MAS221 Analysis (Semester 1)– Solutions to Problems 47 to 54
47. Suppose that limx→a f (x) = l and limx→a f (x) = l0 , with l 6= l0 . Then
given any sequence (xn ) in Df \ {a} that converges to a, we have
limn→∞ f (xn ) = l and also limn→∞ f (xn ) = l0 . But then l = l0 by
Theorem 2.1.1 and we have our desired contradiction.
48. When x > 0,
when x < 0,
x
x
|x|
=
= = 1 = sgn(x),
x
|x|
x
|x|
x
x
=
= − = −1 = sgn(x).
x
|x|
x
lim sgn(x) = −1, lim sgn(x) = 1.
x↑0
x↓0
49. (a) The only point at which left and right limits disagree is x = 1, with
limx↑1 f (x) = 0, limx↓1 f (x) = 1.
(b) In this case, left and right limits disagree at every n ∈ Z, with
limx↑n [x] = n − 1, limx↓n [x] = n.
(c) Here, left and right limits disagree at x = 0, 1 and 2. We have
limx↑0 h(x) = 3, limx↓0 h(x) = −2; limx↑1 h(x) = −2, limx↓1 h(x) = 10,;
limx↑2 h(x) = 10, limx↓2 h(x) = 3.
50. Df = R \ {0}. It has no limit at x = 0. To see this first take θ =
1
π/2, and consider the sequence (xn ) whose nth term is
.
π/2 + 2πn
Then limn→∞ xn = 0, but limn→∞ f (xn ) = 1. Next take θ = 3π/2,
1
and consider the sequence (yn ) whose nth term is
. Then
3π/2 + 2πn
limn→∞ yn = 0, but limn→∞ f (yn ) = −1. Here is how Maple draws the
graph close to zero:
1
Figure 1: Graph of f (x) = sin(1/x)
51. Df = R \ {0}. We have limx→0 x sin(1/x) = 0. To see this let (xn )
be an arbitrary sequence in Df that converges to 0. Then since −1 ≤
sin(1/xn ) ≤ 1 for all n ∈ N, the sequence whose nth term is sin(1/xn ) is
bounded. Then the result follows by Problem 30. Alternatively use the
sandwich rule, together with Problem 23: since −1 ≤ sin(1/xn ) ≤ 1
then −|xn | ≤ xn sin(1/xn ) ≤ |xn |. You should contrast the picture
below with that for the previous question.
Figure 2: Graph of f (x) = x sin(1/x)
52. We’ll just deal with the left limit here, as the right limit argument
is very similar. Assume the (, δ)–criterion for left limits holds, and
consider an arbitrary sequence (xn ) with xn < a for all n ∈ N that
2
converges to a. Then given any η > 0, there exists N ∈ N so that if
n > N , then 0 < a − xn < η. Now choose η to be δ from the criterion,
and argue as in the proof of Theorem 3.3.2.
The converse works in the same way as in that proof, except that this
time the sequence (xn ) is constructed by choosing xn ∈ Df such that
0 < a − xn < 1/n, and |f (xn ) − l| ≥ for each n ∈ N.
53. We’ll just do limx→∞ f (x) here, as limx→−∞ f (x) is so similar.
(a) We say that limx→∞ f (x) = l if whenever (xn ) is a sequence
that diverges to infinity, with xn ∈ Df for all n ∈ N, then
limx→∞ f (xn ) = l.
(b) The analogue of the ( − δ) criterion is ( − K): given any > 0,
there exists K > 0 such that if x > K then |f (x) − l| < .
For the proof, suppose the (−K) criterion holds and limn→∞ xn =
∞. Then given any L > 0, there exists N ∈ N so that if n > N ,
we have xn > L. Now take L to be K from the criterion and we
get that for all n > N, |f (xn ) − l| < . Hence limx→∞ f (xn ) = l,
as required.
For the converse, we again imitate the proof of Theorem 3.2.2.
This time choose successively K = 1, 2, . . . and construct xn ∈ Df
such that xn > n, and |f (xn ) − l| ≥ for each n ∈ N.
(c) Given any > 0, choose K = 1/. Then x > K ⇒ 1/x < , and
the result follows.
54. (a) We have limx→∞ f (x) = ∞ if for any sequence (xn ) (with xn ∈ Df
for all n ∈ N) which diverges to infinity, we also have (f (xn ))
diverges to infinity. The analogue of ( − δ) is: given any K > 0,
there exists L > 0 such that x > L ⇒ f (x) > K. The other cases
are similar.
(b) Given any K > 0, there exists L1 > 0 such that x > L1 ⇒ f (x) >
K, and given any > 0 there exists L2 > 0 so that x > L2 ⇒
l − < g(x) < l + . Choose any < l. Then for x > max{L1 , L2 }
we have f (x)g(x) > K(l − ), and the result follows. In the other
case, we have limx→−∞ h(x) = −∞; for in that case, given K > 0,
there exists L1 > 0 such that x < −L1 ⇒ f (x) < −K, and for
x < min{−L1 , −L2 } we have f (x)g(x) < −K(l + ).
(c) Write m = 2n and let
p(x) = a2n x2n + a2n−1 x2n−1 + · · · + a1 x + a0
3
a2n−1
a1
a0 + · · · + 2n−1 + 2n .
= x2n a2n +
x
x
x
a1
In part (b), take f (x) = x2n and g(x) = a2n + a2n−1
+ · · · + x2n−1
+
x
a0
. Then limx→∞ f (x) = ∞ and limx→∞ g(x) = a2n , and the
x2n
result follows from that of (b). If n is odd, limx→∞ p(x) = ∞, but
limx→−∞ p(x) = −∞.
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