MAS221 Analysis, Semester 2, Week 1 Test
General Comments: The average mark on the test was 9.5/20. I view the
test questions as comparable to the “easy marks” part of the exam; you need
to be getting pretty much all of these marks to pass the exam.
Solution 1
A sequence (an ) is said to converge to a limit l ∈ R if given any ε > 0, there
exists N ∈ N such that for all n > N , we have |an − l| < ε.
(2 marks)
Comment: This is an absolutely crucial definition. You need to know it and
also how to use it in examples. See also the comments on Q2.
Solution 2
a) Since
3n − 7
7
=3− ,
n
n
we guess that the limit is 3.
Given ε > 0, let N ∈ N be such that N > 7/ε. So 7/N < ε. Then for all
n > N,
3n − 7
7
7
7
n − 3 = 3 − n − 3 = n < N < ε.
So the sequence converges to the limit 3.
(5 marks)
Comment: This is a very standard type of example of using the above
definition. Lots of people who correctly wrote out the definition in Q1
were unable to use it correctly in this example.
There’s a standard form to this type of answer: it should have “Given
ε > 0” near the start, followed by “let N...” where some formula for N in
terms of is given. This formula is typically found by rough work. Do not
be tempted to let the rough work stand as your answer - if you do so, you
will almost certainly get the logic wrong. Instead, write out the answer
“forwards”.
b) Since the sequence (bn ) bounded, there is some M > 0 such that |bn | < M
for all n ∈ N.
Given ε > 0, let N ∈ N be such that N > M/ε. So 1/N < ε/M .
Then for all n > N ,
bn
− 0 = |bn | ≤ M < M < M ε = ε.
n
n
n
N
M
Thus the sequence (bn /n) converges to the limit 0.
1
(5 marks)
Comment: As long as you know what bounded means, this is very similar
to the previous part. A distressingly large number of people attempted
to show the conclusion without using that (bn ) is bounded. This was
doomed, since the result is of course not true without this condition. For
example, consider the unbounded sequence bn = n, which has bn /n = 1
for all n and so converges to 1 not 0. Or consider the unbounded sequence
bn = n2 , which has bn /n = n for all n and so does not converge.
There’s a general point here: we do not usually state random conditions
that are not needed!
Solution 3
A function f : R → R is differentiable at a if
lim
x→a
f (x) − f (a)
x−a
exists (and is finite).
(2 marks)
Comment: Again this is a crucial definition you need to know.
Solution 4
a) False. For example, the sequence an = (−1)n is bounded but not convergent.
b) True.
c) True.
d) False. For example, the function f : R → R given by f (x) = |x| is
continuous but not differentiable at 0.
(6 marks)
Comment: This question was generally quite well done.
2