1 Continuity and Limits of Functions

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1 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, we will define continuity and it of functions. Then, we will look at a few eamples 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 Continuity and Limits of Functions. Definitions Eamples Continuity Theorems 5 2. Continuity: Useful Tools Etreme/Intermediate Value Theorems Differentiation 7 3. Big O and Little o Differentiation Continuity and Limits of Functions. 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 R a unique real number y R. If f is a function from the reals to reals, we will write f : R R. Some eamples of functions are: Polynomials e.g. f( = Trig functions e.g. f( = sin Eponential functions e.g. f( = e Constant functions e.g. f( = 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 it. The analog for functions is continuity and it of a function as 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 of

2 Definition (Continuity and Limit Let f : R R be a function. Let a R. Then, L R is called the it of f as approaches a, if for any ε > 0, we can find δ > 0 such that if a < δ, we have f( L < ε. If L is the it of f as approaches a, we will write f( = L Moreover, if L = f(a, we say that the function, f, is continuous at a. If L does not eist or L 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 ε > 0, as we did before. But instead of looking for N 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 ε, L + ε. 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( (L ε, L + ε for all (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 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.. In Figure., we have drawn the graph of a function, say f(. If we wanted to show that f( = L, we would need to show that for any blue interval around L, we can find a small enough green interval around a such that the part of the graph of f( which lies over the green interval sits completely within the blue band. Figure : Idea behind it and continuity.2 Eamples As we already saw with sequences, working directly with definitions involving ε can prove to be quite tricky. Let s look at a few eamples to get some practice with this type of proofs. We start with an appetizer. Eample.. Let f : R R be f( =. Show that f is continuous for all a R. Proof. If we gave a definition of continuity such that f( = (called the identity function is not continuous, we should probably change the definition of continuity! Fortunately, this won t be a problem for our definition. Page 2 of

3 Let ε > 0. Then, we would like to find δ > 0 such that a < δ f( f(a < ε But note that f( = and f(a = a. So we can simply take δ = ε. Then, we have: a < δ f( f(a = a < δ = ε Hence, f satisfies the definition of continuity at each a R. That was too easy. So let s move to a slightly more difficult eample. Eample.2. Show that the function f( = 2 is continuous for all nonzero a R. Proof. As was the case before, we start by eamining the quantity in the absolute value. In our case that is f( f(a = 2 a 2 ( = a 2 2 a 2 2 (2 = (a (a + a 2 2 (3 = a a + a 2 2 (4 = a a + a 2 2 (5 This is the quantity that we want to make small by picking close to a. But note that when is close to a, we can bound the second term. More precisely, if is within a 2 of a, we get: a + a 2 2 So far, we have shown that if is within a 2 of a, i.e. a < a 2, we have f( f(a = a a + a 2 2 a a 3 Eventually, we want to find δ such that f( f(a < ε. It suffices to get = a a 3 < ε a < ε a 3 = ε a3 a + 3a 2 a 2 2 (6 5a 2 a ( 2 a 2 (7 2 5a 2 = (8 We are almost ready to start our proof. Note that so far we have imposed two conditions on with regard to how close it has to be to a. First, we wanted a < a 2. Then, at the end, we had a < ε a3. Our δ will be the minimum of these two values i.e. δ = 2 min( a 2, ε a3. Let s start the proof: a 4 4 a 3 Page 3 of

4 Let ε > 0. Let δ = min( a 2, ε a3. Now, suppose a < δ. Since δ a 2, we know that f( f(a a But note that δ < ε a3, we get a a 3 < ε a3 a 3 = ε Therefore, we conclude that whenever a < δ, we have f( f(a < ε. 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 eamples of continuous functions, let s take a look at a function that is discontinuous. But before we do that, we introduce a lemma that will help us prove that a function is discontinuous. Lemma.. Suppose f : R R is a function. Then, f( = L. if and only if for any sequence of real numbers {a n } such that n a n = a, we have f(a n = L n Conversely, f( L if and only if there eists a sequence {a n } such that n a n = a; and n f(a n L Eample.3. Consider the function, f : R R defined as follows: a 3 { ( sin 0 f( = L = 0 where L is a real number. Show that f( is discontinuous at 0 for any value of L. Proof. To show that f( is discontinuous at 0 for any value of L, we must show that f( f(a = L for any L R. In other words, we must show that f does not have a it as approaches 0. We will try to use Lemma. to show this. Hence, we must find a sequence {a n } converging to 0 such that {f(a n } does not converge to L. Hence it suffices to find two sequences {a n } and {a n}, both converging to 0, but such that {f(a n } and {f(a n} converge to distinct its. Let s first consider a n = 2nπ. Then, we have Now, we look at the sequence {f(a n }: a n = n n 2nπ = 0 ( f(a n = f = sin(2nπ = n n 2nπ 0 = 0 n n For the second sequence, we consider a 2 n = (4n+π. Again, we have 2 n a n = n (4n + π = 0 Page 4 of

5 However, this time the sequence {f(a n} does not converge to 0: ( n f(a n = f 2 n (4n + π Hence, for any L, we can find a sequence {a n } such that (4n + π = sin = n 2 = n n a n = 0; and n f(a n L Therefore, we conclude that no value of L R can be the it of f as approaches 0. discontinuous at 0 for any value of L. Hence, f is 2 Continuity Theorems 2. Continuity: Useful Tools As we saw in the previous section, proving continuity using the definition can be quite tricky. Thankfully, there are a number of useful tools and theorems that will allow us to prove that certain its eist without going through the ε δ business. Theorem 2. (Arithmetic of Limits. Let f, g : R R be functions such that f( and g( both eist. Then, (αf( + βg( = α f( + β g( for all α, β R ( ( (f(g( = f( g( ( f( = g( f( g(, if g( 0. Theorem 2. tells us that its of functions behave as we would like them to. following, very useful, corollary. It also gives us the Corollary 2.. Every polynomial is continuous at a R. Proof. We know that the constant function f( = is continuous everywhere. We also saw in the previous section that the function f( = is continuous everywhere. By multiplying these two functions together and scaling by constant factors, we can make any polynomial (in fact, polynomials are precisely functions that can be made in such way from f( = and f( =. Therefore, by Theorem 2., we conclude that its of polynomials are defined everywhere. To show that polynomials are continuous everywhere, we only have left to show that the its are equal to the value of the polynomial. However, since the it of f( = is everywhere; and the it = a for all a R, we conclude that for any polynomial g(, we have g( = g(a. Hence, every polynomial is continuous everywhere. Theorem 2.2 (Squeeze Theorem. Let f, g, h : R R be functions. Suppose that g( f( h( for all a g( = h( = L Then, f( = L. Page 5 of

6 Eample 2.. Show ( 0 2 sin = 0 Proof. We did a similar eample where we used the Squeeze Theorem for sequences to show that the sequence {a n = sin n n } converged to 0. As before, note that for all 0 R we have ( ( sin 2 2 sin 2 But we know by Corollary = 2 = 0 0 Therefore, by Theorem 2.2, we conclude that ( 0 2 sin = Etreme/Intermediate Value Theorems Theorem 2.3 (Etreme Value Theorem. Let f be a function and suppose f is continuous on the interval [a, b]. Then, f( attains its maimum on this interval i.e. there eists c [a, b] such that for all [a, b], f( f(c. You might ask: Isn t this always true? Well, the answer is not always. For eample, consider the function f( = on the interval (0,. Then, the function f( does not attain its maimum on the interval. The least upper bound for {f( (0, } is which is f(. However, lies outside our interval (0,. But that was cheating, you might say. What if we have a closed and bounded interval, say [0, ]. Shouldn t Theorem 2.3 be true for any function? Let s consider the following function: f( = { 0 0 = 0 It s easy to see that f( is not continuous at 0. But f is defined everywhere on [0, ]. But note that f isn t even bounded from above since f blows up as 0. Well, if you still think I m cheating by giving eamples of unnatural functions, then your intuition is probably telling you natural = continuous. In that case, see Theorem 2.3. Theorem 2.4 (Intermediate Value Theorem. Let f be a function and suppose f is continuous on the interval [a, b]. Suppose f(a < L < f(b. Then, there eists c (a, b such that f(c = L. We can use the Intermediate Value Theorem to prove the following corollary. Corollary 2.2. Let f be a continuous function on [a, b]. Suppose f(a < a and f(b > b. Then, we can find c [a, b] such that f(c = c. Proof. Consider the function g( := f( on [a, b]. By Theorem 2., we know that g( = f( is continuous on [a, b]. Now, g(a = f(a a < 0 and g(b = f(b b > 0 by hypothesis. By the Intermediate Value Theorem, we can find c [a, b] such that g(c = 0. However, that means that Hence, we can find c [a, b] such that f(c = c. g(c = f(c c = 0 f(c = c Page 6 of

7 3 Differentiation 3. Big O and Little o Definition (Big O and little o Let f and g be functions. We say f( is big O of g( as approaches a if we can find a constant C and δ > 0 such that f( C g( for all such that 0 < a < δ. We say that f( is little o of g( as approaches a if f( g( = 0 This is a more general definition than the one given in Ma. The definition given in Ma is in fact a special case of above definition where g( = and a = 0. What s the idea of big O and little o? First, little o is a bit easier to understand. We say that f( is o(g( if the it of f(/g( as approaches a is 0. Obviously, if f(a = 0 and g(a 0, then the statement is satisfied since the it will simply be 0/g(a = 0. But this is not very interesting. If f( does not approach 0 but g( does, then the the it will be ±. If f and g both approach nonzero real numbers, the it will also be nonzero. The interesting case is when f and g both approach 0. In this case, we can t simply say that the it is 0/0 since that isn t defined. To say that the it of f(/g( is 0 means that as approaches 0, f( is a lot smaller than g( at all times near a. In other words f approaches 0 faster than g. Let s look at an eample to see what this means: Eample 3.. Show that f( = 2 is o( as 0. Proof. Well, let s go ahead and compute the it: Hence, 2 is o(. 2 0 = = 0 0 Ok, 2 is a lot smaller than near 0 so maybe this makes sense. How about a slightly more general case? Eample 3.2. n is o( m if and only if n < m Proof. Let s take a look at that it again: n 0 m = 0 n m But this it is 0 if n m > 0; if n = m; and if n < m. Hence, n is o( m if and only if n < m. More generally, we can deduce the following. Proposition 3.. Let f and g be polynomials such that f(a = g(a = 0. approaches a if and only if deg f > deg g. Then, f( is o(g( as The proof is straightforward so I will leave it as an eercise. But this really is everything you need to know about little o. It is telling us whether a function f approaches 0 faster than another function g. In particular, if f doesn t approach 0 at all, f won t be little o of any function. If f does approach 0, then it Page 7 of

8 will be little o of any function that does not approach 0. Finally, if f and g both approach 0, f is o(g if f approaches 0 faster than g does. What about big O? The idea of big O is once again comparing how fast f and g approach 0. Simply put, f is O(g if f approaches 0 as fast as g. Let s do a few eamples to get comfortable with big O. Eample 3.3. Show that is O(. Proof. Well... this is obvious. We can pick δ = (anything, really and C =. Then, for (, we get which is vacuously true. = Clearly approaches 0 as fast as itself. Obviously for any function f, f will be big O of itself. Now, let s look at a very useful theorem. Theorem 3.. Let f( and g( be functions. Suppose that 0 f( = 0 g( = 0. Moreover suppose Then, f( is O(g(. f( 0 g( = L ± Proof. Since f(/g( approaches L as 0, for ε =, we can find δ > 0 such that 0 < δ f( g( L < ε = Hence, we have f( g( L < (9 f( g( L < ( f( g( < + L ( f( < ( + L g( (2 So let δ be the one we found above, and let C = + L. Then, for all < δ, we have f( < C g(. Therefore, f( is O(g(. Corollary 3.. Suppose f( is o(g(. Then, f( is O(g(. Great! This does agree with our description that f is O(g if f approaches 0 as fast as g does; and f is o(g if f approaches 0 faster than g. Put like this, it s obvious that little o is a stronger condition than big O. 3.2 Differentiation Definition (Differentiation Let f be a function. Then, we say that f is differentiable at a if the it f(a + h f(a (a + h a eists. If the it eists, we denote it as f (a and call it the derivative of f at a. Page 8 of

9 Let s look at a few eamples of derivatives for some easy functions. Eample 3.4. The derivative of f( = Proof. Pick any point a 0 in R. Then, we have at any point a 0 is a 2. f(a + h f(a (a + h a = = = a+h a h a (a+h a(a+h h h ha(a + h = a(a + h (3 (4 (5 (6 = a 2 (7 Before we look at the net eample, we define the function f( = e. In particular, we will define it to be the following formal power series: e n = n=0 We will need a couple of properties of e for the net eample. First, we can show using this power series representation and the binomial theorem that for any a, b R, e a+b = e a e b. Second, the series e converges for all R and e 0 =. This, again, is not very hard to prove but we will not present the proof here so we can get to more fun stuff! Eample 3.5. Show that for any a R, the derivative of e at a is e a. Proof. Let s start by eamining the it: f(a + h f(a h e a+h e a = h = e a eh h (8 ( (9 (20 Now, we will use the fact that e h = n=0 h n = + h + h ( = e a ( + h + h 2 2! + h3 3! + h h = e a + h2 2! + h3 3! + h = e ( a h2 + + h2! 3! + ( = e a h n n= ( = e a n= h n (2 (22 (23 (24 (25 Page 9 of

10 Now, we only have left to show that Note that for any h, we have n= h n ( n= h n = n=0 h n Hence, the series n= hn f( can be computed by directly plugging a in. Hence, = e h converges absolutely for all h. For power series, f(, which converges at a, ( n= h n = Therefore, we conclude that the derivative of e at any a R is e a. Page of