Seunghee Ye Ma 8: Week 4 Oct 20 Week 4 Summary This week, we will move on from our discussion of sequences and series to functions. Even though sequences and functions seem to be very different things, they very similar. In fact, we should think of functions as a generalization of sequences. In Section 1, we will define continuity and limit of functions. Then, we will look at a few examples to become familiar with the new definitions. In Section 2, we will go over Theorems that will serve as useful tools in studying continuity of functions. In Section 3, we will talk about big O and little o notation. Then, we will define differentiation. Topics Page 1 Continuity and Limits of Functions 1 1.1 Definitions . .1 1.2 Examples . .2 2 Continuity Theorems 5 2.1 Continuity: Useful Tools . .5 2.2 Extreme/Intermediate Value Theorems . .6 3 Differentiation 7 3.1 Big O and Little o . .7 3.2 Differentiation . .8 1 Continuity and Limits of Functions 1.1 Definitions First, let's define what a function is: Definition (Function) A function, f, from the real numbers to real numbers is a map which assigns to a real number x 2 R a unique real number y 2 R. If f is a function from the reals to reals, we will write f : R ! R. Some examples of functions are: • Polynomials e.g. f(x) = x2 + x + 3 • Trig functions e.g. f(x) = sin x • Exponential functions e.g. f(x) = ex • Constant functions e.g. f(x) = 28 Recall that a sequence converging to a sequence meant that as n goes to infinity, the terms got closer and closer to a number, L, which we called the limit. The analog for functions is continuity and limit of a function as x approaches a particular real number a. How did we learn continuity in school? Or what does \continuous" mean in English? One way to describe a continuous function is being able to draw the graph without taking the pen off the paper. Clearly this isn't a mathematically acceptable definition. So let's first give the precise definition of continuity and then try to unpackage what it really says: Page 1 of 10 Seunghee Ye Ma 8: Week 4 Oct 20 Definition (Continuity and Limit) Let f : R ! R be a function. Let a 2 R. Then, L 2 R is called the limit of f as x approaches a, if for any " > 0, we can find δ > 0 such that if jx − aj < δ, we have jf(x) − Lj < ". If L is the limit of f as x approaches a, we will write lim f(x) = L x!a Moreover, if L = f(a), we say that the function, f, is continuous at a. If L does not exist or L 6= f(a), we say that f is discontinuous at a. Let's take a closer look at the definition and try to understand what the definition is really trying to say. Again, we have " > 0, as we did before. But instead of looking for N 2 N with some magical properties, we are now required to find δ > 0 with some similar magical properties. In words, this is what the definition is saying: Suppose we are given a very tiny interval around L (namely the interval (L − "; L + ")). Then, we are trying to find an interval around our favorite point a (namely the interval (a − δ; a + δ)) such that our function f maps this interval into the given interval i.e. f(x) 2 (L − "; L + ") for all x 2 (a − δ; a + δ). If we can do this for any tiny interval around L that gets thrown to us, we say f approaches L as x approaches a. Pictures are worth thousand words (even though mathematicians would try to argue that it isn't true). Let's take a look at Figure 1.1. In Figure 1.1, we have drawn the graph of a function, say f(x). If we wanted to show that lim f(x) = L, we would need to show that for any blue interval around L, we can find a x!a small enough green interval around a such that the part of the graph of f(x) which lies over the green interval sits completely within the blue band. Figure 1: Idea behind limit and continuity 1.2 Examples As we already saw with sequences, working directly with definitions involving " can prove to be quite tricky. Let's look at a few examples to get some practice with this type of proofs. We start with an appetizer. Example 1.1. Let f : R ! R be f(x) = x. Show that f is continuous for all a 2 R. Proof. If we gave a definition of continuity such that f(x) = x (called the identity function) is not contin- uous, we should probably change the definition of continuity! Fortunately, this won't be a problem for our definition. Page 2 of 10 Seunghee Ye Ma 8: Week 4 Oct 20 Let " > 0. Then, we would like to find δ > 0 such that jx − aj < δ ) jf(x) − f(a)j < " But note that f(x) = x and f(a) = a. So we can simply take δ = ". Then, we have: jx − aj < δ ) jf(x) − f(a)j = jx − aj < δ = " Hence, f satisfies the definition of continuity at each a 2 R. That was too easy. So let's move to a slightly more difficult example. 1 Example 1.2. Show that the function f(x) = x2 is continuous for all nonzero a 2 R. Proof. As was the case before, we start by examining the quantity in the absolute value. In our case that is 1 1 jf(x) − f(a)j = − (1) x2 a2 2 2 a − x = (2) a2x2 (a − x)(a + x) = (3) a2x2 a + x = ja − xj · (4) a2x2 a + x = jx − aj · (5) a2x2 This is the quantity that we want to make small by picking x close to a. But note that when x is close to a a, we can bound the second term. More precisely, if x is within 2 of a, we get: 3a a + x a + 2 ≤ (6) a2x2 a2x2 5a ≤ 2 (7) 2 a 2 a 2 5a 10 2 = 4 = (8) a a3 4 a a So far, we have shown that if x is within 2 of a, i.e. jx − aj < 2 , we have a + x 10 jf(x) − f(a)j = jx − aj · ≤ jx − aj · a2x2 a3 Eventually, we want to find δ such that jf(x) − f(a)j < ". It suffices to get 3 3 10 a "ja j jx − aj · < " ) jx − aj < " · = a3 10 10 We are almost ready to start our proof. Note that so far we have imposed two conditions on x with regard a "ja3j to how close it has to be to a. First, we wanted jx − aj < 2 . Then, at the end, we had jx − aj < 10 . Our 1 a "ja3j δ will be the minimum of these two values i.e. δ = 2 min( 2 ; 10 ). Let's start the proof: Page 3 of 10 Seunghee Ye Ma 8: Week 4 Oct 20 a "ja3j a Let " > 0. Let δ = min( 2 ; 10 ). Now, suppose jx − aj < δ. Since δ ≤ 2 , we know that 10 jf(x) − f(a)j ≤ jx − aj · a3 "ja3j But note that δ < 10 , we get 3 10 "ja j 10 jx − aj · < · = " a3 10 a3 Therefore, we conclude that whenever jx − aj < δ, we have jf(x) − f(a)j < ". Thus, f is continuous at a. Since a was chosen to be an arbitrary nonzero real number, we conclude that f is continuous at any nonzero real number a. Now that we've seen two examples of continuous functions, let's take a look at a function that is discontin- uous. But before we do that, we introduce a lemma that will help us prove that a function is discontinuous. Lemma 1.1. Suppose f : ! is a function. Then, lim f(x) = L. if and only if for any sequence of real R R x!a numbers fang such that lim an = a, we have n!1 lim f(an) = L n!1 Conversely, lim f(x) 6= L if and only if there exists a sequence fang such that x!a • lim an = a; and n!1 • lim f(an) 6= L n!1 Example 1.3. Consider the function, f : R ! R defined as follows: ( sin 1 x 6= 0 f(x) = x L x = 0 where L is a real number. Show that f(x) is discontinuous at 0 for any value of L. Proof. To show that f(x) is discontinuous at 0 for any value of L, we must show that lim f(x) 6= f(a) = L x!a for any L 2 R. In other words, we must show that f does not have a limit as x approaches 0. We will try to use Lemma 1.1 to show this. Hence, we must find a sequence fang converging to 0 such 0 that ff(an)g does not converge to L. Hence it suffices to find two sequences fang and fang, both converging 0 to 0, but such that ff(an)g and ff(an)g converge to distinct limits. 1 Let's first consider an = 2nπ . Then, we have 1 lim an = lim = 0 n!1 n!1 2nπ Now, we look at the sequence ff(an)g: 1 lim f(an) = lim f = lim sin(2nπ) = lim 0 = 0 n!1 n!1 2nπ n!1 n!1 0 2 For the second sequence, we consider an = (4n+1)π .