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MATH 20101 Real Analysis, Exam 2011-
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A1 a) By verifying the ε − δ definition prove that
i) xlim→1+
x^3 − 1
ii) xlim→ 1 −
x^2 − x
(You may assume that a^3 − b^3 = (a − b) (a^2 + ab + b^2 ) for a, b ∈ R.)
b) Assume that a function g is defined on a deleted neighbourhood of a ∈ R.
i) Prove that if limx→a+ g (x) = L and limx→a− g (x) = L then limx→a g (x) = L.
ii) Deduce that the function
f (x) =
x^2 − x for x < 1 , x^3 − 1 for x ≥ 1.
is continuous at x = 1.
c) Assume that a function g is defined on a neighbourhood of a ∈ R. i) State the definition that g is differentiable at a.
ii) Prove that if g is differentiable at a then g is continuous at a.
iii) Prove that the function f in Part b ii) is not differentiable at x = 1.
Solution A1 i) a) Let ε > 0 be given. Choose δ = min (1, ε/7). Assume
1 < x < 1 + δ (equivalently 0 < x − 1 < δ). Consider
|f (x) − 0 | =
x^3 − 1
= |x − 1 |
x^2 + x + 1
< δ
x^2 + x + 1
Since δ ≤ 1 we have 1 < x < 2 and thus |x^2 + x + 1| < 7. Hence
|f (x) − 0 | ≤ 7 δ ≤ 7 (ε/7) = ε,
and we have verified the definition that limx→1+ (x^3 − 1) = 0. [5]
b) Let ε > 0 be given. Choose δ = min (1, ε). Assume 1 − δ < x < 1, in which case 0 < x < 1, since δ ≤ 1. Then
|f (x) − 0 | =
∣x^2 − x
∣ (^) = |x| |x − 1 | ≤ δ = ε.
Thus we have verified the definition that limx→ 1 − (x^2 − x) = 0. [3]
ii) a) Let ε > 0 be given.
lim x→a+ g (x) = L ⇒ ∃δ 1 > 0 : a < x < a + δ 1 ⇒ |g (x) − L| < ε. (1)
xlim→a− g^ (x)^ =^ L^ ⇒ ∃δ^2 >^ 0 :^ a^ −^ δ^2 < x < a^ ⇒ |g^ (x)^ −^ L|^ < ε.^ (2) Choose δ = min (δ 1 , δ 2 ) and assume 0 < |x − a| < δ. There are two cases for these x. Firstly, if x > a then combined with 0 < |x − a| < δ gives a < x < a + δ ≤ a + δ 1 since δ ≤ δ 1. Thus by (1) we have |g (x) − L| < ε.
Secondly, if x < a then combined with 0 < |x − a| < δ gives a − δ < x < a. Now, δ ≤ δ 2 so −δ ≥ −δ 2 , and a − δ 2 < a − δ < x < a. Thus by (2) we have |g (x) − L| < ε.
In both cases we have |g (x) − L| < ε and we have verified the definition of limx→a g (x) = L. Bookwork [5]
b) By Part i we have
xlim→ 1 − f^ (x) = 0 =^ xlim→1+ f^ (x)^. So by Part ii a) we deduce that limx→ 1 f (x) = 0. And since f (1) = 0 we have that f is continuous at x = 1. [1]
iii) a) The function g is differentiable at a iff
xlim→a
g (x) − g (a) x − a
exists. Definition [1]
b) Assume that g is differentiable at a so g′^ (a) exists. Then for x 6 = a,
g (x) − g (a) = g (x) − g (a) x − a
(x − a).
Commonly seen mistakes in Question 1
i) The majority of students did not verify the one-sided limits and so assumed 0 < |x − 1 | < δ in place of either 1 < x < 1 + δ or 1 − δ < x < 1. This lost a number of marks.
Many students didn’t read the question where it reminded you that a^3 − b^3 = (a − b) (a^2 + ab + b^2 ) and they thus incorrectly factorised x^3 − 1.
i) b) Note that you require more that δ < ε. For instance, imagine ε = 103 and δ = 102. Then 1 − δ < x < 1 becomes − 101 < x < 1 and so |x| < 101. Thus |f (x) − 0 | =
∣x^2 − x
∣ (^) = |x| |x − 1 | ≤ 101 δ,
and you would then require δ < ε/ 101. It is for this reason that in the answer above I choose δ = min (1, ε).
ii) a) Most students forgot that you have to first assume ε > 0 is given and then find both δ 1 and δ 2 , i.e. the two δ’s have to be functions of the same ε.
Read the question: I did not ask you to prove that if limx→a g (x) = L then limx→a+ g (x) = L and limx→a− g (x) = L.
ii) b) To show continuity you have to show both that limx→ 1 − f (x) = 0 = limx→1+ f (x) and f (1) = 0. Most students just showed that the two limits are equal.
iii) a) The most important word in the answer is “exists”, without it there was no mark for the answer.
iii) b) A mark was lost if you didn’t say that the Product Rule for Limits was used to justify
xlim→a
g (x) − g (a) x − a (g (x) − g (a)) = lim x→a g (x) − g (a) x − a xlim→a (x^ −^ a)^. c) Some students incorrectly wrote
f ′^ (x) =
2 x − 1 for x < 1 , 3 x^2 for x ≥ 1.
You cannot do this at x = 1.
A2 a) By verifying the definition show that
1 1 + x^2
is differentiable on R and find its derivative.
Any results you use on limits should be stated carefully.
b) State carefully the Mean Value Theorem.
Assume that f is continuous on [a, b], differentiable on (a, b) with f ′^ (x) > 0 for all x ∈ (a, b). Prove that f is strictly increasing on [a, b].
c) State carefully the Intermediate Value Theorem.
Assume that a function g is a strictly increasing continuous function on [a, b]. Explain how to define the inverse function g−^1 , proving that your g−^1 is well-defined.
Prove also that g−^1 is strictly increasing.
Solution A2 i) Let f (x) = 1/ (1 + x^2 ) and let a ∈ R be given. Consider
f (x) − f (a) x − a
1 1+x^2 −^ 1 1+a^2 x − a
a^2 − x^2 (1 + x^2 ) (1 + a^2 ) (x − a)
= − a + x (1 + x^2 ) (1 + a^2 )
2 a (1 + a^2 )^2
as x → a. Say either using product and quotient rules for limits or rational functions are continuous wherever defined. Since a was arbitrary
d dx
1 + x^2
− 2 x (1 + x^2 )^2
for all x ∈ R. On problem sheet [5]
ii) Mean Value Theorem: If f is continuous on [a, b] and differentiable on (a, b) then there exists c ∈ (a, b) for which
f (b) − f (a) = f ′^ (c) (b − a).
Commonly seen mistakes in Question 2
i) Most students failed to give a reason for
− (^) xlim→a a + x (1 + x^2 ) (1 + a^2 )
2 a (1 + a^2 )^2
by quoting either the Product and Quotient Rules for limits or the fact that rational functions are continuous wherever defined.
ii) It is not needed in the Mean Value Theorem to demand f (a) 6 = f (b) as a number of students did.
The function f is defined on [a, b] and a majority of students showed that f ′^ (x) > 0 for all x implies f (b) > f (a). But this does not mean that f is strictly increasing. You need to show that f (w) > f (y) for all possible y, w : a ≤ y < w ≤ b.
Most students failed to prove that f ′^ (x), equal to 1 1 + x − e−x^ + xe−s,
is > 0. Probably because this wasn’t rearranged over a common denominator.
iii) Differentiability is not required in the necessary conditions of the Intermediate Value Theorem.
It is not true that g−^1 =
g
Many students showed that g−^1 is 1-1 instead of showing it is well-defined. That is, they started with g−^1 (k 1 ) = g−^1 (k 2 ) deducing k 1 = k 2. This was not what was required.
A3 a) i) Assume that f is a function whose first n derivatives exist at a ∈ R.
Define Tn,af (x), the Taylor polynomial of degree n for f at a, and Rn,af (x), the remainder.
ii) Calculate T 8 , 0
sin^2 x
b) Assume that f is a function whose first n + 1 derivatives exist on an open interval containing a and x. Prove that
d dt Tn,tf (x) = (x − t)n n! f (n+1)^ (t)
for all t between a and x.
c) Assume Lagrange’s form of the error term, i.e. there exists c between a and x, for which
Rn,af (x) = f (n+1)^ (c) (n + 1)! (x − a)n+^.
i) Prove that (^) ∣ ∣∣ ∣sin
(^2) x − x (^2) + x 4 3
∣ ≤^
|x|^4 ,
for all x ∈ R.
ii) Deduce that
xlim→ 0
sin^2 x − x^2 x^4
(Do not use L’Hopital’s Rule.)
Solution A3 i) a) The Taylor polynomial of degree n for f at a, is
Tn,af (x) = f (a) + f ′^ (a) (x − a) + f ′′^ (a) 2! (x − a)^2 + ... + f (n)^ (a) n! (x − a)n^.
and Rn,af (x) = f (x) − Tn,af (x). Bookwork [2]
a) Start from the observation that we are looking at a remainder term, ∣∣ ∣∣sin^2 x − x^2 + x
4 3
∣∣ = ∣∣R 4 , 0 (sin^2 x)∣∣ (^) =
∣∣^ f^
(5) (^) (c) 5! x^5
|x|^5 =
|x|^5 ,
b) Divide through by x^4 to get ∣∣ ∣∣^ sin
(^2) x − x 2 x^4
|x|.
Quote the Sandwich rule to deduce
xlim→ 0
sin^2 x − x x^4
similar examples in lectures [4]
Commonly seen mistakes in Question 3
a) I suggest you look at the feedback for the 2008-09 paper http://www.maths.manchester.ac.uk/˜mdc/MATH20101.htm#PastExams to see the mistakes for the definition of the Taylor Polynomial that I saw again in the present paper.
There are alternative ways of finding T 8 , 0
sin^2 x
such as
f (1)^ (x) = 2 sin x cos x so f (1)^ (0) = 0 f (2)^ (x) = 2 cos^2 x − 2 sin^2 x so f (2)^ (0) = 2 f (3)^ (x) = −4 cos x sin x − 4 sin x cos x = −8 sin x cos x so f (3)^ (0) = 0 f (4)^ (x) = −8 cos^2 x + 8 sin^2 x so f (4)^ (0) = − 8 f (5)^ (x) = 16 cos x sin x + 16 sin x cos x = 32 sin x cos x so f (5)^ (0) = 0 f (6)^ (x) = 32 cos^2 x − 32 sin^2 x so f (6)^ (0) = 32 f (7)^ (x) = −64 cos x sin x − 64 sin x cos x = −128 sin x cos x so f (7)^ (0) = 0 f (8)^ (x) = −128 cos^2 x + 128 sin^2 x so f (8)^ (0) = − 128.
Students also noted that f (3)^ (x) = − 4 f (1)^ (x) so
f (3)^ (x) = − 4 f (1)^ (x) so f (3)^ (0) = − 4 f (1)^ (0) = 0 f (4)^ (x) = − 4 f (2)^ (x) so f (4)^ (0) = − 4 f (2)^ (0) = − 8 f (5)^ (x) = − 4 f (3)^ (x) = 16f (1)^ (x) so f (5)^ (0) = 16f (1)^ (0) = 0 f (6)^ (x) = − 4 f (4)^ (x) = 16f (2)^ (x) so f (6)^ (0) = 16f (2)^ (0) = 32 f (7)^ (x) = − 4 f (5)^ (x) = − 64 f (1)^ (x) so f (7)^ (0) = − 64 f (1)^ (0) = 0 f (8)^ (x) = − 4 f (6)^ (x) = − 64 f (2)^ (x) so f (8)^ (0) = − 64 f (2)^ (0) = − 128. In part ii) I needed to see some words explaining exactly how the cancel- lation occurred. I.e. something like the first term of one bracket cancelled the second term in the following bracket.
iii) a) It could have been noted that the polynomial x^2 − x^4 /3 is, in fact, T 5 , 0
sin^2 x
so that sin^2 −x^2 + x^4 /3 = R 5 , 0
sin^2 x
This would lead to the bound ≤ 2 |x|^6 /45.
b) Too many students failed to use part a to prove this limit. A few students tried to use L’Hopital’s Rule which the question explicitly states you cannot use.
and mi = glb {f (x) : x ∈ [xi− 1 , xi]}. Bookwork [2]
ii) Fix R and vary Q. We thus see that U (R, f ) is an upper bound for {L (Q, f ) : Q}. But
∫ (^) b a f^ is the^ least^ of all upper bounds, hence^ U^ (R, f^ )^ ≥ ∫ (^) b a f^. Now vary R and see that
∫ (^) b a f^ is^ a^ lower bound for^ {U^ (R, f^ ) :^ R}. But ∫ (^) b a f^ is the^ greatest^ of all lower bounds, so ∫ (^) b
a
f ≤
∫ (^) b
a
f
as required. Bookwork [4]
A bounded function f on [a, b] is Riemann integrable over [a, b] if ∫ (^) b
a
f =
∫ (^) b
a
f.
The common value is called the (Riemann) integral and is denoted by
∫ (^) b a f or
∫ (^) b a f^ (x)^ dx.^ Bookwork^ [1]
iii) Let g : [1, 2] → R, x 7 → 2 x^2 − x and, for every n ≥ 1, define
Pn =
i n : 0 ≤ i ≤ n
We first note that g (x) is increasing on [1, 2], from looking at g′^ (x) = 4 x − 1 > 0. Thus
Mi = 2
i n
i n
n i +
n^2 i^2.
Then, since xi − xi− 1 = 1/n,
U (Pn, g) =
n
∑^ n
i=
n
i +
n^2
i^2
n
n +
n
n (n + 1) 2
n^2
n (n + 1) (2n + 1) 6
19 n^2 + 15n + 2 6 n^2
[8]
Similarly,
L (Pn, g) =
n
∑^ n
i=
n (i − 1) +
n^2 (i − 1)^2
n
∑^ n−^1
i=
n
i +
n^2
i^2
= U (Pn, g) +
n
n
n n +
n^2 n^2
19 n^2 − 15 n + 2 6 n^2 [2]
From the theory,
L (Pn, g) ≤
1
g ≤
1
g ≤ U (Pn, g)
for all n ≥ 1, so
19 n^2 − 15 n + 2 6 n^2
1
g ≤
1
g ≤ 19 n^2 + 15n + 2 6 n^2
Let n → ∞, and since the outer limits are equal we have that the upper and lower integrals are equal and thus g is integrable over [1, 2]. This limit value of 19/6 is the value of the limit, i.e. ∫ (^2)
1
2 x^2 − x
dx =
[3]
