 it is easy to get caught up in circular reasoning if one is not carefuli e have now shown that differentiability implies continuityi The converse is false As a simple example7 consider at z 0 ln fact7 it is possible to construct functions that are continuous everywhere and differentiable nowherei We shall not do so in this course ADDITIONAL EXERCISES AND NOTES FOR 25 7 Exercise 26 a For every real number I there is an integer n depending on I such that n S I lt n i e number n is called the greatest integer of z and is denoted by n Sketch the graph of y for 75 S I S 5 Where is continuous Where is differentiable If y is differentiable at a compute 27 at z a b It can be shown that lim cos 9 1 and that lim sint9 0i Hence7 sin and cos a are continuous at 9 0 Use the sum rules for sine and cosine to show that sin and cos are continuous on the entire real line

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