How to Find Limits. Yilong Yang. October 22, The General Guideline 1

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1 How to Find Limits Yilong Yang October 22, 204 Contents The General Guideline 2 Put Fractions Together and Factorization 2 2. Why put fractions together Formula for factorization Advanced factorization [Optional] Eercises Simplification Tricks 5 3. Polynomial over Polynomial and the division trick Advanced division trick [Optional] Trig functions Conjugate trick for square roots Advanced Conjugate tricks for higher roots [Optional] Eercises Substitution [Optional] 4. Linear subs [Optional] Sub to change the it [Optional] Sub to change ugly epression [Optional] Universal trig formula [Optional] Eercises Sandwich Theorem 3 5. How to find a good sandwich Advanced sandwich [Optional] Sandwich to infinity [Optional] Eercises The General Guideline This article serves to illustrate and to summarize the most important tricks used to find its (without using L Hospital Rule). This was originally written for the preparation of the first midterm of MATH 3A. However, I m afraid that I m not very good at designing the difficulty level of problems. Some fo the eercises in this note might be too hard, and if you can t solve some of them, just don t worry about it. I bet the eam will be easier (or much easier) than those hard problems. For the tricks listed in this article, I would recommand you to use them in the following order. But you can of course use your own best judgement about which trick to use and in what order.

2 Procedure.. I recommand using the following procedure when you are asked to find a it:. See if you can plug-in directly (That is, check if the function is continuous at the point of interest.) 2. (Substitution to change its if necessary) 3. Put fractions together as much as possible, and then factorize if possible. 4. Use appropriate simplifying tricks. (for Polynomials over Polynomials, or for trig functions, or for square roots, etc.) 5. Use substitution wisely. 6. Try Sandwich Theorem for luck. In the following sections, I ll try to eplain each steps, the tricks involved, the formula that you should remember, and provide some eercises or typical problems. When a problem is marked with (HARD), then it is perfectly Okay if you cannot solve it. When a section is marked with [Optional], then it is perfectly Okay if you simply skip the section altogether. As for the answers of these problems, I personally don t think that they are necessary. As long as you know how to solve them, it makes no difference to me if you make any calculation mistakes. But if you must know then feel free to send me an about which answer you would like to see. 2 Put Fractions Together and Factorization 2. Why put fractions together When you see various fractions flying around (adding together, substracting each other, etc.), you should put them together using the following formula: Formula 2.. a b + c ad + bc = d bd a b c ad bc = d bd Sometimes, there might be hidden factors that can be cancelled out, but they only reveal themselves after you put all the fractions together. For eample, we were asked to solve this it in our homework: Problem 2.2. ( ) θ π 2 cos(2θ) cos 2 (2θ) Now we should put these fractions together. Let us do this the foolhardy way, using the formula in 2.. We see the following. 2

3 cos(2θ) cos 2 (2θ) = cos2 (2θ) + 2(cos(2θ) + ) (cos(2θ) + )(cos 2 (2θ) ) = cos2 (2θ) + 2 cos(2θ) + (cos(2θ) + )(cos 2 (2θ) ) (cos(2θ) + ) 2 = (cos(2θ) + )(cos 2 (2θ) ) (cos(2θ) + ) 2 = (cos(2θ) + )(cos(2θ) + )(cos(2θ) ) = cos(2θ) And then since this final epression is continuous at θ = π 2, we can simply plug-in θ = π 2, and see that the answer is Formula for factorization As you can probably see, the most important thing here is how to factorize an epression. Here are some really important formula that you should know. Formula 2.3. ( + ) 2 = ( ) 2 = ( + )( ) = 2 (a + b) 2 = a 2 + b 2 + 2ab (a b) 2 = a 2 + b 2 2ab (a + b)(a b) = a 2 b 2 ( + a)( + b) = 2 + (a + b) + ab ( a)( b) = 2 (a + b) + ab Remember here that in the following formula,, a, b might be changed into any sort of monstrosity. For eample, in our problem 2.2, we see that cos 2 (2θ) + 2 cos(2θ) + = (cos(2θ) + ) 2, and that cos 2 (2θ) = (cos(2θ) + )(cos(2θ) ). It can be super useful if you can factor most degree 2 polynomials by heart. I might or might not write up something about all the tricks of factoring degree 2 polynomials. Here I shall simply give you a bunch of problems to work on at the end of this section. 3

4 2.3 Advanced factorization [Optional] Finally, I d like to remark one more use of factorizations. Any substraction is in fact a factorization. To be more specific, we have the following important formula. Formula 2.4. a b = ( a + b)( a b) This is a fast way to get rid of the square roots sometimes. For eample, you might be asked to solve the following equation. Problem Of course, you can use the conjugate tricks as introduced in Section 3.4, or you can also use substitution as introduced in Section 4.3. But here is another way of showing this. Since 3 is a substraction, therefore there must be a factorization for it. We would of course want to factorize it into something involving + (as in the denominator). So we observe the following: 3 =( + ) 4 =( + 2)( + + 2) With this factorization, we can solve the problem easily. 2.4 Eercises 3 ( = )( + + 2) = 3 ( + + 2) =4 + 2 Find all the its below. Try to use factorization if you can figure out how. Problem 2.6. Problem 2.7. Problem 2.8. Problem e 2 sin(ln ) 2e sin(ln ) + e 2 sin(ln ) 8e sin(ln ) + 7 ( ) 2 4 4

5 Problem 2.0 (HARD). Problem 2.. Problem 2.2. ( π cos cot ) sin 4 π e sin 2 e sin Simplification Tricks There are various simplification tricks for various situations. So although this section strive to be as comprehensive as possible, if you find a situation that is not encompassed by this section, please send me an about it. 3. Polynomial over Polynomial and the division trick After step one where we put all fractions together, we might be looking at a single fraction where both the numerator and the denominator are polynomials, and goes to infinity. For eample, we might see this. Problem In this case, we see that 3 outgrows everything else, so we should only care about the coefficients of 3, and read off the answer 2 5. In this case, there is a trick to make this intuition more rigorous. We can divide both the numerator and the denominator by 3. Then we are left with the epression: 2 3/ 2 + / / + / 2 4/ 3 Then if we take the it as, we see that everything goes to 0 ecept for the 2 in the numerator and the 5 in the denominator (we are implicitly using the it law here). So the answer is Advanced division trick [Optional] When doing mathematics, you should always have a simple faith: whenever you have an intuition, then there must be a corresponding rigorous procedure to prove the intuition. And one of the most fundamental intuition about doing its is that, as goes to, some function will outgrow some other functions. And the division trick combined with the sandwich theorem, as shown in Section 3., is one way to make this rigorous. For eample, we might see the following problem. Problem sin 8 6 sin The intuition here is that as goes to, the function f() = must outgrow the function g() = sin a lot. So, according to our intuition, we are suppose to ignore the sin parts, and only care about the coefficient of. Then we can read off the answer as

6 To make this rigorous, what we do is to first observe the following using the division trick to divide both numerator and the denominator by : sin 8 6 sin = sin 8 6 sin sin Then I claim that we indeed have = 0. Assume that this is true for the moment, then we can see (by using the it law) that we can do the following sin 8 6 sin sin = 8 6 sin = sin sin 8 6 = = 3 8 (The Limit Law) sin Now, the only thing left to do is to figure out. This is sandwich theorem. Consider the inequality: 0 sin Since this inequality is always true, we can take it as goes to infinity, and the inequality should be preserved. So we would have sin 0 = 0 sin So we indeed have = 0, and we are done. The general idea is that, if a function f() outgrows a function g() by a lot, then we can use some sandwich theorem with the following sandwich: 0 g() some smart choice f() From there we can deduce that g() f() = 0. Then with the division trick, we will be able to show the it of the ratio of stuff involving f() and g(). As another illustration, consider the problem. Problem e 4e + 5 Since e will definitely outgrow, we can simply read off its coefficients and get the answer 3 4. We only need to show e = 0. We can use the sandwich theorem in the following way. First, since we only care about, we can assume that > 2 from now on. (The number 2 here is an arbitrary choice. I might as weel use 3,4,5,π, or whatever number, as long as the following inequality holds under such a condition.) Then the following inequality should hold: 6

7 e > 2 Then we see that the following sandwich will be true: 0 e 2 Now take on everything, and we see that e = Trig functions The only useful trig functions ever are sin, cos and tan. Out of the three, the more useful ones are sin and cos. And your best friend here should be sin. What I m trying to stress is this: Change everything into sin, and usually wonder will happen. The reason is the following super important formula that you should remember. Formula 3.4. sin 0 = This formula tells us that sin(blah) and blah are practically the same when this blah goes to 0. For eample, we have the following bunch of formulas, just to name a few. Formula 3.5. sin(2) = 0 2 sin( 2 ) 0 2 = sin( + 3) = sin(e ) (e = ) π 2 sin(cos ) cos Now suppose you transformed everything into sin. Then you will see something like the following problem: = Problem sin 2 () sin(3) sin(5) sin(4) The trick to tackle this is to replace each sin(blah) by sin(blah) blah blah. Then after using the it law, this will simply become blah, which is blah itself. Let do this in detail. We have the following derivations. 7

8 0 sin 2 () sin(3) sin(5) sin(4) = 0 [(sin )(sin )(sin(3)) = [( sin() )( sin() )( sin(3) 0 3 = ( (3) = = 3 20 ( ) ( ) ( ) ] sin(5) sin(4) 3) ( 5 sin(5) ) (This is the it law) 4 5 )( 4 sin(4) Now how eactly should we change all other trig functions into sin? It is simple to change all trig functions in term of cos and sin. So we only need to remember the following formula to change cos into sin. 4 )] Formula cos 2 = sin 2 2. cos(2) = 2 sin 2 3. cos = 2 sin 2 ( 2 ) 4. cos( π 2 ) = sin 5. cos( π 2 + ) = sin For the sake of completeness, here are some other trig identities you should remember. Formula sin( + π) = sin 2. sin( π) = sin 3. sin( ) = sin 4. sin( + π 2 ) = cos 5. sin( π 2 ) = cos 6. cos( + π) = cos 7. cos( π) = cos 8. cos( ) = cos 9. sin(2) = 2 sin cos 0. sin(a + b) = sin a cos b + sin b cos a. cos(a + b) = cos a cos b sin a sin b 2. tan( + π) = tan() 3. tan( π) = tan() 4. tan( + π 2 ) = tan 5. tan( π 2 ) = tan 8

9 6. tan(2) = 2 tan tan 2 7. tan(a + b) = tan a+tan b tan a tan b Finally we have the universal trig formula. Formula 3.9 (Universal trig formula). Let us use the letter t to denote tan( 2 ). Then we have the following:. sin = 2t +t 2 2. cos = t2 +t 2 3. tan = 2t t 2 The full power of the universal trig formula lies in the substitution. We shall talk about this in later sections. 3.4 Conjugate trick for square roots The conjugate trick for square root is essentially a smart use of the formula (a + b)(a b) = a 2 b 2. A typical problem may look like this: Problem To get rid of the square root, we can multiply on the numerator and the denominator. We would have: ( + 9 3)( ) = 0 2( ) ( + 9) 9 = 0 2( ) = 0 2( ) = 0 2( ) = 2 Here we can plug in the value = 0 because the last epression is continuous at = 0. Basically, when you see blah + blee, then multiply blah blee on the numerator and denominator. When you see blah blee, then multiply blah + blee on the numerator and denominator. 3.5 Advanced Conjugate tricks for higher roots [Optional] There is a higher degree generalization of the above formula. The generalization is based on the following two formula: Formula 3... a n b n = (a b)(a n + a n 2 b + a n 3 b ab n 2 + b n ) 2. a n + b n = (a + b)(a n a n 2 b + a n 3 b ab n 2 + b n ) when n is odd. For eample, let us take n = 3. Suppose we see the following problem: 9

10 Problem Recall that we have (a b)(a 2 + ab + b 2 ) = a 3 b 3. So we multiply on both numerator and denominator by We would have: 2 = ( 2)( ) ( )( ) ( 2)( ) = 2 ( + 5) 27 ( 2)( ) = 2 2 ( = 3 2 ) =2 Here we can plug in the value = 2 because the last epression is continuous at = Eercises Problem 3.3. Problem 3.4. Problem 3.5. Problem 3.6 (HARD). Problem 3.7. Problem 3.8. Problem 3.9. Problem e sin 4e π 2 4 tan + 3 π 2 9 tan + 8 π 2 sin(3) sin 2 (5) 0 sin(4) 2 sin(e 2 ) sin 2 () sin(e )e cot 2 (2) + 0 sin 2 (3)

11 Problem 3.2 (HARD). cos Substitution [Optional] 4. Linear subs [Optional] The most common type of substitution is linear substitution. Namely, when you see a it in terms of a variable, you can let t = a + b for some constant a, b, and then transform everything in terms of b. In particular, we have the following fact: Fact 4.. Let t = a + b. Then we have the following: f() = f( (t b)) p t ap+b a In general, suppose we let t = g() for some invertible function g, then we should have the following fact: Fact 4.2. Let t = g() for some invertible function g, and let g denote its inverse. Then we have the following: f() = p t g(p) f(g (t)) 4.2 Sub to change the it [Optional] Sometimes, it might happen that you see a it that you don t like. For eample, when we see lots of sin trig functions, we want to change all trig functions into sin and try to use the fact that 0 =. However, this might be awkward if we don t have 0. For eample, think about the following problem: Problem 4.3. π 2 cos π 2 When we see a it like this, we would like to change all cos into sin. But then we would like to have sin 0 so that we can use the formula 0 =. So the key here is to find something other than that goes to 0, and we substitute all by this something instead. What would go to 0? Well, if goes to a value a, then a will always go to 0. So we do the substitution by letting t = π 2 in our case. Then we would have t = 0, and we would substitute by t + π 2 everywhere. So we have: cos π 2 π 2 cos(t + π 2 = ) t 0 (t + π 2 ) π 2 cos(t + π 2 = ) t 0 t sin(t) = t 0 t = Here are some common way of substitutions and the corresponding change in it. Fact If p and t = p, then t 0.

12 2. If and t = a, then t. 3. If and t = k for some k > 0, then t. 4. If and t = k for some k < 0, then t. 5. If and t =, then t 0+. (The one sided it as t 0 + will be the same as the it ) 6. If and t = e, then t. 7. If and t = ln, then t. 8. If and t = e, then t If 0 and t = ln, then t. 0. If 0 + and t =, then t.. If 0 and t =, then t. 4.3 Sub to change ugly epression [Optional] Now another use of substitution is to get rid of ugly epression you hate. As a simple eample, let us look at this problem. Problem 4.5. sin(ln ) ln Now, we would really like to see sin instead of sin(ln ). So as a result, we can let t = ln and do the substitution. Then we would have: sin(ln ) ln sin t = t 0 t = Another eample is to get rid of square roots. For eample, let us look at problem 3.0 again. Problem The epression we really hate is + 9. So we let t = + 9, (and then we would have = t 2 9). Then we would have the following t 3 = t 3 2(t 2 9) t 3 = t 3 2(t + 3)(t 3) = t 3 2(t + 3) = 2 2

13 The moral of the story is this: whenever you see something you hate, simply substitute it away. 4.4 Universal trig formula [Optional] The universal trig formula refers to the formula 3.9. Namely, if we let t = tan( 2 ), then all trig functions in can be epressed as a polynomial over polynomial type thing. And this is really handy, since we can then use the division trick as introduced in section 3.. For eample, you may see the following problem. Problem 4.7. π 2 cos tan 2 Then we can do this by setting t = tan 2. With this substitution, we shall have: 4.5 Eercises π 2 cos tan( 2 ) (t 2 )/(t 2 + ) = t t t + = t t 2 + = Try to solve the following problem using substitution. Problem 4.8. Problem 4.9. Problem 4.0. Problem 4.. Problem Sandwich Theorem 5. How to find a good sandwich Let us take a good look at an eample first. Problem 5.. π 2 3π 2 sin π 2 sin + ( 3π 2 )2 sin(e ) + (ln(3π) ln 2) (e 3π 2 )2 ( sin( )) sin( ) 3

14 Now, the correct sandwich for this problem is the following inequality. 3 3 sin( ) 3 By taking it of 0, we see that both sides of the inequality would go to 0. So we see that 0 3 sin( ) = 0. The reason why this sandwich works is that sin(blah) always takes value between and. A bounded function is like a tamed little sheep, and they always get pushed around by other functions. In particular, we have the following fact: Fact 5.2. For it of the following two forms: blah sin(blee) somewhere We should always use the following sandwiches: blah cos(blee) somewhere blah blah sin(blee) blah blah blah cos(blee) blah To take things furthrer, we see that we don t really need to have sin or cos. All we need is a function bounded between and. Fact 5.3. Let f() be any function, and suppose g() is a function always bounded between and. For it of the following form: f() g() somewhere We should always use the following sandwich: f() f() g() f() As an eample, let us consider the following problem: Problem 5.4. ( 3) e e 3 3 e e 3 To solve this, we first observe that the function g() = e e 3 e e 3 we let f() = 3, and do the following sandwich: 3 ( 3) e e 3 e e 3 3 And by sandwich theorem we can see that the it we want to solve is 0. is always bounded between and. So 4

15 5.2 Advanced sandwich [Optional] Sometimes it might happen that although we have the form f() g(), and although g() is bounded, we have g() bounded between two number a, b instead of,. What should we do in this case? Let us first look at the following problem. Problem 5.5. ( 4)2cos( 4 ) 4 Now, we observe that since cos(blah), we must have 2 2 cos(blah) 2. So we let g() = 2 cos( 4 ), a bounded function, and let f() = 4, the function that we shall put into absolute value and used as the sandwich. But now we in fact have two situations. Suppose 4 0. Then we shall have 2 4 ( 4)2cos( 4 ) 2 4. On the other hand, if we have 4 0, we would have 2 4 ( 4)2 cos( 4 ) 2 4. To accomodate both situations at once, we should eventually choose to use the following sandwich: 2 4 ( 4)2 cos( 4 ) 2 4 And with this sandwich we can see that the it we want to compute is 0. To sum up, we have the following strategy: Fact 5.6. Let g() be a bounded function such that a g() b, and let f() be any function. Suppose we want to compute the following it: f() g() something Then if a b, we shall use the following sandwich: a f() f() g() a f() If a b, we shall use the following sandwich: 5.3 Sandwich to infinity [Optional] b f() f() g() b f() By sandwich to infinity, we mean to compare the growth rate of different functions. This arises in many different contet, for eample, when you try to do the advanced division trick as in section 3.2. All these advanced sandwiches will be determined by inequalities between functions. And the best way to measure the growth rate of a function is to compare it with the power functions. Let us start with a definition first: Definition 5.7. For two functions f(), g(), we consider the it L = a ] f() g(). If L =, then we say f() >> g() at = a. If L = 0, we say f() << g() at = a. If 0 < L <, then we say f g at = a. Let us first talk about growth rate when. Fact a << b when a < b. 2. e >> a for all a << ln << a for all a > 0. 5

16 4. sin << a for all a > cos << a for all a > a >> b for a > b. Here is a list of growth rate at = 0. Fact a << b when a > b << ln << a for all a > sin. 4. cos. Remember these growth rates will help you tremendously in finding good sandwiches. 5.4 Eercises Problem 5.0. Problem 5.. Problem 5.2. Problem 5.3 (HARD). Problem 5.4 (HARD). Problem 5.5 (HARD). Problem 5.6 (HARD). ( 2 cos(ln ) 9 4)e 5 (e ) ln( 2 sin ) π 2 π sin ln