5.4 Continuity: Preliminary Notions

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1 5.4. CONTINUITY: PRELIMINARY NOTIONS Continuity: Preliminary Notions Definitions The American Heritage Dictionary of the English Language defines continuity as an uninterrupted succession, an unbroken course. Descartes once said that if a function was continuous, one could draw its graph without lifting the pencil. This intuitive definition might be used for very simple functions. However, to prove theorems about continuous functions, as well as to handle more complicated functions, a more precise definition is needed. The attempt to precisely define functions and the notion of continuous functions took place in the mathematics community between 1750 and Among the most famous mathematicians involved in this discussion were Bolzano, Cauchy, Dirichlet, Euler, Fourier, Gauss, and LaGrange. Definition Let f be a function. 1. f is said to be continuous at a point a of its domain if lim x a f (x) = f (a).. f is said to be continuous from the right or right-continuous at a point a of its domain if lim f (x) = f (a). x a + 3. f is said to be continuous from the left or left-continuous at a point a of its domain if lim f (x) = f (a). x a 4. f is said to be continuous on a subset I of its domain if f is continuous at every point of I. Remark 5.4. If a function is both continuous from the left and the right, then it is continuous. Remark In the last part of the definition, at the end points of I, it should be clear to the reader that we are considering left-continuity for the right endpoint, and right-continuity for the left endpoint. We can rewrite the definition of continuity using ɛ and δ in a way similar to the definition of limits. Definition Let f be a function. 1. f is said to be continuous at a point a of its domain if for every ɛ > 0, there exists δ > 0 such that whenever x D (f) and x a < δ, then f (x) f (a) < ɛ.. f is said to be continuous at a point a of its domain if for every ɛ > 0, there exists δ > 0 such that whenever x D (f) (a δ, a + δ) then f (x) f (a) < ɛ.

2 18 CHAPTER 5. LIMIT AND CONTINUITY OF A FUNCTION The reader will notice a fundamental difference between the notion of limits we just studied and the notion of continuity we are starting to study. For limits, we were only concerned about the behavior of a function close to a point a. What happened at a did not matter. In fact, a could have been outside the domain of the function. Continuity takes on the notion of limits, that is the behavior of a function close to a point a. Continuity also requires the point a be in the domain of the function. For a function to be continuous at a, the value of the function at a has to be consistent with the value the function has near a. Let us emphasize that continuity is a pointwise notion. It is a property which is defined at a point. Even when we talk about continuity on a set, we are still talking about a property which has to be satisfied at every point of the set. Continuity preserves limits in the sense that if a sequence converges, then the image of the sequence also converges under a continuous function. More precisely, we have the following theorem. Theorem Let f be a function with domain D (f). 1. If f is continuous at a point a D (f), and {x n } is a sequence of numbers in D (f) converging to a, then f (x n ) converges to f (a).. If f (x n ) converges to f (a) for every sequence {x n } in D (f) converging to a, then f is continuous at a. Proof. Left as an exercise. It is similar to the proof of a similar theorem for limits. The above theorem can be restated as follows. Theorem The following two statements are equivalent. 1. f is continuous at a. f (x n ) converges to f (a) for every sequence {x n } in D (f) converging to a. This theorem provides yet another way to check the continuity of a function at a point. It also provides a way to show a function is not continuous at a point as illustrated in example To establish the continuity of a function at a point, one can either use the definition of continuity involving ɛ and δ. One can also use the fact that continuity is defined in terms of limits, and use known results about limits. One can also use theorem We illustrate the first two possibilities with some examples. The last method will be shown later. Example Show that f (x) = x is continuous at x = 1.

3 5.4. CONTINUITY: PRELIMINARY NOTIONS 183 Method 1: we use limits. Clearly 1 is in the domain of f. We need to show that lim x 1 f (x) = f (1). ( lim f (x) = lim x ) x 1 x 1 = 1 (using our theorems on limits) = f (1) Method : we use the ɛ δ definition. Given ɛ > 0, we need to find δ such that x 1 < δ = f (x) f (1) < ɛ. Proving continuity this way is very similar to computing limits using the definition. We start with the inequality with ɛ. f (x) f (1) < ɛ x 1 < ɛ x 1 < ɛ x 1 x + 1 < ɛ We would like to have an inequality with x 1 on the left, and some constant on the right. This way, the constant would be our δ. Since we are investigating continuity at 1, x will be close to 1. Without loss of generality, we can restrict ourselves to the interval (0, ) that is in the interval (1 h, 1 + h) with h = 1. If x (0, ), then x + 1 < 3, therefore: x 1 x + 1 < 3 x 1 So, it follows that if we make x 1 < ɛ, then we will have f (x) f (1) < ( 3 ɛ ) ɛ. So, given ɛ > 0, δ = min 3, 1 will work. Remark For the above proof, to get a bound on x + 1, we restricted ourselves to the interval (1 h, 1 + h) with h = 1. There is nothing special about h = 1. We could have chosen any other value. We would have gotten a different value for δ. The point is that we can do it. Example Show that f (x) = x is continuous in its domain using the ɛ δ definition. The domain of this function is R. Let a be a fixed but arbitrary point in R. We need to show that f is continuous at a. Let ɛ > 0 be given. We need to find δ > 0 such that x a < δ = f (x) f (a) < ɛ. Since a is arbitrary, if we can do this for a, it means we can do it for any real number, thus f is continuous for every real number. f (x) f (a) < ɛ x a < ɛ x a x + a < ɛ

4 184 CHAPTER 5. LIMIT AND CONTINUITY OF A FUNCTION Once again, we need to get a bound on x + a. We restrict ourselves to the interval (a 1, a + 1) i.e. when x a < 1. Then, in this interval, x < a +1, therefore, It follows that So, we see that if x a < x + a < x + a < a + 1 f (x) f (a) < x a ( a + 1) ɛ, then f (x) f (a) < ɛ. We need to realize a + 1 that we have really imposed two conditions. One is that x a < ( ) ɛ other one is that x a < 1. Thus δ = min 1, will work. a + 1 ɛ a + 1. The Remark Given ɛ > 0, the δ we obtain will of course depend on ɛ. In most cases, as illustrated above, it also depends on a, the point where continuity is being studied. So, when we study continuity on an interval, given ɛ > 0, we have to be able to find a δ at each point of the domain. However, for the same ɛ, the δ we find may be different for every point of the domain. If for a given function, on a given interval, we can find a δ that will work for all the points in the interval, we say that the function is uniformly continuous. This will be studied in another section. Example Show that f (x) = 1 { x is continuous on S = x R such that x 1 } using the ɛ δ definition. Let a be an arbitrary point of S, let ɛ > 0 be given. We need to find δ > 0 such that x a < δ = f (x) f (a) < ɛ. f (x) f (a) < ɛ 1 x 1 a < ɛ a x ax < ɛ x a < ɛ x a Since both x and a are in S, we know that x 1 and a 1. Therefore, x a x a 4 x a. So, we see that if x a < ɛ, the result will follow. There- 4 fore, given ɛ > 0, δ = ɛ will work. You will note here that the δ we found only 4 depends on ɛ.

5 5.4. CONTINUITY: PRELIMINARY NOTIONS 185 Example Show that f (x) = sin x is continuous on its domain. Recall that the domain of sin x is R. Let a R, and let ɛ > 0 be given. We need to find δ > 0 such that x a < δ = f (x) f (a) < ɛ. f (x) f (a) < ɛ sin x sin a < ɛ Using the trigonometric identity sin a sin b = sin a b cos a + b that for every x, sin x x, we have: sin x sin a = sin x a cos x + a x a 1 x a and the fact So, if x a < ɛ, then f (x) f (a) < ɛ. Thus δ = ɛ will work. Since a was arbitrary, sin x is continuous for every x. Note once again that the δ we found only depends on ɛ. A similar proof would work for cos x. Example Most of the functions you have studied so far were continuous, except maybe at a few points. We now give an example { which is discontinuous at every real number except 0. Consider f (x) =. Show that 0 x Q x x / Q f is discontinuous for every x R, x 0. Since lim f (x) = 0 = f (0) so f is continuous at a = 0. Now, suppose a 0. x 0 Then, lim f (x) does not exist so f cannot be continuous there. x p We finish this section by explaining with the use of a picture why, in general, δ depends on both ɛ and a, when establishing the continuity of a function f at a point a of its domain. Figure 5.1 shows that for a given a, δ 1 which corresponds to ɛ 1 is not the same as δ which corresponds to ɛ. Figure 5. shows that for a given ɛ (ɛ 1 and ɛ are the same), the δ we find depends on where a is Some Properties of Continuous Functions Since continuity is defined in terms of limits, we might expect that a lot of the theorems we proved about limits would hold for continuity. This turns out to be true, we list the theorems below. Their proof is left as an exercise, since it is very similar to the proof of the corresponding theorems on limits. Theorem Let f and g be two functions defined in a neighborhood of c, suppose that both functions are continuous at c. Then, the following functions are also continuous at c: 1. f + g

6 186 CHAPTER 5. LIMIT AND CONTINUITY OF A FUNCTION Figure 5.1: How δ depends on ɛ

7 5.4. CONTINUITY: PRELIMINARY NOTIONS 187 Figure 5.: How δ depends on a

8 188 CHAPTER 5. LIMIT AND CONTINUITY OF A FUNCTION. f g 3. fg 4. f g if g (c) 0 Proof. These results can be proven diff erent ways. We can use the ɛ δ definition of continuity. We can also relate continuity to limits, and use the corresponding results on limits. We illustrate the second idea to prove 1. The remaining proofs are left as an exercise. 1. Recall the definition of the sum of two functions. (f + g) (x) = f (x) + g (x). Therefore, if c is in the domain of both f and g, then it is also in the domain of f + g. We need to prove that lim [(f + g) (x)] = (f + g) (c) x c lim [(f + g) (x)] = lim [f (x) + g (x)] x c x c = lim f (x) + lim g (x) (using theorem on limits) x c x c = f (c) + g (c) (since f and g are continuous at c) = (f + g) (c) Corollary Polynomial functions are continuous on R.. Rational functions are continuous in their domain. 3. Trigonometric functions are continuous in their domain. Proof. See exercises at the end of this section Theorem Let f and g be two functions such that f is defined in a neighborhood of c, g is defined in a neighborhood of f (c). In addition, suppose that f is continuous at c and g is continuous at f (c). Then, g (f (x)) is continuous at c. Proof. Let ɛ > 0 be given. We need to find δ > 0 such that x c < δ = g (f (x)) g (f (c)) < ɛ. For the sake of clarity, put y = f (x) and d = f (c). Since g is continuous at d, choose δ 1 such that y d < δ 1 = g (y) g (d) < ɛ. Since f is continuous at c, choose δ > 0 such that x c < δ = f (x) f (c) < δ 1. Now, if x c < δ, then: g (f (x)) g (f (c)) = g (y) g (d) < ɛ Corollary Suppose that f is continuous at c.

9 5.4. CONTINUITY: PRELIMINARY NOTIONS If f 0, then f is continuous at c. f is continuous at c. Proof. See exercises at the end of this section Another important property of continuous functions is that if a function is continuous at a point, then its behavior is predictable near that point. In particular, if a function is continuous at a point, and strictly positive there, then it will remain positive for a while. This is precisely the statement of the next theorem. Theorem If f is continuous at c and f (c) > 0, then there is a neighborhood U of c such that f (x) > 1 f (c) > 0 for every x in U. Proof. Choose δ > 0 such that x c < δ = f (x) f (c) < 1 f (c). Then for these values of x,, 1 f (c) < f (x) f (c) < 1 f (c) 1 f (c) < f (x) < 3 f (c) So, in particular, f (x) > 1 f (c) for all values of x such that x c < δ that is for all values of x in (c δ, c + δ). In particular, U = {x R : x c < δ}. The next two examples examine how to find the largest set on which a function is continuous. One of these examples involves a piecewise function. Example Find where f (x) = tan x is continuous. The domain of this function is D = {x R such that x π } ± nπ for any n = 0, 1,,.... If x D, then sin x is continuous (see example above). So is cos x. Furthermore, cos x 0. Thus, if x D, tan x will be continuous by theorem { x 1 if x 0 Example Find where f (x) = x if x > 0 f is a piecewise function. To study continuity of piecewise functions, we must study the continuity of each branch. In addition, we must study continuity at the breaking points. The breaking point of this function is 0. If x 0. Then, f is a polynomial in both cases, therefore it is continuous. At x = 0. Since f is defined there, we need to check if lim f (x) = x 0 f (0). Since the definition of f changes at 0, we need to compute onesided limits. On one hand, lim (x 1) = 1. On the other = lim x 0 x 0 hand, lim f (x) = lim x = 0. Since the one sided limits are different, x 0 + x 0 + it follows that lim f (x) does not exist, therefore, f is not continuous at 0. x 0

10 190 CHAPTER 5. LIMIT AND CONTINUITY OF A FUNCTION Discontinuities Continuity is the best thing one can hope for in a function. Most of the theorems studied in calculus I or II depended on the fact that the functions involved were continuous. What about when a function is not continuous at some point? Can we still do Calculus? If we think of a function not being continuous at a point as a disease the function has, then one might ask: how serious is the disease? How serious can the disease be for the function to be still usable? Usable for us means being able to do some of the things we do in calculus such as differentiating, integrating. In order to try to answer these questions, we classify discontinuities (points where a function is not continuous) into various categories. Before we do this, let us see what it means for a function not to be continuous at a point. We begin by negating the definition of continuity. Recall that a function f is continuous at a point a of its domain if for every ɛ > 0, there exists δ > 0 such that whenever x D (f) and x a < δ, then f (x) f (a) < ɛ. To see how to negate this, let us think about it the following way: f is continuous at a point a of its domain if for every ɛ > 0, there exists δ > 0 such that some property is true. Negating this is much easier. We see that for the above statement to be false, it is enough if there exists an ɛ > 0 such that no matter which δ > 0 we try, the property is false. What does it mean for the property to be false? The property is: whenever x D (f) and x a < δ, then f (x) f (a) < ɛ. For this to be false, it is enough if for some x D (f) we have x a < δ, and f (x) f (a) > ɛ. Combining the two, we have: Definition A function f is not continuous at a point a D (f) if there exists x D (f) such that for some ɛ > 0 and every δ > 0, we have x a < δ, and f (x) f (a) > ɛ.. If a function f is not continuous at a, then we say that it is discontinuous at a. The number a is called a point of discontinuity for f, or simply a discontinuity for f. We can also think about this in terms of the other definition of continuity. f is continuous at a point a D (f) if lim f (x) = f (a). This statement imposes x a two restrictions. The first being that lim f (x) must exist. The second is that x a this limit has a fixed value, namely f (a). In order for this not to happen, we see that either lim f (x) exists and is not f (a), or lim f (x) does not exist. x a x a This could happen if both one-sided limits exist but are not equal, or one of the one-sided limits (or both) may not exist at all. Discontinuities are classified according to which situation happens. Definition 5.4. (discontinuities) Let f be a function, not continuous at a point a of its domain. 1. If lim x a f (x) exists and is not f (a), then a is called a removable discontinuity.

11 5.4. CONTINUITY: PRELIMINARY NOTIONS 191. If both lim f (x) and lim f (x) exist but lim f (x) lim f (x), then x a x a + x a x a + a is called a discontinuity of the first kind, or a jump discontinuity, or a simple discontinuity. 3. Any other kind of discontinuity, in particular if one of the one-sided limits does not exist, then a is called a discontinuity of the second kind, or an essential discontinuity. { x if x 0 Example What kind of discontinuity does f (x) = 1 if x = 0 have at 0? To answer this question, we need to see why f is not continuous at 0. It should be clear to the reader that lim f (x) = 0 and f (0) = 1. Hence, f has a removable x 0 discontinuity at 0. We call it a removable discontinuity because we can make f continuous there, simply by redefining it to have the value of the limit. { x Example What kind of discontinuity does f (x) = if x 0 x 1 if x = 0 have at 0? To answer this question, we need to see why f is not continuous at 0. It should be clear to the reader that lim f (x) does not exist because lim f (x) = 1, x 0 x 0 lim f (x) = 1. Thus, f has a jump discontinuity. If we look at its graph, we x 0 + can see why such a discontinuity is called a jump discontinuity. { sin 1 Example What kind of discontinuity does f (x) = x if x 0 1 if x = 0 have at 0? This function has a discontinuity of the second kind at 0. We prove it by showing that lim f (x) does not exist. To prove this, we find two sequences u n and v n x 0 + both converging to 0 +, such that lim f (u n ) lim f (v n ) (why will this prove that lim f (x) does not exist?). Let u n = 1 x 0 + nπ, v n =. Clearly, as (n + 1) π n, both sequences converge to 0 +. On one hand, f (u n ) = sin (nπ) = 0. (n + 1) π Therefore, f (u n ) 0. On the other hand, f (v n ) = sin = ±1. Therefore, f (v n ) does not converge to 0. Hence, lim f (u n ) lim f (v n ) Exercises 1. Read about the famous mathematicians mentioned at the beginning of this section.. Prove theorem Prove theorem

12 19 CHAPTER 5. LIMIT AND CONTINUITY OF A FUNCTION 4. Prove theorem Prove theorem Give a complete discussion of the continuity properties (including type of discontinuities) of the functions below in the indicated domain. Prove your answer using definitions only. (a) f (x) = x 3, x R (b) f (x) = x, x [ 3, 0) (0, ] { x if x [0, 1] (c) f (x) = x 3 if x (1, 5] (d) f (x) = x, x R (e) f (x) = x x, x R 7. Decide if the given function is continuous at the given point. If it is, give a convincing argument using the definition and/or theorems studied in class. If it is not, tell what kind of discontinuity there is. { sin x if x 0 (a) f (x) = x at x = 0 0 if x = 0 { 1 e x if x 0 (b) f (x) = 0 if x = 0 at x = 0 (c) f (x) = { x sin 1 x if x 0 0 if x = 0 at x = 0 8. Study the continuity of f (x) = x [x] x it to make it continuous on [ 1, 1]? on ( 1, 0) (0, 1). Can we extend 9. Show that f (x) = { 0 x Q 1 x / Q is discontinuous at every point a R.

13 5.4. CONTINUITY: PRELIMINARY NOTIONS Hints for the Exercises 1. Read about the famous mathematicians mentioned at the beginning of this section. Hint: None.. Prove theorem Hint: Use the theorem we studied about the relationship between the limit of a sequence and the limit of a function. 3. Prove theorem Hint: Use corresponding theorems on limits. 4. Prove theorem Hint: For polynomial and rational functions, use theorem For trigonometric functions, use the fact that sine and cosine are continuous (proven in the notes) and theorem Prove theorem Hint: Use theorem , do not forget to verify all the necessary conditions to apply the theorem hold. 6. Give a complete discussion of the continuity properties (including type of discontinuities) of the functions below in the indicated domain. Prove your answer using definitions only. (a) f (x) = x 3, x R Hint: Use the definition of continuity. (b) f (x) = x, x [ 3, 0) (0, ] Hint: Use the definition of continuity. { x if x [0, 1] (c) f (x) = x 3 if x (1, 5] Hint: Use the definition of continuity, remember how to work with piecewise functions. (d) f (x) = x, x R Hint: Recall that if x [n, n + 1) where n is an integer, then x = n. In other words, f can be though of as a piecewise function, changing definition at each integer. (e) f (x) = x x, x R Hint: Using the hint above, rewrite f (x) as a piecewise function. 7. Decide if the given function is continuous at the given point. If it is, give a convincing argument using the definition and/or theorems studied in class. If it is not, tell what kind of discontinuity there is. { sin x if x 0 (a) f (x) = x at x = 0 0 if x = 0 Hint: Use the definition of continuity.

14 194 CHAPTER 5. LIMIT AND CONTINUITY OF A FUNCTION { 1 e x if x 0 (b) f (x) = 0 if x = 0 at x = 0 Hint: Use the definition of continuity. { x sin 1 (c) f (x) = x if x 0 0 if x = 0 at x = 0 Hint: Use the definition of continuity and the squeeze theorem. 8. Study the continuity of f (x) = x [x] x it to make it continuous on [ 1, 1]? Hint: Use the hint of problem 6d. on ( 1, 0) (0, 1). Can we extend { 0 x Q 9. Show that f (x) = is discontinuous at every point a R. 1 x / Q Hint: See similar example in the section on limit of a function.