MAS221 Analysis, Semester 2 Solutions
Sarah Whitehouse
Revision Problems
Question 1
(a) The sequence (an ) converges to a if for all ε > 0, there exists N ∈ N such
that |an − a| < ε whenever n ≥ N .
(b) Let S be a bounded set of real numbers. The supremum, α = sup S, is the
least upper bound. More precisely, it is the unique real number where:
– If x ∈ S, then x ≤ α
– If β < α, then we have x ∈ S such that x > β.
(c) Let (xn ) be a bounded monotonic increasing sequence. Let S = {xn | n ∈
N}. Let x = sup S.
Let ε > 0. Then by definition of the supremum, we have N such that
xN > x − ε. Since (xn ) is monotonic increasing, whenever n ≥ N , we have
xn > x − ε.
Also, since x is an upper bound, we have xn ≤ x. We see that, for n ≥ N ,
we have |xn − x| < ε. Thus the sequence (xn ) converges to x.
Question 2
(a) The first few terms are, by definition:
a0 = 0, a1 =
1
3
7
, a2 = , a3 = .
2
4
8
It seems reasonable to guess that the sequence (an ) has limit 1.
(b) We claim that
an =
2n − 1
.
2n
This is certainly true when n = 0. Suppose the claim holds when n = k.
Then
2k − 1
.
ak =
2k
1
So
ak+1 =
1
2
2k − 1
+1
2k
=
2k − 1 + 2k
2k+1 − 1
=
.
k+1
2
2k+1
Thus the desired formula holds by induction.
Now
an =
2n − 1
1
=1− n →1
n
2
2
as n → ∞.
Question 3
Let f : [a, b] → R be a function.
(a) For all ε > 0, there exists δ > 0 such that if |x − x0 | < δ then |f (x) −
f (x0 )| < ε, or in other words, f (x) → f (x0 ) as x → x0 .
(b) The limit
f 0 (x0 ) = lim
h→0
f (x0 + h) − f (x0 )
h
exists.
(c) Let us put ε = 1 into the definition of limit. If f is diffentiable at x0 , then
there is some δ > 0 such that if 0 < |h| < δ then
0
f (x0 ) − f (x0 + h) − f (x0 ) < 1
h
Rearranging, and using the triangle inequality, this gives
|f (x0 + h) − f (x0 )| < |h| + |f 0 (x0 )h|.
Let h → 0. Then we see that |f (x0 + h) − f (x0 )| → 0, that is f (x0 + h) →
f (x0 ), and the function f is continuous at x0 .
(d) The function f : [−1, 1] → R defined by the formula f (x) = |x| is continuous but not differentiable at 0.
Question 4
• Certainly, f is differentiable when x > 0, with derivative 2x, and when
x < 0, with derivative −2x. So we only need to check what happens when
x = 0.
Observe that we can write f (x) = x|x|. So
f (0 + h) − f (0)
−h|h| − 0
=
= −|h|
h
h
and this tends to 0 as h → 0. So f is differentiable at 0, with derivative 0.
We see that we have derivative f 0 : R → R given by f 0 (x) = 2|x|.
2
• The function f 0 (x) = 2|x| is not differentiable at 0. So the second derivative of f does not exist.
3