function at the given point. Thus, for our previous example we would find that lim f(x) = 0

Transcription

1 Limits In order to introduce the notion of the it, we will consider the following situation. Suppose you are given a function, defined on some interval, except at a single point in the interval. What, if anything, can you say about the value of the function at that point? Let us suppose we are given the function f(x) = x 2 on [-1,0) (0,1] ie, the function is defined on [-1,1] except for at 0. What can we say about f(0)? Since we know the function is given by x 2 at all points except for 0, it seems that the function should be defined in the same way at 0. Thus, it seems like f(0) should be 0. However, a general function can be defined to have an arbitrary value at any point. Thus, both f(x) = x 2 on [-1,0) (0,1] with f(0) = 0 and f(x) = x 2 on [-1,0) (0,1] with f(0) = 1 are both equally valid functions (as well as any other value for f(0)). Thus, despite whatever intuition we have about what value f(0) should have, there is nothing about our function that demands it have that value, so in fact we cannot conclude anything about the value of f(0). Despite the fact we can conclude nothing about the actual value of f(0), we want to build on the notion that a function should have a certain function value at a certain point, which is determined irrespective of the actual value at that point (ie, what f(x) actually is doesn t determine what f(x) should be, in this notion of what should be). In this way, what a function value should be at a given point is determined only by the values of the function at the points around that point, but not the point itself. Just as we could say that a function should take on a specific value at a specific point based on an algebraic definition of the function, we can also explore the notion of what should be from a graphical standpoint. If we graph the function f(x) = x 2 on [-1,0) (0,1] it looks like a continuous function with a hole in the middle of it. The value that we previously determined should be f(0) is exactly the one which fills the hole. Thus, in some sense the function value that should be defined for f(0) is the one that fits in with the graph. Given that we can define a function in any way we like, it is not difficult to construct a function for which there is no value that fills in the hole. It is worth noting, however, that if there is a value which fills the hole, it will be unique. When we build the it concept from this, it will have both of these important properties - it will not necessarily exist, but if it does, it will be unique. The it is in some sense a refined version of the above concept. Intuitively, the notion of a it is the value a function approaches as its inputs approach a given point. Thus, when we write f(x) = L we mean that as the values x in the domain of the function approach the point x 0, the function values f(x) correspondingly approach the value L. As previously stated, the actual value of a function at a point does not determine in any way what it (if any) the function approaches (the function need not even be defined at the point in question). This is consistent with our previous notion of the function value that seemed like it should be the value of the

2 function at the given point. Thus, for our previous example we would find that f(x) = 0 (the it as x approaches 0 for our previous function is 0). It is important to emphasize that the value of a it is not just a value a function approaches at the given point, but a value it approaches in a very special way (surely this must be so, as an increasing function approaches all larger values, but not all larger values seem like they should be the right value for the function at a given point). The key to describing this special way that the function values approach the it value lies in approximation. Consider the following definition: we say that f(x) = L if for any error tolerance around L, we can restrict the values of f(x) to be within that error tolerance of L by restricting the values of x to be sufficiently close to x 0 ; ie. for all values of x within some interval around x 0 (excluding x 0 ) the distance between f(x) and L is within the error tolerance. Note that this requires we be able to find such an interval no matter how small the error tolerance is (but an error tolerance must be greater than 0, because we are still considering an approximation, just an approximation we can make as accurate as we like); in general, a tighter error tolerance will necessitate a smaller interval around the point x 0 we are interested in. Such an interval cannot contain even a single point for which the function value lies out of the error tolerance around L. While we must be able to find an interval for any error tolerance to show a function approaches a it at some point, we need only to find a single error tolerance for which there is no such interval in order to show a function does not approach a specific it L at some point (and we need to show this for all values L in order to show there is no it). Finally, it follows from this definition that a it is unique; that is, a function can only approach one value at a given point in this special way, if it approaches any value at all. This refined definition is very important, because most cases will not be as simple as the function we previously considered. We will revisit this notion later with the formal definition of a it, but before doing so we will attempt to gain further insight into its using the less daunting, intuitive notion. Our approach will be to begin with very simple functions, and use our understanding of them to calculate its of more complicated functions. Before doing so, it should be emphasized that a function need not have a it at a given point. Consider the function f(x) given by f(x) = x x which is defined for all x, with x 0. This function does not have a it as x approaches 0. Inuititively, the graph is broken with a jump in it, so there is no way for the function values to be approaching a single value at x = 0. The way to see this based on our definition is as follows: for any interval around 0 (an interval must be more than a single point), we can find an x 1 with x 1 < 0 and an x 2 with x 2 > 0. Thus, f(x 1 ) = 1 and f(x 2 ) = 1. For any error

3 tolerance less than 1, it is impossible to find a value L so that both of these function values (-1 and 1) are within the error tolerance around it, yet any interval around 0 must contain such values x 1 and x 2. Thus, this function has no it at x = 0. There are many functions which do not have its at certain points (or perhaps at any point), but fortunately most of these functions are not encountered in physical systems. Now let us consider some functions which do approach specific values in the special way defined by the it. Let us consider the function f(x) = c for all x. I propose that for any point x 0 f(x) = c Consider any error tolerance. Since the value of the function is always c, for any interval around f(x), no matter how large or small, the function value will be within that error tolerance. The above proposition follows. Now let us consider the function f(x) = x. Now I propose that for any point x 0 f(x) = x 0 Consider any error tolerance. If we restrict the values of x within that error tolerance of x 0, then similarly the function values will be within that error tolerance, as they are exactly the same as the values of x. The above proposition follows. Using similar, but more complicated mathematical arguments, in conjunction with the rigorous definition of the it, we can prove the following results. Note that these results follow from the way in which we have defined the it; thus, they are inescapable. By considering this notion of the it, it must have the following properties. Properties of Limits Suppose x x0 f(x) = L and x x0 g(x) = M where L, M R (L, M are real numbers). 1. Constant Multiple Rule: x x0 k f(x) = k L, k R 2. Sum Rule: x x0 [f(x) + g(x)] = L + M 3. Product Rule: x x0 [f(x) g(x)] = L M 4. Quotient Rule: x x0 f(x) g(x) = L M, M 0 5. Power Rule: x x0 [f(x)] m/n = L m/n, m, n Z, L m/n R Using these properties, in conjunction with the its of f(x) = x and f(x) = c at any point, we are now equipped to deal with the it of any polynomial at any point. Let us start

4 with a piecewise linear example, before moving on to higher order polynomials. Example 1 Consider the function f(x) defined below. Find the value of the it as x approaches any point in [0,4] if it exists, and state why it does not, if it does not exist. x 0 x < 1 0 x = < x 2 f(x) = 3 x 2 < x < x 4, x x = 3.5 Solution The first step is to graph this piecewise function, in order to obtain a better idea of what it looks like. Using the it laws and our results about f(x) = x and f(x) = c we can already determine that in the middle of an interval where f(x) is defined as a line the it will exist. Thus, we need to concern ourselves with the junctions of these intervals, and the points where there are holes in the graph. Since the value of the function at a given point does not alter its it, we can immediately conclude that the two holes in the graph are irrelevant, meaning that the it exists as it would at those points if there were no holes. At a junction where the function lines up (ie. there are no jumps in the graph) the it exists, as both sides are approaching the same value, in the fashion required by the it. Thus, the it exists at all points other than x = 3 where there is an actual jump in the graph of the function. Example 2 Find x 1 (3x 2 + x 4) Solution First we apply the sum rule to spread the it to each of the individual terms, and then we can evaluate it of the second two terms immediately. For the first, we use the constant product rule to separate 3 from x 2, and finally the product rule (using x x) or power rule to evaluate the it for x 2. Thus, x 1 (3x2 + x 4) = 3 x 2 + x + ( 4) = = 2 x 1 x 1 x 1 3x 2 5x 2 Example 3 Find x 2 x 2 Solution The first thing to notice is that the denominator is 0 when x = 2, so we cannot immediately apply the quotient rule. If we evaluate the numerator at x = 2 we see that = 0, so the numerator also has a 0 at x = 2. If a polynomial has a 0 at some point, it means that the polynomial factors with a linear term that is 0 at that point. Thus, we know there is a factor of (x 2) in the numerator. Now we need to find another linear term that will factor the original polynomial. (3x 2 5x 2) = (a + b)(x 2) Since we know a x = 3x 2 it follows a = 3x, and similarly b = 1. Thus, 3x 2 5x 2 x 2 x 2 = x 2 (3x + 1)(x 2) x 2 = x 2 (3x + 1) = 7

5 Note that above we were able to cancel the factors of (x 2) precisely because when we are considering the it, we are looking at points where x 2, so that term is nonzero (if we were faced with something of the form 0 we could not cancel the 0 s and conclude that it is 1). 0 Example 4 Find f(x) if f(x) = Solution We cannot use the quotient rule, because x = 0. However, if we simplify the expression to modify the factor of in the denominator, then we will be able to apply the quotient rule. We do this by multiplying by 1, where 1 is conveniently chosen to be the conjugate of the numerator over itself (which will remove the square root from the numerator) = = x ( ) = With this simplified expression, we can now take the it of the numerator and denominator, to find that f(x) = 0 There is one final theorem to discuss, which may be used to calculate a it of a function, even if we do not have an expression for that function. Theorem: The Sandwich Theorem Suppose g(x) f(x) h(x) for all x in an open interval about x 0, not necessarily containing x 0. If g(x) = h(x) = L x x0 Then f(x) = L Example 5 Find f(x) if for every x 0 Solution Since it follows from the sandwich theroem that This above analysis shows, for instance, that x 2 f(x) x 2 x2 = 0 = x 2 f(x) = 0 x2 sin( 1 x ) = 0