Solving epsilon-delta problems

1 Solving epsilon - delta problemsMath 1A, 313,315 DISS eptember 29, 2014 There will probably be at least one epsilon - delta problem on the midterm and the kind of problems ask you to show1thatlimx af(x) =Lfor some particularfand particularL,using the actual definition of limits in terms of sand srather than the limit laws. For example , there might be a question asking you toshow thatlimx a7x+ 3 = 7a+ 3(1)orlimx 5x2 x 1 = 19,(2)using the definition of a The rules of the gameNormally, the answer to this kind of question will be of the following form:Given >0, let = [something positive, usually depending on anda]. If0<|x a|< then [some series of steps goes here], so|f(x) L|< .Some examples of this are Examples 2-4 of section Note that [some series of steps goeshere] should consist of a proof that|f(x) L|< , from the assumptions that >0 is whatever we said it was, and 0<|x a|< . , prove1In these kind of problems , much of the work goes into figuring out what should of this work is shown in the actual answer.

2 To clarify: in examples 2-4 of section ,each solution consists of two parts. Part 2 ( showing that this works ) is the actualanswer what you would turn in if asked this question on a homework or an exam. Part 1( guessing a value for ) is the bulk of the work done to produce this there s a sense in which you don t have to show your work in this kind of problem ; itsuffices to just write down the final answer. This is a little strange because for most mathproblems itisnecessary to show your work. For example , if there was a problem asking youto evaluatelimx 1x4 1x 1,it would not be acceptable to just write down 4. This would be unacceptable becausethere s no way for the person reading your answer to see why the limit should be 4. But ifthe answer to a question is a proof, rather than a number or an expression, then the readercan see directly whether or not the answer is correct, because the correctness of a proof isself-evident.

3 In problems where the answer is a number or an expression, when we say showyour work we really mean show that the answer is correct. For example , a more correctanswer to limx 1(x4 1)/(x 1) would belimx 1x4 1x 1= limx 1(x3+x2+x+ 1)(x 1)x 1= limx 1(x3+x2+x+ 1)= 13+ 12+ 1 + 1 = first step is just rewriting the thing whose limit is being taken. The second step is usingthe fact that limx 1only looks at values ofxthat aren t 1, for which we can cancel out thefactors of (x 1). The third step is the direct substitution principle for polynomials, andthe last step is basic Common mistakesFrom looking through people s homework, I got the impression that the following mistakeswere common: Dividing by zero, or treating as if it were an actual number. Writing things likelimx 1x4 1x 1=x3+x2+x+ 1 = limx 1x4 1x 1, the variablexis a bound variable. To paraphrase Wikipedia, thereis nothing calledxon which limx 1(x4 1)/(x 1) could depend.

4 It doesn t makesense to say that the limit is equal tox3+x2+x+ 1, because what isx? Not specifying what you chose to be! If you don t do this, it s really unclear whatyou re ultimately trying to Confusing the preliminary analysis to figure out , with the actual answer (the proof),or flat out omitting the actual answer. Making depend onx. Perhaps you re trying to show thatlimx 0xx2+ 1= 0and so you need to show that for every >0 there is a >0 such that|x|< implies|x/(x2+ 1)|< . You note xx2+ 1 < |x||x2+ 1|< |x|<|x2+ 1| ,so you would like to take to be|x2+ 1| .But you can t, since the rules of the - game say that you have to specify beforebeing told whatxis. In this case, you need to find a which will be guaranteed tobe less than|x2+ 1| . Since|x2+ 1|is always at least 1, you could take = /2 orsomething Strategies for finding deltaOne general strategy is to try Solving |f(x) L|< forx. Once you know what values ofxwill work, you choose so that the interval (a ,a+ ) sits inside the set of example , suppose you re trying to prove that limx 83 x= 2.

5 Given >0, you needto find >0 such that0<|x 8|< = |3 x 2|< .One approach is to just solve the inequality|3 x 2|< forx, as follows:|3 x 2|< 2 <3 x <2 + (2 )3< x <(2 + )3In order for (8 ,8 + ) to sit inside the interval from (2 )3to (2 + )3, one needs(2 )3 8 and 8 + (2 + )3,or equivalently 8 (2 )3and (2 + )3 the biggest value of that would work is = min{8 (2 )3,(2 + )3 8}.3 Iff(x) is a polynomial or a nice enough rational function, so thatL=f(a), then anotherapproach is to look atf(x) f(a)x you can find some constantCand guarantee that f(x) f(a)x a C,then it s safe to take = /C, because then|x a|< = |f(x) L|=|f(x) f(a)|=|x a| f(x) f(a)x a < C= .In practice, one usually can t find such aCwithout assuming thatxis bounded. But this isokay, because we can always take to be the smaller of two numbers. IfConly works whenxis within 1/2 ofa, we just take to be the minimum of 1/2 and example , suppose you re trying to show that limx 1x3 2x= 1.

6 Look atx3 2x+ 1x the numerator, this is(x 1)(x2+x 1)x 1which is the same thing asx2+x 1, sincexis not 1. Nowx2+x 1 could be pretty if we decide thatxwill be within 1/2 of 1, then|x|is at most 3/2. So|x2+x 1| |x|2+|x|+|1| 9/4 + 3/2 + 1 = 19/4< it turns out that we can take to be min(1/2, /5).This kind of approachalwaysworks for polynomials, and often works for rational taking limits of rational functions, it helps to remove any discontinuities that example , the first step in analyzinglimx 1x2 1x 1is to replace it with the equivalent expressionlimx 1x+ way of thinking about these problems is to keep track of what things can bemade small (because they have limit 0), and what things can be bounded (because they havesome finite limit, or at least don t have limit infinity).4 For example , if you re trying to prove using - that limx 0x(cosx)(x2+ 1) = 0, then thegoal is to makex(cosx)(x2+ 1) be really small. This is a product of three things.

7 The first,x, can be made arbitrarily small, because limx 0x= 0. On the other hand limx 0cosxand limx 0(x2+ 1) are nonzero, so we shouldn t expect to make those small. But theydo approach finite limits, so we can at least make them be bounded: by choosing smallenough, we can ensure that cosxwill be at most 2 (duh), and thatx2+ 1 will be at most 2,because 2>limx 0(x2+ 1).So of the three factors inx (cosx) (x2+ 1), we can make the first one as small as welike, and the second and third be as small as 2. We want the product to be smaller than ,so we should make the first one be as small as now we just need to choose to ensure that|x|< /4, that|cosx|<2, and that|x2+ 1|<2. The first condition is ensured by /4. The second is ensured by anything;it s always true. The third is ensured by, I guess, taking 1/2. We need to take thesmallest of these three values of , so we take = min( /4,1/2).

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