Download Homework 4 Solutions - Math-UMN

Math 4603 Advanced Calculus I
Homework 4: Sections 2.2, 2.3
Due date: Wednesday, July 3
Short Answer
State the answer to each of the following questions. It is not necessary to write down
any justification (though of course one would want to be able to explain how each
solution was obtained).
1. Use Theorem 2.1 to prove that each function f : D → R does not have a limit at
∞
x0 . That is, find a sequence {xn }∞
n=1 with xn ∈ D and {xn }n=1 converging to x0 ,
for which the sequence {f (xn )}∞
n=1 diverges.
(a) f : (0, 1) → R, f (x) = sin(1/x); x0 = 0
nπ ∞
2 ∞
∞
Solution: {xn }∞
n=1 = { nπ }n=1 ; {f (xn )}n=1 = {sin 2 }n=1 = {1, 0, −1, 0, 1, . . .}
(b) f : R → R, f (x) = 1 when x is rational and f (x) = 0 when x is irrational;
x0 = 0
Solution: {xn }∞
n=1 defined by xn = 1/n for n even and xn = π/n for n odd;
{f (xn )}∞
=
{0,
1, 0, 1, . . .}
n=1
2. Find an example of functions f, g : D → R and x0 an accumulation point of D,
for which neither f nor g has a limit at x0 , but
(a) f + g has a limit at x0
Solution: f, g : (0, ∞) → R, x0 = 0; f (x) = 1/x, g(x) = −1/x
(b) f g has a limit at x0
Solution: f, g : R → R, x0 = 0; f (x) = 1 for x rational and f (x) = −1 for x
irrational; g(x) = −1 for x rational and g(x) = 1 for x irrational
(c) fg is well-defined and has a limit at x0
Solution: f, g : (0, ∞) → R, x0 = 0; f (x) = 1/x and g(x) = 1/x
3. Show that each function has a limit at x0 = 0 and find that limit.
(a) f : (0, 1) → R, f (x) =
x3 +6x2 +x
x2 −6x
2
Solution: For x ∈ (0, 1), we have f (x) = x +6x+1
. This function is a quotient of
x−6
two functions which each have limits at 0, and for which the function in the
denominator does not take the value 0 nor have limit 0. Therefore, by Theorem
2.4, the limit of f at 0 exists, and limx→0 f = − 61 .
√
(b) g : (0, 1) → R, g(x) =
1+x−1
x
x
1
√
Solution: For x ∈ (0, 1), we have g(x) = x((1+x)−1
= x(√1+x+1)
= √1+x+1
. This
1+x+1)
function is a quotient of two functions which each have limits at 0, and for
which the function in the denominator does not take the value 0 nor have limit
0. Therefore, by Theorem 2.4, the limit of g at 0 exists, and limx→0 g = 21 .
√
(c) h : (0, 1) → R, h(x) =
9−x−3
x
−x
−1
√
Solution: For x ∈ (0, 1), we have h(x) = x((9−x)−9
= x(√9−x+3)
= √9−x+3
. This
9−x+3)
function is a quotient of two functions which each have limits at 0, and for
which the function in the denominator does not take the value 0 nor have limit
0. Therefore, by Theorem 2.4, the limit of h at 0 exists, and limx→0 h = − 61 .
Proofs
For the remaining problems, a proof is required. In particular, a complete solution
must be stated in sentence form with appropriate justification for each step.
1. Suppose f, g, and h : D → R where x0 is an accumulation point of D,
f (x) ≤ g(x) ≤ h(x) for all x ∈ D, and f and h have limits at x0 with
limx→x0 f (x) = limx→x0 h(x). Prove that g has a limit at x0 and
lim f (x) = lim g(x) = lim h(x).
x→x0
x→x0
x→x0
Solution: Write L = limx→x0 f (x) = limx→x0 h(x).
We will apply Theorem 2.1. Let {xn }∞
n=1 be a sequence converging to x0 with
xn ∈ D and xn 6= x0 for all n. Since f (x) and h(x) both have limit L at x0 , the
∞
sequences {f (xn )}∞
n=1 and {h(xn )}n=1 both converge to L. Because
f (xn ) ≤ g(xn ) ≤ h(xn ) for all n, we may apply Proof 7 from Homework 2 to
conclude that {g(xn )}∞
n=1 converges to L as well. By Theorem 2.1, it follows that
g must have a limit at x0 , and the only possible limit is L.
2. Suppose f : D → R has a limit at x0 . Prove that g : D → R defined by
g(x) = |f (x)| has a limit at x0 and that limx→x0 g(x) = | limx→x0 f (x)|.
Solution: Write L = limx→x0 f .
We will use the definition of the limit of a function. Let > 0 be given. Since f
has limit L at x0 , there exists δ such that if 0 < |x − x0 | < δ and x ∈ D, then
|f (x) − L| < . By Theorem 0.25, we have
g(x) − |L| = |f (x)| − |L| ≤ |f (x) − L| < ,
so the same δ works to show that g has limit |L| at x0 .
3. Prove Theorem 2.5: Suppose f : D → R and g : D → R with x0 an accumulation
point of D, and suppose f and g have limits at x0 . If f (x) ≤ g(x) for all x ∈ D,
then
lim f (x) ≤ lim g(x).
x→x0
x→x0
Solution: Write F = limx→x0 f and G = limx→x0 g. Suppose that f (x) ≤ g(x) for
. Since f has limit F at x0 ,
all x ∈ D, and suppose that F > G. Choose = F −G
2
there exists δf such that if |x − x0 | < δf , then |f (x) − F | < F −G
. Since g has limit
2
G at x0 , there exists δg such that if |x − x0 | < δg , then |g(x) − G| < F −G
.
2
Choose δ = min{δf , δg }. Since x0 is an accumulation point of D, there exists
x ∈ (x0 − δ, x0 + δ) for which x ∈ D. Since |x − x0 | < δ ≤ δf , we know that
, so f (x) > F − F −G
= F +G
. Since |x − x0 | < δ ≤ δg , we know
|f (x) − F | < F −G
2
2
2
F −G
F −G
that |g(x) − G| < 2 , so g(x) < G + 2 = F +G
. But now we have
2
g(x) <
F +G
< f (g),
2
which contradicts the original assumption that f (x) ≤ g(x). It follows that, if
f (x) ≤ g(x) for all x, then F ≤ G.
4. Prove Theorem 2.4 (i) directly. That is, using the definition of a limit, show that
for f, g : D → R with x0 an accumulation point of D at which both f and g have
limits, the function f + g has a limit at x0 and
lim (f + g) = lim f + lim g.
x→x0
x→x0
x→x0
Solution: Suppose f and g have limits at x0 , and write F = limx→x0 f and
G = limx→x0 g.
Let > 0 be given. Since f has limit F at x0 and g has limit G at x0 , we know that
there exists δf such that if 0 < |x − x0 | < δf and x ∈ D, then |f (x) − F | < /2;
and there exists δg such that if 0 < |x − x0 | < δg and x ∈ D, then |g(x) − G| < /2.
Take δ = min{δf , δg }. Then for x ∈ D with 0 < |x − x0 | < δ, we have
|(f +g)(x)−(F +G)| = |f (x)+g(x)−F −G| ≤ |f (x)−F |+|g(x)−G| < /2+/2 = .
This shows that the function f + g has limit F + G at x0 .
5. Prove Theorem 2.4 (ii) using sequences. That is, by applying Theorems 2.1 and
1.9, show that for f, g : D → R with x0 an accumulation point of D at which
both f and g have limits, the function f g has a limit at x0 and
lim (f g) = ( lim f )( lim g).
x→x0
x→x0
x→x0
Solution: Suppose f and g have limits at x0 , and write F = limx→x0 f and
G = limx→x0 g.
Let {xn }∞
n=1 be a sequence converging to x0 with xn ∈ D and xn 6= x0 for all n.
Since f and g have limits at x0 , by Theorem 2.1 we know that the sequence
∞
{f (xn )}∞
n=1 converges to F and the sequence {g(xn )}n=1 converges to G. By
Theorem 1.9, the sequence {f (xn )g(xn )}∞
n=1 then converges to F G. By Theorem
2.1, the function (f g)(x) = f (x)g(x) converges, and it must converge to F G.