Limits and Continuity Philippe B. Laval Kennesaw State University January 2, 2005 Contents Abstract Notes and practice problems on its and continuity. Limits 2. Introduction... 2.2 Theory:... 2.2. GraphicalMethod... 3.2.2 NumericalMethod... 3.2.3 AnalyticalMethod... 3.3 Eamples... 4.3. NumericalMethod... 4.3.2 Direct application of the it rules or of one of the two theoremsabove... 5.3.3 Rational function for which both the numerator and denominator approach 0... 5.3.4 Fraction whose denominator and numerator go to 0 and whose numerator or denominator contains a radical... 6.3.5 Function whose behavior changes depending on whether >aor <a... 7.3.6 SqueezeTheorem... 8.4 LimitsInvolvingInfinity... 8.4. If f getsarbitrarilylarge... 8.4.2 Limits at infinity that is its when approaches or... 0.5 Sampleproblems:... 2 Continuity 3 2. Theory:... 3 2.2 Eamples... 5 2.3 Sampleproblems:... 6
3 Limit Laws 7 Limits. Introduction We are trying to study the behavior of a given function f ) near a given value = a. This is denoted by the symbol f). Itisread"itoff ) as approaches a". As you know, given a function y = f ), to each value of corresponds a unique value y = f ). As we plug in various values of, we obtain corresponding values for y. We are interested in knowing that if the values of follows a certain pattern, do the corresponding values of y follow a similar pattern? More specifically, the pattern we are interested in is values of such that approaches a given value a. This is denoted by a. As we will see, there are several possibilities. If y = f ) and a, then we could have:. The corresponding values of y also approach some number we will call L. In this case, we will write f) =L. Wealsosayy L as a y approaches L as approaches a). 2. The corresponding values of y get arbitrarily large go to infinity) or arbitrarily small go to minus infinity). In this case, we write f) = or f) =. 3. The corresponding values of y do not seem to follow any pattern. In this section, we will learn how to determine which situation we are in. Since a function can be given different ways graph, table, formula), we must learn how to do this in all these cases..2 Theory: Definition Limit of a function at a point) f) means that we are looking at the values of f ) for values of close to a but not equal to a. In fact, f may not even be defined at a. If we can make f ) as close as we want to a number L simply by taking closeenoughtoa, then we say that f) =L. Remark 2 When we say close to a, we must realize that can get closer to a from the right i.e. is getting closer to a but is larger than a). But, can also get closer to a from the left i.e. is getting closer to a but is smaller than a). Thus, we have the two additional its, which are called one-sided its Definition 3 One-sided its) When we say approaches a, we need to realize that can get close to a from the right as well as from the left. If we wish to consider only one side, we have the following one-sided its: 2
. f) means that we are looking at values of f ) for close to a + and larger than a to the right of a). 2. f) means that we are looking at values of f ) for close to a and smaller than a to the left of a). One condition for f) toeististhatboth f) and + f) eist and are equal. There are three methods used to evaluate the it of a function at a point:.2. Graphical Method In this case, the graph of the function is used to estimate the values of f ) as gets closer and closer to a. See your book on pages 02, 03 for nice pictures..2.2 Numerical Method In this case, we use the formula defining the function to compute the values of f ) as gets closer and closer to a. When I want students to use this method, I will ask the students to guess the it. It is a good idea to use two tables. One table will have the values of close to a but less than a. This will allow us to guess f). The other will have the values of close to a but larger than a. This will allow us to guess Eample 4 Guess 3 2 We construct the two tables mentioned above..9.99.999.9999 3 6.859 7.8806 7.988 7.9988 and f). Consider the following eample: + 2. 2.0 2.00 2.000 3 9.26 8.206 8.02 8.002 From the table, we see that as gets closer to 2, from either the right or the left, 3 is getting closer to 8. We conclude that 3 =8 2 However, you should use caution with this method. See eamples 4 and 5 on pages 04, 05..2.3 Analytical Method In this case, the it rules are used. See a list of the it rules at the end of this document). Using the it rules, the following two theorems can be proven Theorem 5 Limit of a polynomial) If f ) is a polynomial, then f) = f a). In other words,we just plug in the value of a in the polynomial. 3
Theorem 6 Limit of a rational function) If f ) is a rational function that is if there eists two polynomials p ) and q ) such that f ) = p ) q ) then f) =f a) providing q ) 0 In other words, we just plug in the value of a in the rational function as long as q a) 0..3 Eamples We look at several eamples and how to handle them. The way its are evaluated depends a lot on the type of function involved. The eamples below are classified into various categories. You should first establish which category the function falls under, then follow the steps described for that category. The cases below assume we are trying to evaluate f)..3. Numerical Method Students are often confused regarding the values of to try and how many to try. There is not a definite rule. The key is to try values which follow the given pattern. Enough velues should be tried to allow seeing a pattern if there is one. Let us consider a specific eample. Suppose that we are computing the it of some function as 2. For values of approaching 2 from the right, one could try 2., 2.0, 2.00, 2.000. These values are clearly getting closer and closer to 2. The values should get very close to 2, which these do. 3 or 4 values should be tried. If there is a pattern in the corresponding y values, 4 values will allow to see it. For values of approaching 2 from the left, one could try.9,.99,.999,.9999. Let us now look at specific eamples. 2 Eample 7 Guess We are supposed to try values of which get closer and closer to from both the left and the right. To do so, we build the following tables..0.00.000 2 2. 2.0 2.00 2.000 and.9.99.999.9999 2.9.99.999.9999 Thus, it appears that as, 2 write 2 =2. is getting closer and closer to 2. We 4
Eample 8 Guess sin ) π 0 We are supposed to try values of which get closer and closer to 0 from both the left and the right. To do so, we build the following tables..0.00.000 sin ) π and 0 0 0 0 -. -.0 -.00 -.000 sin ) π 0 0 0 0 So, we might be tempted to conclude that sin ) π 0 =0. This is not correct however. This it is not defined. See eample 4 on page 04. A good way to avoid this is to break the pattern in the values of we try. For eample, if we π ) had tried =.005 then we would have seen that sin =0.866 03 which.005 is not zero..3.2 Direct application of the it rules or of one of the two theorems above Eample 9 Find 3 5 +2 ) 0 This is a polynomial, to find the it, we simply plug in the point. Thus, 0 3 5 +2 ) =2 2 + Eample 0 Find 2 5 This is a rational function, its denominator is not zero when =2,thus,by the theorem on it of a rational function, we evaluate the it by plugging in the point. Thus, 2 + 2 5 = 22 + 2 5 = 5 3 = 5 3 Eample Find 3 22 This looks like rule #. Since 3 2 2 ) =7> 0, it follows that 3 22 = 7.3.3 Rational function for which both the numerator and denominator approach 0 In this case, none of the rules work. The theorem on rational function does not work either since the denominator goes to 0. Since both the numerator 5
and denominator go to 0 as approaches a, we can factor a from both the numerator and denominator. 2 Eample 2 Find This is indeed a rational function, both the numerator and denominator are 0 when is. We proceed as follows: 2 = ) +) = +) =2since + is a polynomial) factor the numerator) 2 + 6 Eample 3 Find 2 2 2 2 + 6 2 2 2 = 2) +3) 2 2) +) +3) = 2 +) = 5 3.3.4 Fraction whose denominator and numerator go to 0 and whose numerator or denominator contains a radical In this case, try to rationalize the part which contains the radical. This means multiplying by the conjugate. Recall, the conjugate of an epression of the form a + b is a b. We will use the identity a + b)a b) =a 2 b 2. 2 Eample 4 Find 4 4 4 2 4 2 +2 = 4 4 +2 2) +2) = 4 4) +2) ) 2 4 = 4 4) +2) 4 = 4 4) +2) = 4 +2 = 4 6
.3.5 Function whose behavior changes depending on whether >a or <a In this case, the one sided its f) and + f) must be evaluated. This is the case when we have a piecewise function and we are evaluating the it at the breaking point. It is also the case if we have absolute values and the point at which we are evaluating the it is the point where the epression inside the absolute values changes sign. The eamples below illustrate this: 2 Eample 5 Find 2 2 We note that 2 changes sign at 2, the point where we are evaluating the it. So, instead, we compute 2 2+ 2 = 2 2 2 if 2 + then >2, so 2 > 0 hence 2 = 2) =) 2 = and also 2 2 2 = 2) 2 2 = ) 2 = 2 Since the one-sided its are not equal, we conclude that 2 2 eist. does not Remark 6 If we had computed the it at any number other than 2, there would not have been a problem because in that case, 2 would not have changed sign. { + if >2 2 if <2 Eample 7 Find f ) where f ) = 2 Onceagain,thefactthatwewanttheitas approaches 2 is crucial. 2 is the point where the definition of f changes. So, we have to compute and f ) = +) 2+ 2+ =3 f ) = 2) 2 2 =0 7
Since the one-sided its do not agree, we conclude that 2 f ) does not eist..3.6 Squeeze Theorem Theorem 8 If f) g) h) for all in an open interval containing a, ecept possibly at a, and if f) = h) =L then g) =L This is used as follows. Supposed we are trying to evaluate the it of some function g, i.e. wewanttofind g). The function g issuchthatitdoes not fit any of the cases described above. We may try to find two functions f and h which satisfy the conditions of the theorem. If we do, then we will know g). ) Eample 9 Find 2 sin 0 Our goal is to find two functions, one larger, one smaller than 2 sin the same it as approaches 0. We start by noticing that ) sin ) 2 2 sin 2 since 2 is positive) ) having Clearly, ) 2 = ) ) 2 =0. By the squeeze theorem, 2 sin = 0 0 0 0.4 Limits Involving Infinity There are two cases to consider. First, the function could go to infinity i.e. get arbitrarily large) when approaches a finite number. Second, could go to infinity..4. If f gets arbitrarily large This happens when we have a fraction whose numerator approaches a finite number and whose denominator approaches 0. In this case, you will have to look at the one-sided its. These one-sided its will be ±, the sign being determined by the sign of the fraction. Also, remember that ln approaches as 0 + therefore, ln =. 0+ Eample 20 Find 0 2 The numerator of this fraction is non zero while the denominator approaches 0. 8
Thus, we look at and Thus 0 0+ 0 0 2 = 2 = 2 = Eample 2 Find The numerator of this fraction is non zero while the denominator approaches 0. Thus, we look at 0+ = and 0 = Thus does not eist 0 Eample 22 Find 2 2 The numerator of this fraction approaches a non-zero number while the denominator approaches 0. Thus, we look at 2+ 2 = and 2 2 = Thus does not eist 2 2 Definition 23 Vertical asymptote) We say that the line = a is a vertical asymptote of the curve y = f ) if one of the conditions below is satisfied:. f ) =± 2. f ) =± + 3. f ) =± Remark 24 The function ln has a vertical asymptote at =0. The function tan has vertical asymptotes at ± π 2, ±3π 2, ±5π 2,... See your book on page 32 for pictures of asymptotes. 9
.4.2 Limits at infinity that is its when approaches or In this case, the it could be finite, infinite or may not eist. Definition 25 Horizontal asymptote) We say that the line y = a is a horizontal asymptote of the curve y = f ) if one of the conditions below is satisfied:. f ) =a 2. f ) =a Theorem 26 The following its at infinity are important:. =0if n is a positive integer. ± n 2. ± n = if n is an even positive integer. 3. ± n = ± if n is an odd positive integer. 4. e = 5. e =0 6. ln = 7. The its of the trigonometric functions do not eist at infinity. Remark 27 When finding the it at infinity of a polynomial or a rational function, factor the term of highest degree. See eamples below. Remark 28 When evaluating its, if you find one of the forms below, you won t be able to conclude. You will have to try to simplify further. These forms are called indeterminate forms because their behavior is not predictable.. 2. 0 3. 0 4. 0 Eample 29 Find 2 ) If we try to simply evaluate, we get. Remember, this is not 0. The correct way to do it is as follows: 2 ) = 2 )) = 2 = 0
2 +5 3 Eample 30 Find 3 2 +2 2 2 + 5 +5 3 3 2 +2 = 3 ) 2 3 2 + 2 ) 3 2 = 3 2 = 3 Eample 3 Find the vertical and horizontal asymptotes of f ) = + Vertical asymptotes. We know these happen if f goes to infinity when approaches a finite number. Since f is a fraction, it will go to infinity if its denominator goes to 0 and not its numerator. It is easy to see that the denominator of f goes to 0 when approaches. The line =is a vertical asymptote. Horizontal asymptotes. These are found by computing f ) and f ) + f ) = = = ) + ) Similarly, f ) = + = ) + ) = So, the line y =is a horizontal asymptote..5 Sample problems:. Do#3,5,7,3onpages08,09
2. Do#,2,3,7,9,,5,24onpages7,8 3. 2 + Guess use table of values) answer: does not eist) 0 4. 3 Guess use table of values) answer: ) 2+ 2 5. Evaluate 4 2 +2 8 2 2 +2 8 6. Evaluate 2 2 3 2 7. Evaluate 2 4 3 2 8. Evaluate 2 2 4 answer: 8.0) answer: 6.0) answer: 0.0) answer: 9. Evaluate answer:.0) 3 ) 6 0. Evaluate answer: does not eist) if <0. Let f ) = 2 if 0 < 2 8 if >2 a) Evaluate f) answer.0) b) Evaluate f) answer: 0.0) 0 c) Evaluate f) answer: 0.0) 0+ d) Evaluate f) answer: 0.0) 0 e) Evaluate 2 f) answer: 4.0) f) Evaluate f) answer: 6.0) 2+ g) Evaluate f) answer: does not eist) 2 3 2 2. Find 4 5 answer: 0.0) 4 3 2 3. Find 3 answer: 4.0) 5 + 4. Find the asymptotes of f ) = 2 answer: horizontal asymptotes: y =0, vertical asymptotes: = and =) 2
2 Continuity 2. Theory: Definition 32 Continuity) Afunctionf is said to be continuous at = a if the three conditions below are satisfied:. a is in the domain of f i.e. f a) eists ) 2. f ) eists 3. f ) =f a) Definition 33 Continuity from the left) A function f is said to be continuous from the left at = a if the three conditions below are satisfied:. a is in the domain of f i.e. f a) eists ) 2. f ) eists 3. f ) =f a) Definition 34 Continuity from the right) A function f is said to be continuous from the right at = a if the three conditions below are satisfied:. a is in the domain of f i.e. f a) eists ) 2. f ) eists + 3. f ) =f a) + Definition 35 Continuity on an interval) Afunctionf is said to be continuous on an interval I if f is continuous at every point of the interval. If a function is not continuous at a point = a, wesaythatfis discontinuous at = a. When looking at the graph of a function, one can tell if the function is continuous because the graph will have no breaks or holes. One should be able to draw such a graph without lifting the pen from the paper. In general, functions are continuous almost everywhere. The only places to watch are places where the function will not be defined, or places where the definition of the function changes breaking points for a piecewise function). Thus, we should watch for the following: Fractions: watch for points where the denominator is 0. Be careful that functions which might not appear to be a fraction may be a fraction. For eample, tan = sin cos. 3
Piecewise functions: When studying continuity of piecewise functions, one should first study the continuity of each piece by using the theorems above. Then, one must also check the continuity at each of the breaking points. Functions containing absolute values: Since an absolute value can be written as a piecewise function, they should be treated like a piecewise function. Theorem 36 Assume f and g are continuous at a and c is a constant. Then, the following functions are also continuous at a:. f + g 2. f g 3. cf 4. fg f 5. if ga) 0 g Theorem 37 Any polynomial function is continuous everywhere, that is on, ). Theorem 38 Any rational function is continuous everywhere it is defined Theorem 39 If n is a positive even integer, then n is continuous on [0, ) Theorem 40 If n is a positive odd integer, then n is continuous on, ) Theorem 4 The trigonometric functions and their inverses, the natural logarithm functions and the eponential functions are continuous wherever they are defined. One of the important properties of continuous functions is the intermediate value theorem. Theorem 42 Intermediate Value Theorem) Suppose f is continuous in the closed interval [a, b] and let N be a number strictly between fa) and fb). Then, there eists a number c in a, b) such that fc) =N This theorem is often used to show an equation has a solution in an interval. Eample 43 Show that the equation 2 3=0has a solution between and 2. Let f ) = 2 3. Clearly, f is continuous since it is a polynomial. f ) = 2, f 2) =. So, 0 is between f ) and f 2). By the intermediate value theorem, there eists a number c between and 2 such that f c) =0. 4
Another important property of continuous functions is that it makes computing its easier. By the third condition in the definition of its, we know that if f is continuous at a then f) =fa). This means that if f is continuous at a,to compute f), we just plug in a in the function. 2.2 Eamples. Is f ) = 2 +5 continuous at =2? f is a rational function, so it is continuous where it is defined. Since it is defined at =2,itiscontinuousthere. 2. Is f ) = 2 +5 continuous at =? f is not defined at =, hence it is not continuous at. { + if 2 3. Let f ) =.Isfcontinuous at =2, =3? 2 if <2 Continuity at =2 Since f is a piecewise function and 2 is a breaking point, we need to investigate the continuity there. We check the three conditions which make a function continuous. First, f is defined at 2. Net, we see that f ) =0and 2 2+ f ) =3. Thus, f ) does not 2 eist. f is not continuous at 2. Continuity at 3. When is close to 3, f ) = +, which is a polynomial, hence it is continuous. 4. Find where ln 5) is continuous. The natural logarithm function is continuous where it is defined. ln 5) is defined when 5 > 0 >5 So, ln 5) is continuous on 5, ). 5. Find where f ) = is continuous. +2 Since f is a rational function, it is continuous where it is defined that is for all reals ecept = 2. { + if >2 6. Find where f ) = is continuous. 2 if <2 When >2, f ) = + is a polynomial, so it is continuous. When 5
<2, f ) = 2 is also a polynomial, so it is continuous. At =2, f is not defined, so it is not continuous. Thus, f is continuous everywhere, ecept at 2. 2.3 Sample problems:. Do # 3, 5,, 25 on pages 28, 29 if <0 2. Is f ) = 2 if 0 < 2 continuous at 0,, 2. Eplain. 8 if >2 answer: not continuous at 0 and 2, continuous at ) 2 + 3. Find where f) = 2 is continuous. answer: Continuous for + 6 all reals ecept at =2and = 3) 4. Find where ln 2 ) is continuous. answer: continuous on, 2)) 5. Use the intermediate value theorem to show that 5 2 4 3=0has asolutionin2, 3). 6. Use the intermediate value theorem to show that there is a number c such that c 2 =3. This proves the eistence of 3. 6
3 Limit Laws Suppose that c is a constant, n is a positive integer, and the its f) and g) eist. Then:. [f)+g)] = f) + g) 2. [f) g)] = f) g) 3. cf) =c f) 4. [f)g)] = [ ][ f) ] g) f) f) 5. g) = if g) 0 g) 6. [f)] n = [ ] n f) 7. c = c 8. = a 9. n = a n 0. n = n a If n is even, we must also have a 0. n f) = n f) If n is even, we must also have f) 0 7