Definition of a Limit

Definition of a Limit Suppose you have decided to volunteer (or are being paid to do so, if you need money as a motivator) to help with a juvenile sports team (either baseball or cricket - your sport of choice). Because of your adept mathematical and reasoning skills, your task is to help out with technical and strategical issues. During practice the coach of the team asks you to setup the pitching machine to 27 m/s (about 60 mph). Let the pitching machine be represented by the function f (x) = x3 , x = 6 3, where the domain of this function is the dial that sets the speed, and the speed of the pitch is the output of the function. Now you have a problem here, because the dial is broken, insofar that it skips over the value f (3) = 27, which is exactly the value you need! Rather than disappointing the coach and the team, you try to find a s olution t o t his problem. Realizing that by sight it is pretty difficult to judge the speed of a soaring projectile, you decide to set the knob to 3.1, presuming that the coach and team members won’t be able to tell the difference (30 m/s is close to 27 m/s after all). This might work at first, but the coach might bring out an instrument to measure the speed of the pitch. Also realizing that there is a limited accuracy to this device, you set the dial to 3.001 (so the output is about 27.03). If the coach brought out a more accurate instrument, you could just crank the dial closer and closer to 3 (granted that there would be physical limitations after some point). The idea here is that even though you cannot make the machine pitch at exactly 27 m/s, you can approximate 27 m/s as accurately as you would like by turning the nob close enough (but not equal to) 3. This is the basic idea of the limit - approximation. In this case we would write lim f (x) = 27 x→3 to mean that for this function f , we can approximate the output of 27 as accurately as we like by considering x values approaching (but not equal to) 3. Another useful observation is that when we look at the graph of x3 , x 6= 3, in a sense we can see what the value of the function ought to be at x = 3, even though the function is not defined there. In this sense, the limit as x → 3 is the value that the function ought to have at x = 3, regardless of whether or not it actually has the value at that point. We gain one advantage by thinking of a limit in this way. Intuitively, it only makes sense that a function ought to take on a single value at a given point (after all, if it were to take on multiple values it would not be a function). Due to this observation, we demand that a limit be unique, and if it is not possible to find a unique value so that a function ought to have that value at a given point x0 , then we would say the limit as x → x0 does not exist. A key aspect of mathematics is precision of language. Having developed some notion of a limit, we need to formulate a definition that encapsulates the properties a limit should have. Let’s examine a number of possible definitions, where they fall short, and how they lead us to the actual definition of a limit. Remember that we could define a limit differently, but then it would have different properties. The definition we want is the one that will give the limit the properties we want it to have (which may not yet be obvious at this time). In the following discussion, we will move from the most coarse, to most refined definition of the limit. Proposed Definition 0.1.1 (Limit). When we write limx→x0 f (x) = L we mean that as the input values x approach the point x0 , the output values of the function f (x) approach L. This is in a sense the most crude understanding of what a limit is. It is based on a perception that for a reasonable function, it seems like there is a value that the function ought to have at the point x0 , which is the limit L - whether or not the function is actually defined so that f (x0 ) = L. 1 However, there are some major problems with this definition. First of all, this definition does not indicate the fact that the value of the function at the point x0 , f (x0 ), has no bearing on whether or not the limit exists, or what it exists as. Second, the terms approach and approaches are much too vague for us to really work with them with any amount of precision. It is reasonable to assume that when we say approach, we mean get closer to. Thus, this definition says as the input values x get closer to x0 , the function values f (x) get closer to the limit value L. However, there is no mention of how close the function values need to get to the limit. Consider the example f (x) = −x2 , which is a parabola opening downward, with its vertex at the origin of the Cartesian coordinate plane. From the above discussion about the value the function ought to have, we know that lim (−x2 ) = 0, x→0 yet based on our current definition, we could just as well conclude that lim (−x2 ) = 5, x→0 or any other positive number. As x → 0, clearly f (x) gets closer to 5, which means that the limit is 5, according to the above definition. In this way we have an infinite number of different values for the limit, yet we want the limit to be unique. This is a serious problem with our definition, because there is nothing about the value 5 (or any other positive number) that makes it seem it ought to be the value of f (0). We will need to keep searching for a better definition. Proposed Definition 0.1.2 (Limit). When we write limx→x0 f (x) = L we mean that we can force the output values f (x) to be arbitrarily close to, but not necessarily equal to, the limit value L, by evaluating f at x arbitrarily close to, but not equal to x0 . This definition is a definite improvement upon the previous one. First, it illustrates the fact that the value of f (x0 ) has no bearing on the existence or value of the limit, because we don’t care about the point x = x0 . Second, it clears up the problem with the previous example, insofar that lim (−x2 ) 6= 5, x→0 or any other positive value that one might decide to choose (the function values don’t become arbitrarily close to 5). Nevertheless, this definition still does not guarantee uniqueness of a limit. For instance, consider the function |x| f (x) = , x which is 1 for x > 0 and -1 for x < 0. Using this definition we can conclude both lim f (x) = 1 x→0 and lim f (x) = 1, x→0 since we can choose x values arbitrarily close but not equal to 0, so that f (x) is both arbitrarily close (in fact equal to) 1 or -1. Although this definition is an improvement, it is still insufficient, as it does not guarantee uniqueness of a limit. Proposed Definition 0.1.3 (Limit). When we write limx→x0 f (x) = L we mean that we can force the output values f (x) to be arbitrarily close to, but not necessarily equal to, the unique limit value L, by choosing x arbitrarily close to, but not equal to x0 . 2 This is exactly the same definition as above, with a single addition - the word unique. Now we require that the limit value L be uniquely related to the function f and the point x0 in the way described by the definition - it cannot be related to another x1 in the same way. This is more promising, as we can conclude that limx→0 |x|/x does not exist, as there is no unique value that satisfies the relationship (there are two). This is a perfectly good definition of a limit, as it encapsulates all of the properties we have decided a limit ought to have. Nevertheless, we will work with another definition instead of this one, in order to simplify the language used (avoiding the term arbitrarily close), and to emphasize that limits are about approximation. Definition 0.1.4 (Limit). When we write limx→x0 f (x) = L we mean that L is the unique number that can be approximated as accurately as desired by evaluating the function f using x-values close enough to, but not equal to x0 . This definition is equivalent to the one above, insofar that they classify limits in the same way. Here we replace the term arbitrarily close with approximated as accurately as desired, but the two phrases have essentially the same meaning. For any distance away from L, we can force the function values to be within that distance; ie. approximate the limit value L with any degree of accuracy. When we use the words close enough to, we have the exact same meaning as sufficiently close or arbitrarily close to. At this point we could be satisfied with our definition of the limit, and agree to let this be our definition of the limit. However, if one does a little bit of reading he or she will probably not encounter this definition of the limit. The reason for this lies in the way mathematics is developed. Although we have a perfectly good definition of the limit, this definition is not ideal for making deductive arguments. Instead, we favor to recast this verbal definition in terms of variables, to arrive at the formal definition of a limit, which is found in most calculus texts. Definition 0.1.5 (Formal ε-δ Definition of the Limit). Let f be defined on an open interval containing x0 (except possibly at x0 ), and L ∈ R. We say that L is the limit of f as x approaches x0 , written lim f (x) = L x→x0 if for every number ε > 0, there exists a corresponding number δ > 0 such that for all x with 0 < |x − x0 | < δ we have |f (x) − L| < ε. Written in English, this definition reads: for any error tolerance of approximation (ε > 0), we can restrict the values f (x) to be within that error tolerance of L (|f (x) − L| < ε) by restricting the values of x to be sufficiently close but not equal to x0 (by choosing δ > 0 and considering x with 0 < |x − x0 | < δ). If we are able to meet any error tolerance of approximation, it follows that we are able to approximate our function as accurately as we like. Here the introduction of δ makes precise the notion of sufficiently close - we only care about the points that fall within the given δ-interval around x0 . We require 0 < |x − x0 | in order to exclude the point x0 from the interval of interest, because the limit does not depend on the function value at that single point. The way this definition classifies a limit as existing or not existing differs from our previous definition. Consider |x| lim . x→0 x According to our previous definiton the limit does not exist because the values we can approximate are not unique (we can approximate both 1 and −1). For the ε-δ definition of the limit, the limit 3 does not exist because if we try and put an error tolerance of ε < 2 around either of the limit values L = −1 or L = 1, any δ-interval around x0 = 0 will contain function values that fall outside of the required error tolerance. In our previous definition we needed to explicitly state uniqueness in order to properly classify limits; this is not required with the ε-δ definition. Uniqueness follows from the way we have formulated the definition. Our two definitions of the limit generally classify the same limits as existing or not existing, except for in very esoteric examples. Consider ( 1/x x ∈ Q f (x) = 0 x∈ / Q. This is a function that looks like 1/x when the input is a rational number, and 0 when the input is an irrational number. According to the first definition lim f (x) = 0, x→0 because 0 is the unique number we can approximate as accurately as desired. However, according to the ε-δ definition of the limit the limit does not exist, because for any error tolerance around 0, the function values on the rational numbers fall outside of the error tolerance. For an example such as this, our guiding principle of what value the function ought to have doesn’t really make much sense anymore. Thus, it is not entirely clear whether or not we would want a definition of the limit that classifies this limit as existing. Either way, the ε-δ definition of the limit is the definition we will work with, so for this function the limit as x → 0 does not exist. The formal ε-δ definition of the limit requires that the function be defined on an interval around the point x0 , which ensures that x0 is a reasonable point to even consider the notion of a limit. Consider a point in the domain of f with no other points around it. In such a case we could construct a small enough δ-interval so that it would only contain x0 , and the limit would trivially exist, even though it would be completely unrelated to the notion of the limit we are interested in. In figure 0.1 we have a graphical representation of the ε-δ definition of the limit. The point x0 is the point we are looking at the limit at, and the point L is the actual value for the limit. Here L + ε and L − ε signify the bounds of the error tolerance - any function value must be within these two bounds in order to satisfy the error tolerance. In the first figure the vertical lines denote the positions on the x-axis where the function values transition between satisfying and not satisfying the given error tolerance. Thus, for any choice of x in between them, the function value satisfies the error tolerance, signified by the shaded region. In the second figure a specific δ has been chosen so that all function values for points within the δ-interval satisfy the error tolerance. Note that this is just a sample interval, and that any smaller (and some larger) intervals would work as well. In order to make sense of the notion of approximation, we need to have a notion of distance. When we are looking at real numbers, the distance between two numbers x and y is simply |x − y|, the magnitude of their difference. Thus, both |f (x) − L| and |x − x0 | should be interpreted as the distance between points. When we are working with distances, the fundamental result we will build from is the triangle inequality. Theorem 0.1.1 (Triangle Inequality). For x, y ∈ R we have |x + y| ≤ |x| + |y|. Proof. |x + y|2 = (x + y)2 = x2 + 2xy + y 2 ≤ |x|2 + 2|x| · |y| + |y|2 ≤ (|x| + |y|)2 4 L+ε f (x) f (x) L+ε L L L−ε L−ε x0 x0 − δ x x (a) The shaded region represents points where the function satisfies the error tolerance. x0 x0 + δ (b) One possible choice for δ satisfying the given ε. Figure 1: Graphical interpretation of a limit. The result follows upon taking square roots of both sides of the equation. As stated before, it follows directly from the definition that a limit is unique. Theorem 0.1.2 (Uniqueness of Limits). Let f : R → R. limx→x0 g(x) = M , where L, M ∈ R. It follows that L = M . Suppose limx→x0 f (x) = L and Proof. Fix ε > 0, arbitrary. Since limx→x0 f (x) = L and limx→x0 g(x) = M there exist δL and δM so that for x with |x − x0 | < δL we have |f (x) − L| < ε/2 and for x with |x − x0 | < δM we have |f (x) − M | < ε/2. Choose δ > 0 so that δ < δL and δ < δM . Consider x with |x − x0 | < δ. It follows that |L − M | = |L + f (x) − f (x) − M | = |L − f (x) + f (x) − M | ≤ |f (x) − L| + |f (x) − M | < ε/2 + ε/2 = ε. Thus, for any ε > 0, |L − M | < ε. It follows that L = M . Suppose L 6= M . This would imply that |L − M | is a positive number, but then it could not be smaller than any arbitrary positive number ε. Although proof is not our emphasis here, it is worth taking a minute to look at how the above proof works. To begin, we fix an arbitrary ε. By fixing the ε we are saying that we want to look at an arbitrary value, but one that does not change throughout the course of the proof. If we can show that we meet this arbitrary error tolerance, then it easily follows we can meet any error tolerance, because the proof doesn’t depend on ε having a specific value - the proof works no matter what value is substituted for ε. Next, we know that limx→x0 = L, which means that for any arbitrary  > 0, we can find a δ so that |x − x0 | < δ =⇒ |f (x) − L| < . Namely, we know that if we let  = ε/2, which is our fixed value, we can find such a δ. Since we want to work with multiple δ-intervals, we call this one δL . Using the exact same logic we find a 5 δM . Now we choose a δ smaller than both δL and δM . This means that any x in this δ-interval will fall in both the δL and δM intervals, so it will satisfy the corresponding inequalities in f (x). Now we want to show L = M , and one way of doing so is showing that |L − M | is less than any positive number. If this is true, then the only possible conclusion is |L − M | = 0. Since we are working with approximations of arbitrary accuracy, this is going to be the best way to proceed, as trying to show directly that L = M will not allow us to use the information we already have. Now our goal is to write |L − M | in terms of quantities we already know are small, and if we overestimate in order to do so, it is no problem. The trick here is to add 0, or f (x) − f (x), which lets us use the triangle inequality to overestimate by known quantities, eventually allowing us to reach the desired conclusion. Although these techniques may seem a bit unfamiliar, they are very common and very standard to proving results involving limits. With time they will feel much more familiar. Example 1. Given f (x) = 2x + 1, is it possible to find a δ-interval around x0 = 2 so that for x with 0 < |x − 2| < δ it follows that |f (x) − 5| < ε, if ε = 0.5? Using this information can you conclude whether or not lim f (x) = 5 x→2 is a true statement? Solution We will begin with the inequality we want to satisfy, and try to deduce a restriction on x in order to satisfy it. |f (x) − 5| < ε |2x + 1 − 5| < 0.5 |2x − 4| < 0.5 2|x − 2| < 0.5 |x − 2| < 0.25 Since the inequality |f (x) − 5| < ε is true for |x − 2| < 0.25, it follows that δ = 0.25 will define an interval sufficiently small. Note that any smaller value for δ will also work, so we could just as well let δ = 0.1. Either way, we conclude that it is possible to find such a δ-interval. Although lim f (x) = 5 x→2 is in fact a true statement, being able to find δ for a single error tolerance (as we have just done) is insufficient information to make this conclusion; we would need to find the width of the δ-interval for every ε. In order to do so we would need to work with an arbitrary ε. Example 2. Given 1 f (x) = sin( ), x is it possible to find a δ-interval around x0 = 0 so that for x with 0 < |x| < δ it follows that |f (x) − 0| < ε, 6 if ε = 0.1? Using this information can you conclude whether or not lim f (x) = 0 x→0 1 1 0.5 0.5 sin(1/x) sin(1/x) is a true statement? Solution If we approach this problem algebraically, we run into difficulty trying to manipulate the inequality 1 | sin( ) − 0| < 0.1 x in order to reach a restriction for x. Instead, let us consider the problem graphically (see figure 2). 0 −0.5 −1 −2π 0 −0.5 −π 0 x π −1 −0.1 2π (a) Zoomed out view. −0.05 0 x 0.05 0.1 (b) Zoomed in view. Figure 2: Graph of the sewing machine function. Figure 2 shows that the function values oscillate between 1 and −1, and that oscillations become more rapid as x becomes closer to 0. Because these oscillations continue indefinitely close to x = 0, no matter how small of a δ-interval we define around x = 0 we will be able to find a point x within it such that f (x) = 1, which is outside of the error tolerance around 0. Recalling that we find that sin(x) = 1 when x = 1 sin( ) = 1 x when x = π + 2nπ, 2 n∈Z 1 , + 2nπ n ∈ Z. π 2 No matter how small of a δ-interval we consider, we will always find a point x as defined above, by considering sufficiently large values of n. Thus, in any interval around x0 = 0 there will be values of x such that 1 | sin( ) − 0| = 1 > 0.1. x We conclude that it is impossible to find a δ-interval so that the inequality | sin( x1 ) − 0| < 0.1 is satisfied for all x in the interval. Using this information we can conclude that 1 lim sin( ) 6= 0, x x→0 because we have found a single error tolerance ε = 0.1 for which we cannot find a corresponding δ-interval to satisfy the definition of the limit. In fact, it is not just that the limit does not equal 0, but that there is no value L that satisfies the limit (so the limit does not exist), which can be shown using a technique similar to the one we used above; we would simply fix a given error tolerance, 7 such as ε = 0.1, and show that in any δ-interval we can find two points where the function values are further than ε apart. It is probably easiest to choose points where the output values are 1 and −1, but other choices would work as well. Now let’s consider some examples where we want to prove the existence or nonexistence of a limit for a given function. In order to do so, we need to work with an arbitrary ε. The standard means of doing so is fixing an arbitrary ε, and then performing the analysis with that ε. For this fixed ε, we need to find a δ so that if we have any x with 0 < |x − x0 | < δ it follows that |f (x) − L| < ε, meaning that the values of the function within this δ-interval are within the error tolerance ε of the limit L. In general, δ will depend on ε (except for in some cases like f (x) = c). Example 3. Prove that lim f (x) = 3 if f (x) = 2x + 1. x→1 Solution Fix ε > 0, arbitrary. We must find a corresponding δ > 0 such that for 0 < |x − 1| < δ we have |f (x) − 3| < ε. We find that |f (x) − 3| = |(2x + 1) − 3| = |2x − 2| = 2|x − 1| and 2|x − 1| < ε ⇐⇒ ε |x − 1| < . 2 ε If we let δ = , then for an arbitrary x with 0 < |x − 1| < δ it follows 2 |f (x) − 3| = 2|x − 1| < 2 · ε = ε. 2 Thus, from the definition of the limit, it follows that lim f (x) = 3. x→1 Example 4. Prove that lim f (x) = 9 for x→3  f (x) = x2 x 6= 3 3 x=3 Solution Fix ε > 0, arbitrary. We must find a corresponding δ > 0 such that for 0 < |x − 3| < δ we have |f (x) − 9| < ε. Since 0 < |x − 3|, it follows x 6= 3, so f (x) = x2 . Thus, |f (x) − 9| = |x2 − 9| = |(x + 3) · (x − 3)| = |x + 3| · |x − 3|. We can control the value of |x − 3| directly with δ, but we will have to control |x + 3| indirectly. If we knew that δ would be less than 1, then we’d know from 0 < |x − 3| < δ that 3 < x < 5. Now the expression |f (x) − 9| = |x + 3| · |x − 3| < ε consists of two terms we can control. If δ < 1, then the maximum value of x is 5, so the maximum value of |x + 3| is 8. It follows, granted δ < 1, that if |x − 3| < ε/8, then |f (x) − 9| < ε. Thus, we choose δ = min(1, ε/8). Consider arbitrary x with 0 < |x − 3| < δ. For such an x, it follows |f (x) − 9| = |x + 3| · |x − 3| < 8 · 8 ε = ε. 8 1 6= 3. x→0 x Solution When we want to prove a limit is equal to a certain value, we need to work with an arbitrary ε, in order to show that our function values are close to the limit value for any error tolerance. In contrast, in order to show a limit does not equal a certain value, we need to find a single value for ε, so that for any δ-interval there is at least one point x within the interval with |f (x) − L| > ε, meaning that f (x) is outside of the error tolerance. This shows that it is impossible to restrict the function values to be as close as we want to the limit value by choosing input values close enough to x0 , by exhibiting a bad error tolerance, that we cannot meet. This particular function f (x) grows without bound as x → 0 from the right and decreases without bound as x → 0 from the left, so clearly the limit does not exist (note that we are not trying to show that the limit does not exist, only that it does not have a specific value). Now it is just our job to come up with a value for ε, and show that no matter how small someone chooses δ to be, they cannot meet this error tolerance. One possibility is to simply guess a very small number, such as ε = 10−10 and check if it will work. This strategy is not ideal however, because it complicates the mechanics of the problem. Since our function in this case grows without bound, we know it will exceed any error tolerance. Thus, we might as well choose a simple value for the error tolerance, and see if we can make progress from there. Let ε = 1. Now we consider δ > 0, arbitrary. At this point we need to use a little trick. We want to specify an x value within the δ-interval around 0, but we can’t simply specify a point like x = 0.2, because we don’t know how large δ is. Instead, we need to specify x in terms of δ. If x < 0, then we know that f (x) < 0, so we let δ x=− 2 which ensures both that x < 0, and that it falls in our δ-interval. For this x we have Example 5. Prove that lim 0 < |x − 0| = with | δ <δ 2 1 1 2 − 3| = | − 3| = 3 + > 1. x −δ/2 δ This proves that 1 6= 3, x because we have shown that no matter how close to x0 = 0 one creates a δ-interval, we can find points within where the function does not meet the error tolerance of 1 around the limit value L = 3. lim x→0 1 6= 10. x→0 |x| Solution We need to find a value for ε so that for any δ > 0 we can find an x with 0 < |x−0| < δ and |f (x) − 10| > ε. Looking at this function graphically we see this must be possible, but because we do not have the same separation of positive and negative values for f (x) as we cross the y-axis, the mechanics of this problem are a little bit more difficult. For the sake of illustration, let us solve this problem using ε = 0.5, rather than ε = 1. Consider δ > 0, arbitrary. Suppose we try to use the same trick as last time and define Example 6. Prove that lim δ x= . 2 9 Since δ is arbitrary, we don’t know what value it might have. If it happens that δ = 0.2, making x = 0.1, we would find 1 |f (x) − 10| = | − 10| = 0 < ε = 0.5, 0.1 so we would have failed in our task to show the limit value is not 10. There is nothing special about choosing δ x= , 2 and in fact we will run into the same problem no matter how we define x directly in terms of δ; if x = k · δ, then we will run into problems if it happens that δ = 0.1/k. No matter how we define x, there is a corresponding range of δ values that will cause us trouble. However, if we can guarantee that x will not be in this problematic range of values, and 0 < |x| < δ will be satisfied, then we can overcome this hurdle. We will need to restrict x close to 0 in order to accomplish this, but how small must we make x be? This depends on our choice of ε. Since we let ε = 0.5, in order to have |f (x) − 10| = | we will need to have |x| < 1 − 10| > 0.5 |x| 1 . 10.5 After considering arbitrary δ > 0, let us set x = min( |f (x) − 10| = | 1 δ , ). For such an x we find 10.5 2 1 − 10| > 0.5 = ε, |x| which proves 1 6= 10. x→0 |x| lim 10