School of Economics, Australian National University February 1, 2010
Continuous Functions. A continuous function is one we can draw without taking our pen off the paper Definition. Let f be a real-valued function whose domain is a subset of R. The function f is continuous at x 0 in dom(f ) if, for every sequence (x n ) in dom(f ) converging to x 0, we have lim f (x n ) = f (x 0 ). If f is continuous at each point of a set S dom(f ), then f is said to be continuous on S. The function f is said to be continuous if it is continuous on dom(f ).
Continuous Functions. y x 0 x Figure: A continuous function. For any sequence of points (x n) n N converging to x 0, the sequence (f (x n)) n N converges to f (x 0 ).
Continuous Functions. Example Let f : R R be given by f (x) = 2x 2 + 1 for all x R. Prove f is continuous on R. Proof. Suppose we have a real-valued sequence (x n ) converging to x 0 i.e. lim x n = x 0. Then we have lim f (x n ) = lim[2x 2 n + 1] = 2[lim x 2 n ] + 1 = 2x 2 0 + 1 = f (x 0), where the second equality follows by application of the limit theorems. We have shown that for any sequence (x n ) converging to x 0, we have lim f (x n ) = f (x 0 ).This proves that f is continuous at each x 0 R and so f is continuous on R.
Continuous Functions. Example Let f : R R be given by f (x) = (1/x) sin(1/x 2 ) for x 0 and f (0) = 0. Show that f is discontinuous at 0. Proof. It is sufficient to find a sequence (x n ) converging to 0 such that (f (x n )) does not converge to f (0) = 0. To find such a sequence, we will rearrange (1/x n ) sin(1/x 2 n ) = 1/x n where x n 0. Thus, we want sin(1/x 2 n ) = 1. For this we need 1/x 2 n = 2nπ + π/2, and it follows that x n = 1 2nπ + π. 2 Then lim x n = 0, while lim f (x n ) = lim(1/x n ) = + 0.
Continuous Functions. The sequential definition of continuity implies that the values f (x) are close to f (x 0 ) when the values x are close to x 0. The following theorem provides an alternative definition of continuity. Theorem (ε-δ definition of continuity) Let f be a real-valued function whose domain is a subset of R. Then f is continuous at x 0 dom(f ) iff for each ε > 0 there exists δ > 0 such that x dom(f ) and x x 0 < δ imply f (x) f (x 0 ) < ε. If we draw two horizontal lines, no matter how close together, we can always cut off a vertical strip of the plane by two vertical lines in such a way that all that part of the curve which is contained in the strip lies between the two horizontal lines.
Continuous Functions. y Epsilon Delta Applet f (x 0 ) + ε f (x 0 ) f (x 0 ) ε x 0 δ x 0 x 0 + δ Figure: ε-δ definition of continuity x
Continuous Functions. Example Let f : R R be given by f (x) = x 2 sin(1/x) for all x 0 and f (0) = 0. Prove f is continuous at 0. Proof. Let ε > 0. Clearly f (x) f (0) = f (x) x 2 for all x. We want this to be less than ε, so set δ = ε. Then x 0 < δ implies x 2 < δ 2 = ε, and so x 0 < δ implies f (x) f (0) < ε. So f is continuous at 0.
Continuous Functions. Example Let f : R R be given by f (x) = 2x 2 + 1 for all x R. Then f is continuous on R. Proof. We now use the ε-δ property. Let x 0 R, and let ε > 0. We want to show that f (x) f (x 0 ) < ε provided x x 0 is sufficiently small, i.e. less than some δ. First note that f (x) f (x 0 ) = 2x 2 + 1 (2x 2 0 + 1) = 2x 2 2x 2 0 = 2(x x 0 )(x + x 0 ) = 2 x x 0 x + x 0. So, what we need is to find a bound for x + x 0 that does not depend on x.
Continuous Functions. Now, if x x 0 < 1 then x < x 0 + 1 and hence x + x 0 x + x 0 < 2 x 0 + 1. So, we have f (x) f (x 0 ) < 2 x x 0 (2 x 0 + 1) if x x 0 < 1. To arrange for 2 x x 0 (2 x 0 + 1) < ε it is sufficient to have x x 0 < ε/[2(2 x 0 + 1)] and also x x 0 < 1. So let { } ε δ = min 1, 2(2 x 0 + 1) The working above shows that x x 0 < δ implies f (x) f (x 0 ) < ε.
Continuous Functions. We can form new functions from old functions in several ways. Definition. Let A R and let B R. Consider two functions f : A R and g : B R and let k R. We define the functions into R as follows. f given by f (x) = f (x) for all x A; kf given by (kf )(x) = kf (x) for all x A; f + g given by (f + g)(x) = f (x) + g(x) for all x A B; fg given by (fg)(x) = f (x)g(x) for all x A B; f /g given by (f /g)(x) = f (x)/g(x) for all x A B such that g(x) 0.
Continuous Functions. Theorem Let f and g be real-valued functions that are continuous at x 0 in R and let k R. Then (i) f is continuous at x 0 ; (ii) kf is continuous at x 0 ; (iii) f + g is continuous at x 0 ; (iv) fg is continuous at x 0 ; (v) f /g is continuous at x 0 if g(x 0 ) 0. Theorem If f is continuous at x 0 and g is continuous at f (x 0 ), then the composite function g f is continuous at x 0.
Properties of Continuous Functions. Theorem (Intermediate Value Theorem) If f is a continuous real-valued function on an interval I, then f has the intermediate value property on I: Whenever a, b I, a < b and y lies between f (a) and f (b), there exists at least one x (a, b) such that f (x) = y. This theorem can be used to establish that a continuous function f has a fixed point, i.e. a point x 0 dom(f ) such that f (x 0 ) = x 0.
Properties of Continuous Functions. y 0 f (x) f (b) y 0 f (a) [ a x 1 x 2 x 3 b ] I Figure: By the intermediate value theorem, for any f (a) < y < f (b), we can find an x such that f (x) = y. In the case of y 0, we can find three i.e. y 0 = f (x 1 ) = f (x 2 ) = f (x 3 ). x
Properties of Continuous Functions. Example Show that the continuous function f : [0, 1] [0, 1] has a fixed point x 0 [0, 1]. y 1 y = x f (x) x 0 Figure: Fixed point. Kinda obvious? 1 x
Properties of Continuous Functions. We will use the IVT. We will define a new function g on I, and consider a = 0 and b = 1. Solution. Consider the function g(x) = f (x) x, which is also continuous on [0, 1] by (iii). Now g(0) = f (0) 0 = f (0) 0 and g(1) = f (1) 1 1 1 = 0. So, by the intermediate value theorem, we have that g(x 0 ) = 0 for some x 0 [0, 1]. Then clearly f (x 0 ) = x 0.
Definition. Let S be a subset of R, let a be a real number or symbol or that is the limit of some sequence in S, and let L be a real number or symbol or. We write lim x a S f (x) = L if f is a function defined on S, and for every sequence (x n ) in S with limit a, we have lim n f (x n ) = L. The expression lim x a S f (x) is read limit, as x tends to a along S, of f (x). Using this definition, we see that a function f is continuous at a in dom(f ) = S iff lim x a S f (x) = f (a).
Definition. For a function f and a R we write (i) lim x a f (x) = L provided lim x a S f (x) = L for some set S = J\a where J is an open interval containing a. lim x a f (x) is called the (two-sided) limit of f at a. (ii) lim x a + f (x) = L provided lim x a S f (x) = L for some open interval S = (a, b). lim x a + f (x) is called the right-hand limit of f at a. (iii) lim x a f (x) = L provided lim x a S f (x) = L for some open interval S = (c, a). lim x a f (x) is called the left-hand limit of f at a. (iv) lim x f (x) = L provided lim x S f (x) = L for some interval S = (c, ). (v) lim x f (x) = L provided lim x S f (x) = L for some interval S = (, b).
y t f (x) s x a Figure: The right hand limit at a is t, while the left hand limit is s.
Example (1) We have lim x 4 x 3 = 64 and lim x 2 (1/x) = 1/2 because the functions x 3 and 1/x are continuous at 4 and 2 respectively. One can easily show that lim x 0 +(1/x) = + and lim x 0 (1/x) =. It follows that lim x 0 (1/x) does not exist (see theorem 7). (2) Consider lim x 2 [ x 2 ] 4. x 2 The function we are finding the limit of is not defined at x = 2. We can rewrite the function as x 2 4 x 2 = (x 2)(x + 2) x 2 = x + 2 for x 2.
Now we can see that lim x 2 [ x 2 4 x 2 ] = lim x 2 (x + 2) = 4. It is important to note that the functions given by (x 2 4)/(x 2) and (x + 2) are not identical. The domain of the first is (, 2) (2, + ), while the domain of the second is R. (3) Consider lim x 1 [ ] x 1. x 1 To find the limit, we multiply the numerator and denominator of the function by x + 1 to get x 1 x 1 = x 1 (x 1)( x + 1) = 1 for x 1. x + 1
Now we can see that lim x 1 [ x 1 x 1 ] [ ] 1 = lim = 1 x 1 x + 1 2. This in fact shows that if h(x) = x then h (1) = 1/2 (as you will see when we look at differentiation). (4) Let f be a real-valued function given by f (x) = 1/(x 2) 3 for all x 2. Then limx + f (x) = 0, limx f (x) = 0, limx 2 + f (x) = +, limx 2 f (x) =.
Proof. We will show that lim x f (x) = 0. By definition, it is enough to show that lim x S f (x) = 0 for S = (2, + ). So consider any sequence (x n ) with x n S for all n N, such that lim n + x n = + and show that lim f (x n) = n + lim n + [ ] 1 (x n 2) 3 = 0. To prove the above assertion we could use our limit theorems. Instead we will prove it directly. Let ε > 0. We need to find an N such that for n > N, we have (x n 2) 3 < ε. This inequality can be rearranged as ε 1 < (x n 2) 3 or ε 1 3 < x n 2. We need x n > ε 1 3 + 2, for the previous inequality to be satsified.
Now we use the fact that lim n + x n = +. By definition of an infinite limit of a sequence, for any M > 0 we can find an N such that n > N implies x n > M. Thus, if we set M = ε 1 3 + 2, there exists an N such that n > N implies x n > ε 1 3 + 2 Then reversing the steps above, we find that n > N implies (x n 2) 3 < ε. This shows that lim n + f (x n ) = 0 for any sequence (x n ) in S such that lim n + x n = + and the result follows by definition.
The following result allows us to avoid sequences and provide ε-δ definitions of a function s limits. Theorem Let f be a function defined on a subset S of R, let a be a real number that is the limit of some sequence in S, and let L be a real number. Then lim x a s f (x) = L iff for each ε > 0 there exists δ > 0 such that x S, x a < δ imply f (x) L < ε. This theorem has a number of corollaries, which are listed in the next theorem. These give us alternative definitions for the limit of a function, its lateral limits, and its limits at infinity.
Theorem (i) Let f be a function defined on J\{a} for some open interval J containing a, and let L be a real number. Then lim x a f (x) = L iff for each ε > 0 there exists δ > 0 such that 0 < x a < δ implies f (x) L < ε. (ii) Let f be a function defined on some interval (a, b), and let L be a real number. Then lim x a + f (x) = L iff for each ε > 0 there exists δ > 0 such that a < x < a + δ implies f (x) L < ε. (iii) Let f be a function defined on some interval (c, a), and let L be a real number. Then lim x a f (x) = L iff for each ε > 0 there exists δ > 0 such that a δ < x < a implies f (x) L < ε.
y t s + ε s s ε a δ a a + δ f (x) Figure: The (two-sided) limit at a is s, while f (a) = t. (This function is not continuous at a.) x
y t + ε t t ε f (x) s + ε s s ε a δ a a + δ Figure: The right hand limit at a is t, while the left hand limit is s. x
(iv) Let f be a function defined on some interval (c, + ), and let L be a real number. Then lim x + f (x) = L iff for each ε > 0 there exists δ < + such that x > δ implies f (x) L < ε. (v) Let f be a function defined on some interval (, b), and let L be a real number. Then lim x f (x) = L iff for each ε > 0 there exists δ > such that x < δ implies f (x) L < ε.
y 1 ε ε ε 1 ε f (x) x Figure: The function f : R \ {0} R is given by f (x) = 1/x for all x R \ {0}. The limits at + and are both 0.
Example Let f : R \ {0} R be given by f (x) = 1/x for all x R \ {0} (this is the function drawn in the previous figure). Then lim x + = lim x = 0. Proof. We will prove the first limit. First note that the function is defined on some interval (c, + ) we could take any c > 0. Also the limit L = 0 is a real number. So we use definition iv. Let ε > 0. We need to find a finite δ such that x > δ implies 1/x < ε. If we take δ = 1/ε, then the desired condition is satisfied.
(vi) Let f be a function defined on J \ {a}. Then lim x a f (x) = iff for each M > 0 there exists δ > 0 such that 0 < x a < δ implies f (x) > M. (vii) Let f be a function defined on J \ {a}. Then lim x a f (x) = iff for each M < 0 there exists δ > 0 such that 0 < x a < δ implies f (x) < M. We can also mix the previous limit concepts, and define the following limits. lim x a + f (x) = +, lim x a + f (x) =, limx a f (x) = +, limx a f (x) =, lim x + f (x) = +, limx + f (x) =, limx f (x) = +, limx f (x) = +.
y 1 M M M 1 M f (x) x Figure: The function f : R \ {0} R is given by f (x) = 1/x for all x R \ {0}. Here lim x 0 + f (x) = +, while lim x 0 f (x) =. The two-sided limit at a does not exist. It cannot be + because there is no δ such that x ( δ, +δ) implies f (x) > M.
The next result tells us that if the left-hand and right-hand limits at a of a function exist and are equal, then the two-sided limit at a exists and is equal to these two limits. It also says the converse is also true if the limit exists, then both the left-hand and right-hand limits exist and are equal to the two-sided limit. Theorem Let f be a function defined on J\{a} for some open interval J containing a. Then lim x a f (x) exists iff the limits lim x a + f (x) and lim x a f (x) both exist and are equal, in which case all three limits are equal.
Theorem Let f 1 and f 2 be functions for which the limits L 1 = lim x a S f 1 (x) and L 2 = lim x a S f 2 (x) exist and are finite. Then (i) lim x a S(f 1 + f 2 )(x) = L 1 + L 2 ; (ii) lim x a S(f 1 f 2 )(x) = L 1 L 2 ; (iii) lim x a S(f 1 /f 2 )(x) = L 1 /L 2 if L 2 0 and f 2 (x) 0 for x S. Proof of (iii). blah
The following result provides conditions under which we can change the order of taking the limit and applying the function g. Theorem Let f be a function for which the limit L = lim x a S f (x) exists and is finite. If g is a function defined on {f (x) x S} {L} that is continuous at L, then lim x a S g f (x) exists and equals g(l). Example Compute lim x 1 x 3 + 1. Proof. Here the function g : R + R, given by g(x) = x for all x R + is continuous. Thus lim x 1 x 3 + 1 = lim x 1 x 3 + 1 = 1 + 1 = 2.