Part 1.2 Limits of Functions Divergence Or “When the Limit Is Infinite”

Part 1.2 Limits of functions v1 2020-21

Divergence or “when the limit is inﬁnite”.

In the theory of sequences we say that a sequence {an}n≥1 diverges if it fails to converge. There are a number of ways it could fail to converge, e.g. n it could oscillate, i.e. an = (−1) , n ≥ 1, or it could be unbounded, i.e. an = n, n ≥ 1. For certain unbounded functions there is a type of limit that can still be defined. The first definition below encapsulates the situation in which 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 positive large K. Presumably the larger the K the smaller the deleted neighbourhood. Then, if limx→a f (x) is to be assigned a value connected in some way with the values taken by the function on deleted neighbourhoods of a, this value should be larger than every positive large K. There is no such possible real value! Instead we assign the symbol +∞ to limx→a f (x).

Deﬁnition 1.2.1 1. Let f : A → R be a function whose domain contains a deleted neighbourhood of a ∈ R. We write lim f(x) = +∞, x→a or say “f(x) tends to + ∞ as x tends to a” if, and only if, for all for all K > 0 there exists δ > 0 such that if 0 < |x − a| < δ then f(x) > K. That is: ∀K > 0, ∃ δ > 0, ∀x ∈ A, 0 < |x − a| < δ =⇒ f(x) > K. (1)

2. Similarly, we write lim f(x) = −∞, x→a or say “f(x) tends to − ∞ as x tends to a” if, and only if, for all for all K < 0 there exists δ > 0 such that if 0 < |x − a| < δ then f(x) < K. That is: ∀K < 0, ∃ δ > 0, ∀x ∈ A, 0 < |x − a| < δ =⇒ f(x) < K.

1 Note that here we have f(x) < K < 0 and so, because the numbers are negative, |f(x)| > |K|, i.e. f(x) will be larger, in magnitude, than K!

As an illustration of the K - δ deﬁnition of the limit being +∞ we have

x 1 δ 1 1+δ −

Example 1.2.2 Verify the K - δ deﬁnition to show that x lim = +∞. x→1 (x − 1)2

Graphically, this function is very much like that used in the ﬁgure above:

1 x

Solution Rough Work Assume 0 < |x − 1| < δ with δ > 0 to be chosen. If we demanded that δ ≤ 1 then 0 < |x − 1| < δ would imply 0 < x < 2, which gives x 0 > = 0, (x − 1)2 (x − 1)2

2 which is of no use. Thus we instead demand that δ be less than a number strictly less than 1. The ‘simplest’ positive number strictly less than 1 is 1/2. If we demand that δ ≤ 1/2 then 0 < |x − 1| < δ ≤ 1/2 which opens out as −1/2 ≤ x − 1 ≤ 1/2, i.e. 1/2 < x < 3/2. Thus x is no smaller than 1/2 in which case x 1 > . (x − 1)2 2 (x − 1)2 Also, 0 < |x − 1| < δ implies (x − 1)2 < δ2 and so x 1 1 > ≥ . (x − 1)2 2 (x − 1)2 2δ2 √ We demand this is ≥ K, which rearranges as δ ≤ 1/ 2K. We put these two demands on δ together as 1 1 δ = min , √ . 2 2K End of rough Work. Solution left to students

Note To show limx→a f(x) = L, with L ﬁnite you have to show that |f(x) − L| < ε. This is normally done by ﬁnding a simpler, upper bound for |f(x) − L| and then demanding this upper bound is < ε.

To show limx→a f(x) = +∞ you have to show that f(x) > K. This is normally done by ﬁnding an simpler, lower bound for f(x) and then demanding this lower bound is > K.

To show limx→a f(x) = −∞ you have to show that f(x) < K. This is normally done by ﬁnding an simpler, upper bound for f(x) and then demanding this upper bound is < K. Advice for the exams It is necessary to be able to deal with inequalities concerning negative numbers. For example x < y < 0 implies |x| > |y| ; x < y implies − x > −y; x < y and u < 0 imply ux > uy. And, as long as x and y are of the same sign, the direction of the inequality is reversed on inverting, i.e. 1 1 if either x > y > 0 or 0 > x > y then > . y x

3 For the function illustrated in the following ﬁgure it would appear that, as x tends to +∞, the function f(x) also tend to +∞. And that as x tends to 1 from above, i.e. x → 1+, that f(x) tends to +∞.

x 0 1

It should not be hard for the student to supply deﬁnitions for

lim f(x) = +∞ or − ∞, and lim f(x) = +∞ or − ∞. x→+∞ x→a+ Further, the student should be able to also deﬁne

lim f(x) = +∞ or − ∞ and lim f(x) = +∞ or − ∞. x→−∞ x→a− In fact you will be asked to do just this in a question on the Problem Sheets.

Note In none of the cases above do we say that the limit exists. If we say “lim f(x) exists” 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 limx→a f(x) = +∞ is to say that f satisfies (1), and there is no use of the symbol +∞ in that definition.

4 A slightly more diﬃcult example discussed in the Tutorial.

Example 1.2.3 Verify the K - δ deﬁnition to show that x lim = −∞. x→−1 (x + 1)2

Hint Recall that given K < 0 we want to ﬁnd x for which f(x) < K. We might attempt this by ﬁnding a simpler upper bound for f(x) and then demanding this upper bound is < K . Graphically:

1 x

Solution Rough Work Assume 0 < |x − (−1)| < δ with δ > 0 to be chosen.

If δ ≤ 1 then 0 < |x + 1| < δ ≤ 1 would imply −2 < x < 0, which gives the upper bound x 0 < = 0. (x + 1)2 (x + 1)2 There is no way of demanding this is < K < 0. Instead demand that δ ≤ 1/2. Then

0 < |x + 1| < 1/2 =⇒ −1/2 < x + 1 < 1/2 =⇒ −3/2 < x < −1/2.

Use the upper bound on x to give x 1 < − . (x + 1)2 2 (x + 1)2

Then 0 < |x + 1| < δ implies 0 < (x + 1)2 < δ2 so 1 1 1 1 > and − < − . (x + 1)2 δ2 2 (x + 1)2 2δ2

5 We now demand this is < K. This rearranges as

r 1 δ ≤ − . 2K Remember that K is negative so we are not taking the square root of a negative number. So δ = min 1/2, p−1/2K will suﬃce. End of Rough Work

Proof Let K < 0 be given. Choose δ = min 1/2, p−1/2K . Assume 0 < |x + 1| < δ. Then ﬁrst |x + 1| < δ ≤ 1/2 implies x < −1/2. Secondly

r 1 1 1 |x + 1| < δ ≤ − =⇒ (x + 1)2 ≤ − =⇒ ≥ −2K. (2) 2K 2K (x + 1)2

Combine these inequalities as

x 1 1 1 < − ≤ − (−2K) , (3) (x + 1)2 2 (x + 1)2 2 where the direction of the inequality in (2) has changed in (3) because of being multiplied by the negative −1/2. Hence x < K. (x + 1)2

2 Thus we have veriﬁed the K - δ deﬁnition of limx→−1 x/ (x + 1) = −∞.

6 Limit Rules An important result says that if the limit of f(x) as x → a exists and is non-zero then, for x suﬃciently close to a, the values of the function f (x) cannot be too large nor too small.

Lemma 1.2.4 If limx→a f (x) = L exists then there exists δ > 0 such that  L 3L  < f(x) < if L > 0  2 2   if 0 < |x − a| < δ then 3L L < f(x) < if L < 0  2 2    |f(x)| < |L| + 1 for all L.

The ﬁrst two cases can be summed up as

|L| 3 |L| < |f(x)| < 2 2 if L 6= 0. A deduction from this is that if the limit at a is non-zero then there exists a deleted neighbourhood of a in which f is non-zero. Proof Assume L > 0. Choose ε = L/2 > 0 in the ε - δ deﬁnition of limx→a f(x) = L to ﬁnd δ > 0 such that L 0 < |x − a| < δ =⇒ |f(x) − L| < 2 L L =⇒ − < f(x) − L < 2 2 L 3L =⇒ < f(x) < . 2 2

Assume L < 0. Choose ε = −L/2 > 0 in the ε - δ deﬁnition of limx→a f(x) = L to ﬁnd δ > 0 such that L 0 < |x − a| < δ =⇒ |f(x) − L| < − 2 L L =⇒ − − < f(x) − L < − 2 2 3L L =⇒ < f(x) < . 2 2

7 Given any L choose ε = 1 in the ε - δ deﬁnition of limx→a f(x) = L to ﬁnd δ > 0 such that if 0 < |x − a| < δ then |f(x) − L| < 1. Then, by the triangle inequality,

|f(x)| = |(f(x) − L) + L| ≤ |f(x) − L| + |L| < 1 + |L| .

Theorem 1.2.5 Suppose that f and g are both deﬁned on some deleted neighbourhood of a and that both limx→a f(x) = L and limx→a g(x) = M exist. Then: Sum Rule: the limit of the sum exists and equals the sum of the limits:

lim [f(x) + g(x)] = L + M, x→a Product Rule: the limit of the product exists and equals the product of the limits: lim f(x) g(x) = LM, x→a Quotient Rule: the limit of the quotient exists and equals the quotient of the limits: f(x) L lim = , provided that M 6= 0. x→a g(x) M

Proof of the Sum Rule is left to student.

Proof of the Product Rule. Consider

|f(x) g(x) − LM| = |f(x) g(x) − Lg(x) + Lg(x) − LM| ,

“adding in zero”, ≤ |f(x) g(x) − Lg(x)| + |Lg(x) − LM| by the triangle inequality, = |f(x) − L| |g(x)| + |L| |g(x) − M| (4)

Let ε > 0 be given. From the Lemma above limx→a g(x) = M means there exists δ1 > 0 such that

0 < |x − a| < δ1 =⇒ |g(x)| < |M| + 1. (5)

8 From the deﬁnition of limx→a f(x) = L we ﬁnd that there exists δ2 > 0 such that, if 0 < |x − a| < δ2 then ε |f(x) − L| < . (6) 2 (|M| + 1)

Finally, the deﬁnition of limx→a g (x) = M means there exists δ3 > 0 such that, if 0 < |x − a| < δ3 then ε |g(x) − M| < , (7) 2 (|L| + 1) (where we have put a “+1” in the denominator, 2 (|L| + 1) , in case L = 0).

Choose δ = min (δ1, δ2, δ3) > 0. Assume 0 < |x − a| < δ. For such x all the three bounds (5), (7) and (6) hold. Then, returning to (4) , |f(x) g(x) − LM| ≤ |f(x) − L| |g(x)| + |L| |g(x) − M| , ε ε < (|M| + 1) + |L| 2 (|M| + 1) by (5) 2 (|L| + 1) | {z } | {z } by (6) by (7)   1 1 |L|  =  + ×  ε < ε. 2 2 (|L| + 1) | {z } <1

Thus we have veriﬁed the deﬁnition that limx→a f(x) g(x) = LM. Proof of the Quotient Rule for limits.

Assume limx→a g(x) = M and M 6= 0.

By the Lemma 1.2.4 we also have δ1 > 0 such that if 0 < |x − a| < δ1 then |M| |g(x)| > . (8) 2 In particular g (x) 6= 0 and so, for such x we can consider

1 1 1 |g(x) − M| 2 |g(x) − M| − = < × . (9) g(x) M |g(x)| M |M| M |{z} by (8)

Let ε > 0 be given. From the deﬁnition of limx→a g(x) = M there exists δ2 > 0 such that if 0 < |x − a| < δ2 then ε |M|2 |g(x) − M| < . (10) 2 9 Let δ = min (δ1, δ2) and assume 0 < |x − a| < δ. For such x both (10) and (9) hold and we have

2 ! 1 1 2 2 |M| ε − ≤ |g(x) − M| < × = ε. g(x) M M 2 M 2 2 | {z } by (10)

Hence we have veriﬁed the deﬁnition of limx→a 1/g (x) = 1/M

I leave it to the Student to use the Product Rule to deduce f(x) L lim = x→a g(x) M

Advice for the exam. If asked to evaluate a limit by verifying the ε − δ deﬁnition, do not use the limit laws. If asked to evaluate a limit without restriction on the method and you use a limit law tell me the rule being a used.

10 Example 1.2.6 For a power xn, n ∈ N we have, by the Product Rule, n lim xn = lim x = an. x→a x→a

n n−1 For any polynomial p(x) = cnx + cn−1x + ... + c1x + c0, we have, by the Sum Rule, n n X i X i lim p(x) = ci lim x = cia = p(a) . x→a x→a i=0 i=0

This says: The limit of a polynomial at a point is the value of the polynomial at that point.

Example 1.2.7 A rational function is the quotient of polynomials, so r(x) is a rational function if, and only if, it can be written as p(x)/q(x) for some polynomials p(x) and q(x). Then

p(x) lim p(x) lim r(x) = lim = x→a x→a x→a q(x) limx→a q(x) by the quotient rule, provided limx→a q(x) = q(a) is non-zero. Thus, since the limits of these polynomials equal their values at the limit point, p(a) lim r(x) = = r(a) . x→a q(a)

This says The limit of a rational function at a point is the value of the rational function at that point, provided that value is deﬁned.

Example 1.2.8 As particular examples we deduce

lim (x + 3) = 2 + 3 = 5, x→2 and lim x2 + 2x + 2 = 4 + 4 + 2 = 10. x→2 Then, since 5 6= 0, we can use the Quotient Rule to deduce,

x2 + 2x + 2 lim (x2 + 2x + 2) 10 lim = x→2 = = 2, x→2 x + 3 limx→2 (x + 3) 5 as has been proved earlier by verifying the ε − δ deﬁnition.

11 Note 1 We can not use the Quotient Rule to calculate x2 − 1 lim . x→1 x2 − x

2 This is because limx→1 q(x) = limx→1 (x − x) = 0, and so the necessary con- ditions of the Theorem 1.2.5 are not satisﬁed.

Note 2 The Rules for Limits also hold if x → a is replaced by either of the one-sided limits x → a+, x → a− or limits at inﬁnity x → +∞ or x → −∞. It would be useful for the student to modify the proof I have given to show that it holds in these cases.

Recalling limx→+∞ 1/x = 0, proved by verifying the deﬁnition, means that by the Product Rule for limits at inﬁnity 1 1 n lim = lim = 0, x→+∞ xn x→+∞ x for all n ≥ 1. This simple result has applications as in the following.

Example 1.2.9 4x2 + 2 lim = 2. x→+∞ 2x2 + 4x Solution Divide top and bottom by the largest power of x, namely x2 to get

2 2 2 4x + 2 4 + 2/x limx→+∞ (4 + 2/x ) lim 2 = lim = , x→+∞ 2x + 4x x→+∞ 2 + 4/x limx→+∞ (2 + 4/x) by the Quotient Rule, allowable since both limit top and bottom both exist and the bottom one is non-zero. Thus

2 2 4x + 2 limx→+∞ (4 + 2/x ) 4 lim 2 = = = 2. x→+∞ 2x + 4x limx→+∞ (2 + 4/x) 2 Note We cannot say

2 2 4x + 2 limx→+∞ (4x + 2) lim 2 = 2 , x→+∞ 2x + 4x limx→+∞ (2x + 4x) because neither of the limits on the right hand side exist.

12 Theorem 1.2.10 Sandwich Rule: Suppose that f, g and h are three functions such that

h(x) ≤ f(x) ≤ g(x) for all x in some deleted neighbourhood of a.

If limx→a h(x) = L and limx→a g(x) = L then limx→a f(x) = L.

Proof By the assumption in the Theorem there exists δ0 > 0 such that if 0 < |x − a| < δ0 then h(x) ≤ f(x) ≤ g(x). Let ε > 0 be given.

From the deﬁnition of limx→a h(x) = L there exists δ1 > 0 such that

0 < |x − a| < δ1 =⇒ |h(x) − L| < ε =⇒ L − ε < h(x) < L + ε =⇒ L − ε < h(x) .

From the deﬁnition of limx→a g(x) = L there exists δ2 > 0 such that

0 < |x − a| < δ2 =⇒ |g(x) − L| < ε =⇒ L − ε < g(x) < L + ε =⇒ g(x) < L + ε.

Let δ = min (δ0, δ1, δ2) > 0 and assume 0 < |x − a| < δ. For such x we have all of h(x) ≤ f(x) ≤ g(x), L − ε < h(x) and g(x) < L + ε. Combine as in L − ε < h(x) ≤ f(x) ≤ g(x) < L + ε, i.e. |f(x) − L| < ε.

Thus we have veriﬁed the deﬁnition of limx→a f(x) = L. Note The Sandwich rule also holds if x → a is replaced throughout by x → a+ or a−, or x → +∞ or x → −∞.

Example 1.2.11 Let

f(x) = (x + 1)2 sin (10 (x + 1)) − 1.

Find limx→−1 f(x).

13 Solution Start from the simple fact that −1 ≤ sin θ ≤ 1 for all θ. Hence

−1 ≤ sin (10 (x + 1)) ≤ 1.

Thus

− (x + 1)2 − 1 ≤ (x + 1)2 sin (10 (x + 1)) − 1 ≤ (x + 1)2 − 1.

By the product and sum rules for limits we have

2 lim − (x + 1)2 − 1 = − lim x + 1 − 1 = −1 x→−1 x→−1 and lim (x + 1)2 − 1 = −1. x→−1

So, by the Sandwich rule,

lim (x + 1)2 sin (10 (x + 1)) − 1 = −1. x→−1

14 Example 1.2.12 Prove that π lim θ sin = 0. θ→0 θ

Solution Start from the fact that, for any α ∈ R we have − |α| ≤ α ≤ |α| .

In fact more is true, either α = |α| or α = − |α| but the inequality is all we require. Apply this with α = θ sin (π/θ) , θ 6= 0, to get π π π − θ sin ≤ θ sin ≤ θ sin . θ θ θ

Then since |sin (π/θ)| ≤ 1 we deduce π − |θ| ≤ θ sin ≤ |θ| , θ for θ 6= 0. Finish oﬀ quoting the Sandwich Rule along with limθ→0 |θ| = 0.

15 Perhaps this ﬁgure will show what is happening:

16 Appendix 1.2

1. Do not mix up the deﬁnitions for limx→a f(x) = c with a ﬁnite limit c ∈ R and limx→a f(x) = +∞ or −∞.

For limx→a f(x) = +∞ we do not look at |f(x) − ∞| for x close to a since ∞ is not a real number and thus f(x) − ∞ has no meaning. Do not use the quotient rule when trying to prove that a limit is inﬁnite.

Example 1.2.13 We cannot say 1 1 lim = x→1 2 2 (x − 1) limx→1 (x − 1)

2 because limx→1 (x − 1) = 0 which is excluded from the Quotient Rule. If you continue you get 1 1 1 lim = = x→1 2 2 (x − 1) limx→1 (x − 1) 0 which is not deﬁned (and in particular it is not equal to +∞).

2. Proof of the Sum Rule for limits. If limx→a f(x) = L and limx→a g(x) = M then lim (f + g)(x) = L + M. x→a Let ε > 0 be given.

From the deﬁnition of limx→a f(x) = L we ﬁnd δ1 > 0 such that

0 < |x − a| < δ1 =⇒ |f(x) − L| < ε/2. (11)

From the deﬁnition of limx→a g(x) = M we ﬁnd δ2 > 0 such that

0 < |x − a| < δ2 =⇒ |g(x) − M| < ε/2. (12)

Let δ = min (δ1, δ2). Assume 0 < |x − a| < δ. For such x both (11) and (12) hold. Thus |(f(x) + g(x)) − (L + M)| = |f(x) − L + g(x) − M| ≤ |f(x) − L| + |g(x) − M| by triangle inequality, < ε/2 + ε/2 = ε.

17 Thus we have veriﬁed the deﬁnition of

lim (f + g)(x) = L + M. x→a

3. Limits of Rational Functions. In the lectures we have shown that the limit of a rational function at a point is the value of the rational function at that point, provided that value is defined. So assume we are given a rational function r(x) = p(x)/q(x) and a point a for which q(a) 6= 0 (so r(a) is defined). If we are required to verify the definition that limx→a r(x) = r(a) we need examine

p(x) p(a) p(x) q(a) − p(a) q(x) |r(x) − r(a)| = − = . q(x) q(a) q(x) q(a)

The numerator here, p(x) q(a) − p(a) q(x), is a polynomial that is zero when x = a. This means is has a factor of x − a, i.e.

p(x) q(a) − p(a) q(x) = (x − a) m(x) ,

for some polynomial m(x). Then

m(x) |r(x) − r(a)| = |x − a| . q(x) qv (a)

The verification proceeds by assuming 0 < |x − a| < δ, which is applied to the first term. Then assume δ ≤ C for some constant C for which the second term m(x)/q(x) q(a) is defined on |x − a| < C. Then an upper bound M > 0 is found for this term, i.e.

m(x) ≤ M q(x) q(a)

for |x − a| < C. It suﬃces then to choose δ = min (C, ε/M).