211 Real Analysis. f (x) = x2 1. x 1. x 2 1

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1 Part. Limits of functions. Introduction 2 Real Analysis Eample. What happens to f : R \ {} R, given by f () = 2,, as gets close to? If we substitute = we get f () = 0 which is undefined. Instead we 0 might try to substitute numbers close to : It appears that if is very close to then f () is very close to 2. In fact, 2 = ( ) ( + ) = +, though, as an equation, this only holds for, since we are never allowed to divide by zero. Graphically we have: 2 where the point (, 2) is omitted from the graph. It is obvious from the graph that as approaches then f () approaches 2.

2 Eample The function sin ( π ) is defined for all 0. What happens as gets close to 0? We can calculate a few values: sin ( ) π and, in fact, sin ( π ) = 0 when = 0 n for all n. It is important not to derive the wrong conclusion from this little information. For we also have sin ( ) π and, in fact, sin ( π ) = when = 2/ ( + 4n) for all n Z. From this table we might draw a completely different conclusion to that drawn from the first. In fact the function oscillates wildly near the origin. 2

3 .2 Limit of a function.2. Limit of a function at a finite point. First, recall from course 53 the definition of the limit of a sequence: Definition The limit of the sequence {a n } n N is L if, and only if, for all ε > 0 there eists an N N such that if n > N then a n L < ε. This definition epresses the thought that for all sufficiently large n the terms of the sequence a n are close, and stay close, to the limit L. For a limit of a function we want the same idea that for all sufficiently close to a R, the values of f () get close, and stay close, to the limit. So we need to change the above definition by replacing talk of n large by close to a, which is the same as a small. But as we have seen in eamples above we do not want to require that f (a) be defined, just that f () be defined for a close to a, i.e. satisfying a > 0. This motivates the net definition. First recall that in course 53 we defined the open intervals (a, b) = { : a < < b} and the closed intervals [a, b] = { : a b} along with the obvious open-closed and closed-open intervals. Definition A neighbourhood of a is an open interval (a δ, a + δ) for some δ > 0. (Often in lectures I will write nbhd instead of the longer word, neighbourhood). ( ) In fact, we can go further and say that a neighbourhood of a is any open interval (a δ, a + δ 2 ) for some δ, δ 2 > 0. Then an open interval would be a neighbourhood of any one of it s points.( ) Definition A deleted neighbourhood of a is a neighbourhood of a with a deleted, i.e. removed. So it is really the union of two open intervals, i.e. for some δ > 0. (a δ, a) (a, a + δ) = { : 0 < a < δ} Assumption Given a function f : A R and a R, which may or may not lie in A, we will assume that there eists a deleted neighbourhood of a in A. Eample If A = R \ {} and a = then the assumption is satisfied (even though a / A). If A = [0, ] and a = then the assumption is not satisfied (even though a A). ( ) This is a stronger assumption than we actually need. It would suffice to assume that every deleted neighbourhood of a contains a point of A. 3

4 This would be the definition that a is a limit point, cluster point or point of accumulation of A. ( ) Definition Let f : A R be a function whose domain contains a deleted neighbourhood of a R. The limit of a function f at a is L if, and only if, for all ε > 0 there eists a δ > 0 such that if 0 < a < δ then f () L < ε. That is: ε > 0, δ > 0,, 0 < a < δ f () L < ε. (Recall that in logic we read p q as if p holds then q will hold.) We write lim f () = L. a But we may also write f () L as a, which we read as f () tends to L as tends to a. (Often in lectures I will write L a f () = L). Recall from course 53 that we used the definition of convergence to test a given sequence by assuming that an ε > 0 is given and then trying to find an appropriate N. Similarly, we will check a given function converges at a point a to a value L by assuming that some ε > 0 is given and then trying to find an appropriate δ > 0. L + ε L L - ε ( )( ) a (a-δ, a)(a, a+δ) Eample (i) Use the ε δ definition to prove that 2 lim = 2. (ii) Use the ε δ definition and proof by contradiction to show that lim 0 sin (π/) does not eist. 4

5 (iii) Use the ε δ definition to prove that lim 2 2 = 4. See also question 2 (As in the proofs of convergence of given sequences that we gave in a number of eamples in course 53, it is often necessary first, to do rough work. Also remember, that proofs start with what you know and end with what you want to prove, not the other way round!).2.2 One sided limits. Can we calculate lim 0? We cannot verify the definition because any deleted neighbourhood of 0 will contain negative for which is not defined. So we must conclude that lim 0 is not defined. Yet, from the graph, it seems that we could talk meaningfully of the limit of as approaches 0 from the right. Definition Right hand limit: Suppose that f : A R is defined for to the right of a R, i.e. in an interval (a, a + α) for some α > 0. Then lim a + f () = L if, and only if, for any ε > 0 there eists δ > 0 such that if a < < a + δ then f () L < ε. That is: ε > 0, δ > 0,, a < < a + δ f () L < ε. We sometimes write f (a + ) for lim a + f (). Left hand limit: Suppose that f : A R is defined for to the left of a R, i.e. in an interval (a β, a) for some β > 0. Then lim a f () = L if, and only if, for any ε > 0 there eists δ > 0 such that if a δ < < a then f () L < ε. That is: ε > 0, δ > 0,, a δ < < a f () L < ε. 5

6 We sometimes write f (a ) for lim a f (). Eample Using the definition calculate the two one-sided limits for (the signum function, also written as sgn) at = 0. The graph for this is: where the value of / is left undefined at 0. Eample Verify the definition to prove that lim 0 + = 0. See Question 2 Eample Consider + 2 if f () = 2 if < < if 3. Evaluate, if possible, the following: lim f () = lim + f () = lim f () = f () = lim 3 f () = lim 3 + f () = lim 3 f () = f (3) = lim 2 f () = lim 2 + f () = lim 2 f () = f (2) = Give full details of the calculation of lim 3 + f (). Theorem We have if, and only if, lim f () = L a lim f () = L and lim a f () = L. a + 6

7 So a limit eists if and only if both one-sided limits eist and are equal. Eample The limit lim 0 does not eist..2.3 Limit of a function at infinity. Here we want to know whether f () approaches a number L as takes arbitrarily large positive values or arbitrarily large negative values. This is similar to the convergence of sequences seen in course 53. To relate the definition from course 53 with the present situation assume we have a real sequence {a n } n, when we can then define a function f : N R by f (n) = a n for all n. The definition of convergence given in 53 becomes: for all ε > 0 there eists an N N such that if n > N then f (n) L < ε. Changing f : N R tof : A R, N N to X R and n > N to > X leads to the following definitions. Definition Assume f : A R is defined for all sufficiently large positive. Then lim f () = L + if, and only if, for all ε > 0 there eists an X R such that if > X then f () L < ε. That is: ε > 0, X R,, > X f () L < ε. Definition Assume f : A R is defined for all sufficiently large negative. Then lim f () = L if, and only if, for all ε > 0 there eists an X R such that if < X then f () L < ε. That is: ε > 0, X R,, < X f () L < ε. Definition If either one of the limits eists, then the line y = L is called a horizontal asymptote to y = f (). Note The symbols + (also simply known as ) and are not real numbers. You cannot say = or = 0. The symbols are used in this 0 course simply as a shorthand notation. So + should be read as takes arbitrarily large positive values, similarly for. 7

8 Eample: Use the definition to calculate the limits at infinity, if they eist, of the function f () =.2.4 Divergence In the theory of sequences we said that a sequence diverges if it didn t converge. There were a number of ways it could diverge, e.g. it could oscillate or it could be unbounded. For certain unbounded functions there is a type of limit that can still be defined. The first definition encapsulates the situation that given a function defined on a deleted neighbourhood of a R, and given any real number K, which we might think of as positive and large, there is some deleted neighbourhood of a on which the function is greater than K. This can be repeated for each and every real K. Presumably the larger the K the smaller the deleted neighbourhood. Then, if lim a f () is to be assigned a real value connected in some way with the values taken by the function on deleted neighbourhoods of a, this value should be larger than every real K. There is no such possible real value! Instead we assign lim a f () the symbol +. Definition Let f : A R be a function whose domain contains a deleted neighbourhood of a R. We write lim f () = +, a or say f () tends to + as tends to a if, and only if, for all for all K R there eists δ > 0 such that if 0 < a < δ then f () > K. That is: K R, δ > 0,, 0 < a < δ f () > K. () (Though we only demand K is real it is sufficient to think of K as large and positive.) Similarly, we write lim f () =, a or say f () tends to as tends to a if, and only if, for all for all K R there eists δ > 0 such that if 0 < a < δ then f () < K. That is: K R, δ > 0,, 0 < a < δ f () < K. (This time it suffices to think of X negative and large.) ( ) Since sequences can be written as a function, f : N R, presumably there is, for some unbounded sequences, a type of limit that can be defined. 8

9 This was not done in course 53 but it will be necessary to do it net year, in course 34, Measure Theory. ( ) Eample Verify the definition to show that lim ( ) 2 = +. It should not be hard for the student to supply definitions for lim a a f () = + or and lim f () = + or. + Further, the student should be able to define See Question 6. lim f () = + or, and lim f () = + or. + Eample Let Again, see Question 6. p () = a n n + a n n + a n 2 n a + a 0 be a polynomial of odd degree with real coefficients and a n > 0. Prove that lim p () = + and lim p () =. + See Question. Note In none of the cases above do we say that the limit eists. If we say that lim f () eists we are implicitly assuming that it is finite. This is because, as in the definitions of limits at infinity, the symbols + and are not real numbers. They are used simply as shorthand. To say lim a f () = + is to say that f satisfies (), and there is no use of the symbol + in that definition. Definition In any of the cases above (i.e. limit is + or as a, a+ or a ) we say that the line = a is a vertical asymptote to y = f (). Eample Sketch the graph for y = tan, showing the vertical asymptotes. 9

10 Question Sheet. Without detailed proofs evaluate the following limits. 2 2 a) lim + c) lim e) lim 2 { } b) lim d) lim t 9 f) lim t t 3 t. 8 t 2 3 t. Hint: In part d) recall a result oft used in course 53, namely that ( a ) ( a ) a b = b + b for all a, b 0. For part f) try to use a similar result where a b is divided by 3 a 3 b. 2. Use the ε δ definition to prove the following limits. a) lim = 24, b) lim = 2, c) lim 9 = a) Show by means of an eample that lim a {f () + g ()} may eist even though neither lim a f () or lim a g () eist. b) Do the same for lim a f () g (). Hint: First ask yourself what functions h, do you know of for which lim a h () does not eist for some a. Then try to construct f and g from this h. 4. Let For what is this well-defined? F () = 2. 0

11 a) Find lim + F () and lim F (). b) Does lim F () eist? c) Sketch the graph of F (), R. 5. Sketch the graph of 2 2, < 2 f () = 3, = 2 8 2, > 2 Use the ε δ definition to evaluate the following limits. Does lim 2 f () eist? lim f () and lim f () Write out the symbolic definitions for each of the following limits. lim a +f () = +, lim lim f () = +, lim + a f () = f () = 7. (i) Let G () = 2 Evaluate, without detailed proofs, for or. if they eist. lim G (), lim G (), lim G () and lim + (ii) Sketch the graph of G. G () What are the vertical and horizontal asymptotes for this function? (iii) Sketch the graphs of H () = 2 and J () = 2 2 both defined for ±.

12 8. Prove that 9. (i) Let lim f (t) = L if, and only if, lim t + f 0 + ( ) = L. f () = ( + ) By looking at very small guess the value of lim 0 f (). Prove that your guess is correct by verifying the ε δ definition. (ii) Show that Hence use Question 8 to evaluate f ( ) = f(). 0. Prove, in two different ways, that lim f (). + does not eist. lim sin (π) + Hint: First assume it does eist, choose ε = /3 in the definition and then find two sufficiently large values of for which sin (π) has quite different values. This should lead to a contradiction. The second way is also by contradiction but you should use Question 8 along with an appropriate result from the lectures.. Let p () = a n n + a n n + a n 2 n a + a 0 be a polynomial with real coefficients and a n 0. 2

13 a) Prove that lim + p () p () = and lim a n n a n =. n b) Use part a) to show that if p is of odd degree and a n > 0 then there eists X such that > X, p () > a n n 2. c) Use part b) to show that if p is of odd degree and a n > 0 then lim p () = +. + d) Similarly show that if p is of odd degree and a n > 0 then X 3, < X 3, p () < a n n 2 and lim p () =. 3

14 Additional Questions No solutions will be given to these questions, hopefully any hints given will be sufficient. If you need further help, please see me. 2. Verify the definition to prove that lim 0 + = 0. Hint: Given an ε > 0 show that δ = ε 2 will satisfy the definition. 3. Sketch the graph of the function y = , showing any horizontal and vertical asymptotes. 4