Some problems and solutions in calculus
Vladimir Jakovljevic
February 2018
1
Problems
1. Find x1 , x34 and x55 in the sequences:
n
− π))
(a) xn = arctan (tan ( n+1
(b) x1 = 1, xn = 3 + xn−1
(c) xn = bsin n + 1c

n

n = 2k
 n+3
n
(d) xn = e
3|n, n = 2k − 1


0
otherwise.
2. Find limits by definiton:
2n+1
n+2
limn→+∞ b π4 2n+1
n+2 c
1
limn→∞ sin n
(a) limn→+∞
(b)
(c)
(d) limn→+∞
(e)
(f)
(g)
√
n+1−
√
n
2
+n+1
limn→+∞ n n−2
x+2
limx→2+ x−2
x+1
limx→−∞ x−1
(h) limx→0 sin (x + 2008π).
3. Find limits:
2
n3 −n2 +1 2n +1
n3 +1
limx→0 arctan x1
1
limn→+∞ 1+bxc
n,x ∈ R
(a) limn→+∞
(b)
(c)
xn
(d) limn→+∞ supx∈[0,1) 1+x
2
qP
10
n
(e) limn→+∞ n
k=1 bk · (ak ) , ak > 0, bk > 0, ak 6= al for k 6= l.
1
(f) limn→+∞ (n + a)α − (n + b)α for real a, b, α.
(g)
lim
n→+∞
(arcsin n1 − sin n1 −
(cos
1 sin
n)
1
n2
−
1
3n3 )(n +
(1 + n13 )β
3
3) 2
1
·√
n
for real β.
√
n
4. (a) Prove that limn→+∞ nn! = 1e .
√
n+1
(n+1)!
√
(b) If an =
, calculate limn→+∞ an and limn→+∞
n
n!
an −1
ln an .
5. Let f be a continuous function on its domain Df , and let xn ∈ Df be a
Cauchy sequence.
(a) For Df = [a, b] prove that f (xn ) is also Cauchy sequence.
(b) Is f (xn ) Cauchy sequence in case that Df = (a, b)?
(c) Let x be a real number and an sequence of it’s decimal approximations. Examine the convergence of a sequence bn = arctan ean .
6. Let f : R → R be a continuous function and let |f (x) − f (y)| ≥ |x − y| for
all x, y ∈ R. Prove that f is bijection.
7. Let f : R → R be twice differentiable, concave upward function. Prove
that f (x + f 0 (x)) ≥ f (x) for all x ∈ R.
8. Let an be sequence that satisfies:
(∃ε > 0)(∀m, n ∈ N)(m 6= n ⇒ |am − an | ≥ ε).
(a) Can an be convergent sequence?
(b) Can an be bounded sequence?
9. Observe next function: f (x) = 2x3 + x3 sin x1 .
(a) Prove that f can be differentiably extended on R.
(b) Does exist some neighbourhood of x = 0 in which f is monotone.
10. Let f ∈ C 1 [0, +∞) and let exists real T > 0 for which equality f 0 (x) =
f 0 (x + T ) stands for all x ∈ (0, +∞). Suposse that f (0) > 0 and f 0 (x) < 1
for all x ∈ (0, +∞). Prove that exists some ξ ∈ [0, +∞) that satisfies
f (ξ) = ξ.
11. Let f : R → R be a continous function that satisfies next condition: if
a < b and f (a) = f (b) then exists c ∈ (a, b) such that f (a) = f (c) = f (b).
Prove that f is monotone.
12. Let an be a sequence such that limn→+∞ nan = 0.
(a) Calculate limn→+∞ (1 +
1
n
+ an )n .
2
(b) Prove that for a 6= −1, sequence bn = np cos nπ (1 + a + an )−n is well
defined starting form some n0 . Calculate lim inf bn and lim sup bn .
π
13. Show that for x > 2 inequality (x + 1) cos x+1
− x cos πx > 1 stands.
14. Let F, G : R → [0, +∞) be continuous functions such that inf x∈[0,1] F (x) =
inf x∈[0,1] G(x). Prove that exists some ξ ∈ [0, 1] such that F (ξ) = G(ξ).
15. Let f, g : R → [0, +∞) be continuous functions such that inf x∈[0,1] f (x) =
inf x∈[0,1] g(x). Prove that exists some ξ ∈ [0, 1] such that f (ξ)3 − g(ξ)3 =
(ξ) log 1 + g(ξ)−f
.
1+f (ξ)
16. Let’s observe recurrently defined sequence
x1 > 0
xn+1 =
3x2n
.
(1 + xn )3 − 1
Find limn→+∞ xn and limn→+∞ nxn .
17. Let’s observe recurrently defined sequence
x1 > 0
xn+1 = −
xn
.
2 + x2n
(a) Show that x2k and x2k−1 are monotone.
(b) Examine convergence of xn and find its limit.
18. Observe the function
(
f (x) =
x+b
a+1
2
x ln (a + x)
x<1
x ≥ 1,
where a > −1.
(a) Find a and b such that f is differentiable.
(b) Examine uniform continuity of f for that a and b.
19. Find a, b, c and d for which equality
√
3
c
d
x2
1
= ax + b + + 2 + o 2
x x
x
x3 + 8
when x → +∞, stands.
20. Let f : (a, +∞) be a differentiable function such that limx→+∞ f 0 (x) =
+∞.
3
(a) Prove that limx→+∞ f (x) − f (x − 1) = +∞.
(b) Prove that limx→+∞ f (x) = +∞.
21. Let f : R → R be upward concave function and an = f (n) + f (−n).
(a) Show that f (k + 1) − f (k) ≤ f (k + 2) − f (k + 1) for all k ∈ N.
(b) Show that an and an+1 − an are monotone.
22. Let m be a natural number. Calculate limn→+∞
n
(m
)
nm
.
23. Let (an )n∈N0 be an arithmetic sequence with non-zero common difference
d. Prove that equality
n
X
a3k =
k=1
(an an+1 )2 − (a0 a1 )2
4d
stands for all n ∈ N.
24. Let f : (0, +∞) → R be a differentiable function. Let there exist different
points a and b such that ab = ef (a)−f (b) . Show that there exists some
positive c such that cf 0 (c) = 1.
25. Let f : [0, 1] → R be a continuous function.
Prove that there exist α and
β such that β − α = 13 and 3 f (β) − f (α) = f (1) − f (0).
26. Find accumulation points of a sequence
(√
k
3k − 2 ln k + k
q
k
an =
(−1)k 2k( 1 + k1 − 1)
27. Let an be a sequence such that an+1 − an ∼
n = 2k
n = 2k − 1.
1
n2 .
(a) Prove that an is increasing starting from some n0 .
(b) Prove that an+1 − an <
2
n(n+1)
starting from some n0 .
(c) Show that there exists some n0 such that if m > n ≥ n0 then am −
an < n2 .
(d) Prove that an converge.
28. Let f : R → R be continuous, bounded function. If a 6= 0 then the
equation f (x) = ax has at least one solution. Prove it.
29. Let a be some positive constant. Observe the function f (x) = x ln (1 + xa )
for x > 0.
(a) Show that f is strictly increasing.
x
(b) Show that 1 + xa ≤ ea for all x > 0.
4
30. Let f : R → [0, 1] be a differentiable function such that |f 0 (x)| < 1. Prove
that equation f (x) = x has unique solution.
31. Let an be a sequence such that an+1 − an = √1n + o √1n . Calculate
limn→+∞
an
√
n
and limn→+∞
√
3 a
√ n.
n
32. Observe the function f (x) = (x2 + 3x)e−2x . Find f (n) (0).
33. Show that equation 3x + 6 = tan x has infinite number of solutions.
34. Prove that xn = cos n diverge.
35. Prove inequality tan x > x +
x3
3
for x ∈ (0, π2 ).
36. Calculate the limit
lim
n→+ inf
where Hn =
ln n!
nHn
Hn
Pn
1
k=1 k .
37. Let f : (0, 1) → [0, 1] be surjective, two times differentiable function.
Prove that there exists some c ∈ (0, 1) such that f 00 (c) = 0.
38. Let f : R → R be two times differentiable function and let f 00 (x) < 0.
Prove inequality f (b) > f (a) + (b − a)f 0 (b) for a < b.
39. Suppose that f is a continuous function on [0, 2] such that f (0) = f (2).
Show that there is a real number ξ ∈ [1, 2] with f (ξ) = f (ξ − 1).
40. Let f : [0, +∞) → R be differentiable function such that f 0 (x) > x, for
x ∈ (0, +∞).
(a) Prove that (∀n ∈ N)(∀c > 0)f (n + c) − f (n) > nc.
(b) Examine uniform continuity of f .
41. Let an be a positive sequence. If limn→+∞
+∞. Prove.
an+1
an
> 1 then limn→+∞ an =
42. Consider the function f ∈ C 1 [a, b] that satisfies condition f (a) = f (b) = 0.
Prove that for any real λ there exists some c ∈ (a, b) such that f 0 (c) =
λf (c).
43. Let’s consider the function f : (a, b) → R that satisfies the condition
|f (x1 ) − f (x2 )| ≤ c|x1 − x2 |p for all x1 , x2 ∈ (a, b) and for some constants
c > 0 and p ∈ R. Prove that if p is a constant greater then 1 then function
f is constant.
44. Find all functions f : R → R which satisfy f (x + y) + f (y + z) + f (z + x) ≥
3f (x + 2y + 3z) for all x, y, z.
5
45. Does there exist a bounded real-valued function f such that f (1) = 1 and
2
f x + x12 = f (x) + f x1 for all non-zero x?
46. Show that set {cos n|n ∈ N ∪ {0}} is dense in [0, 1].
47. Let’s observe function f : R → R that satisfies next condition: ∀y ∈ f (R)
the set f −1 ({y}) has two elements. Prove that f isn’t continuous.
48. Let f, g : [0, 1] → [0, 1] are continuous functions such that f ◦ g = g ◦ f .
Prove there exists some x0 such that f (x0 ) = g(x0 ).
49. Let f : [0, 1] → [0, 1] be continuous function so there exists some n ∈ N
such that f ◦ · · · ◦ f (x) = x. Suppose that f (0) = 0 and f (1) = 1. Prove
| {z }
that
n times
f (x) = x.
50. Let f : [0, 1] → R be continuous function so f (0) = f (1) and let n > 1 bea
natural number. Prove there exists some x0 such that f (x0 ) = f x0 + n1 .
51. Find all continuous functions f : R → [1, +∞) for which there exist a ∈ R
and k ∈ N such that f (x)f (2x) . . . f (nx) ≤ ank for all x ∈ R and all
n ∈ N.
52. Let f : R → R be differentiable function with bounded derivative f 0 .
Calculate
x2 + f (x)
lim √
.
x→±∞
1 + x2
Pn
53. Examine the convergence of an = k=1 arctan k1 .
54. Let an be the sequence which satisfies 0 ≤ am+n ≤ am +an for all m, n ∈ N.
Prove that ann is convergent sequence.
55. Let f : R → R be a non-constant, continuous function. Prove that
|Im(f )| = |R|.
56. What is greater: eπ or π e .
57.
58. Prove that there exists infinitely many pairs of positive real numbers α
and β such that αα = β β but α 6= β.
59. Let a0 , a1 . . . an be real numbers. If
a1
a2
an
a0 +
+
+ ... +
=0
2
3
n+1
then the equation
a0 + a1 x + a2 x2 + . . . + an xn = 0
has at least one soulution in the interval (0, 1). Prove.
6
60. Let f [0, 1] → R be differentiable function such that f (0) = 0 and f (x) > 0
for all x ∈ (0, 1]. Prove that there exists some c ∈ (0, 1) which satisfies
0
0
(1−c)
(c)
the equality ff (1−c)
= 2ff (c)
.
61. Let the function f : (−1, 1) → R be continuous at x = 0. If f (x) = f (x5 )
show that f is constant.
62. Let f : (a, b) → R be differental function which satisfies conditions: f 0 (x)+
f 2 (x) ≥ −1, limx→a+ f (x) = +∞ and limx→b− f (x) = −∞. Prove that
b − a ≥ π.
63. Calculate next integrals:
R +∞
dx
(a) 0
(x2 +1)2 (x+9)
Rπ
sin(x)dx
(b) 04 √
4
cos4 x−sin2 x+1
(c)
R1
0
2 arctan x−ln 1+2x
dx.
x2
64. Examine convergence and absolute convergence of next integrals
R +∞ x sin x3
(a) 0
1−e−x dx
R +∞ ln x cos xπ
6
(b) 1
dx
x
R +∞ ln x 1
xβ − 1)dx
(c) 1
xα (e
R +∞
√ 2 dx2
(d) 1
.
(x −α )(x−1)
(e)
R +∞ R x
1
cos2 t
1
5
t2
dt dx.
65. Examine the convergence of next series:
P+∞ 1
(a)
n=1 ln nln n
P+∞
1
(b)
,α > 0
n=1 1 − cos(1 − cos nα
(d)
R
1 n+1
n=1 cos n n
P+∞
1
n=1 n1+e−n
(e)
P+∞
(f)
P+∞
(g)
P+∞
(c)
P+∞
cos x2 dx
2n (n!)2
n=1 4·11·...·(2n2 +n+1)
α
n=1 n
n=1
n
1
3
q
−
√1
n
q
R n+1
1
− ln 1 + n
n
ln 1 +
66. Observe the sequence In =
R1
0
1
n
√1
n
1
ln x dx.
xn cos( πx
2 )dx
(a) Find recurrent relation between In and In+2 .
7
(b) Show inequalities 0 < In <
1
n+1 .
P+∞
(c) Examine the convergence of the series
67. Find minimum of the function F (x) =
R x2
x
n=1 In .
1
t
ln t−1
9 dt for x ∈ (1, +∞).
68. Observe the function f : (0, +∞) → R defined in next way:
Z 2
sin t ln t2
dt.
f (x) =
t2
x
(a) Determine the sign of f (3).
(b) Find f 0 (3).
(c) Find limn→+∞ f (3n ) (if exists).
P+∞
P+∞
69. (a) If the series n=1 an absolutely converges show that n=1 a3n absolutely converges.
(b) Is this statement true if we speak about regular convergence?
70. Let f : [a, b] → R be a concave upward function. Prove inequality
Rb
(b)
f (x) ≥ f (a)+f
(b − a).
2
a
Rx q 1
1
√1
71. Calculate arc length of the curve y(x) = √1
t6 + t4 − 1dt for 3 ≤ x ≤
3
1.
72. Let ω(n) be the number of digits of natural number n. Examine convergence of next series:
P+∞ (−1)n
(a)
n=1 ω(n)
P+∞ 1
(b)
n=1 nω(n)
P+∞ 1
(c)
n=1 nω 2 (n) .
P+∞
73. Let un be a decreasing sequence such that n=1 un converges.
P+∞
(a) Show that n=1 n(un − un+1 ) converges.
(b) Shove that limn→+∞ nun = 0.
74. Let f : R → R be continuous function such that f (x) =
that f vanishes on R (f (x) = 0).
Rx
0
f (t)dt. Prove
75. Observe the function f ∈ C 1 [1, +∞). Let’s suppose that f has bounded
R +∞
derivative and that integral 1 |f (x)|dx converges.
R +∞
(a) Prove that 1 f (x)f 0 (x)dx converges.
(b) Show that limx→+∞ f 2 (x) exists.
(c) Calculate limx→+∞ f (x).
8
76. Observe the functions f (x) =
defined on [1, +∞).
R x+5
x
cos t2
t8 dt
and g(x) =
R x+4
x
arctan
t3
√
t
dt
(a) Prove that limn→+∞ g(x) = 0.
(b) Show that f (x) = o(g(x)) when x towards +∞.
R +∞
(c) Examine the convergence of 1 f (x)dx.
R +∞ dx
√
.
77. Observe the sequence In = 1
xn x−1
(a) Calculate In .
(b) Examine the convergence of the series
P+∞
3
n=1 In .
78. Let f : [0, 1] be continuous and increasing function. Prove that g(x) =
R1
1
x−1 x f (t)dt is decreasing on [0, 1).
79. Find the set of convergence for next power series:
(
P+∞ αn xn
1
n = 2k
(a)
n=1
n , where αn =
c
n = 2k − 1
P+∞ (n!)2
n
(b)
n=1 (2n)! (x − 1)
P+∞ (n!)α
(c)
n=1 n((2n)!)β .
80. P
Let an be decreasing sequence that towards zero such that the series
+∞
n=1 an diverges.
n
P+∞
an
.
(a) Examine regular and absolute convergence of the series n=1 (−1)
ean
P+∞ an n
(b) Find the set A ⊂ R where power series n=1 ean x converges.
81. Calculate the sum
(−1)n
n=2 n2 +n−2 .
P+∞
82. Observe the function f ∈ C 1 (R) such that f (1) = 0. Prove that there
R1
exists some ξ which satisfies equality 0 xf (x)dx = − 16 f 0 (ξ).
83. Find volume of the body√ which is obtained by rotating part of the graphic
arctan x
of the function f (x) = xe1+x2 , 0 ≤ x ≤ 1 around x-axis.
84. Examine the convergence of the series
P+∞
n=1
n!
qR
n
n
ex2 dx
0
.
nn ln n
85. Let f ∈ C 1 [0, 1] be the function that satisfies the condition: f (1) − f (0) =
R1
1. Prove inequality 0 (f 0 (x))2 dx ≥ 1.
86. (a) Let f ∈ C 1 [a, b] be strictly decreasing function such that f (a) = b and
Rb
Rb
f (b) = a. If g is the inverse of f prove that a g(x)dx = a f (x)dx.
R1
R1
1
1
(b) Show equality 0 (1 − xq ) p dx = 0 (1 − xp ) q dx for any p, q > 0.
9
87. Let f : R → R be integrabile on each segment of real line, and let the
Rx
R x+ 1
equality 0 n f (t)dt = 0 f (t)dt + n1 f (x) stands for all x ∈ R and all
n ∈ N. Show that f (x) = c, for some real c.
88. Let f : (a, b) → R be continuous function such that f (x) ≥ 0 . Show that
R b−ε
f is integrable if and only if limε→0+ a+ε f (x)dx exists.
89. Consider a continuous function f : [0, +∞) → R such that limx→+∞ f (x) =
1. Calculate
Z
1 1
f (x)e−x xn dx.
lim
n→+∞ n! 0
2
Solutions
π
1. (a) Equality arctan (tan x) = x + π stands for all x ∈ (− 3π
2 , − 2 ). Since
π
3π
n
n
0 < n+1 < 2 we have − 2 < −π < n+1 − π < 1 − π < − π2 .
n
n
n
Thus, arctan (tan( n+1
− π)) = n+1
− π + π = n+1
. Finally, x1 = 12 ,
34
55
x34 = 35 , and x55 = 56 .
(b) Here, we actually have arithmetic progression where a = x1 = 1 and
d = 3. Thus, x1 = 1, x34 = 1 + 33 · 3 = 100 and x55 = 1 + 54 · 3 = 163.
(c) First, notice that sin x ∈
/ {−1, 0, 1} if x is rational. Since 0 < 1 < π
we have that 0 < sin 1 < 1. Since 31.4 ≈ 10π < 34 < 11π ≈ 34.6
we have that 0 < sin 34 < 1. Since 53.4 ≈ 17π < 55 < 18π ≈ 56.5
we have that −1 < sin 55 < 0. Finally, x1 = bsin 1 + 1c = 1, x34 =
bsin 34 + 1c = 1 and x55 = bsin 55 + 1c = 0.
(d) Since 3 6 |1 and 1 6= 2k we conclude that x1 = 0. However, 34 = 2 · 17,
34
so x34 = 34+3
= 34
37 . Finally, 3 6 |55 and 55 isn’t even, so x55 = 0.
2. Let ε be any constant greater then zero.
(a) We know that our solution must be 2. Let’s prove it: 2n+1
n+2 − 2 <
−3
3
ε ⇐⇒ | n+2 | ⇐⇒ n+2 < ε. Solving this by n we have n > 3ε − 2.
Thus, we find that n0 = max{1, b 3ε − 2c}.
(b) We proved in a) that limn→+∞ 2n+1
n+1 = 2. Thus, there exists some
2n+1
n1 ∈ N such that n+2 − 2 < 0.1 ⇐⇒ 1.9 < 2n+1
n+2 < 2.1 for n > n1 .
4 2n+1
4·2.1
Therefore, 2.42 ≈ 4·1.9
<
<
≈
2.67
⇒ b π4 2n+1
π
π n+2
π
n+2 c = 2 for
4 2n+1
n > n1 . Thus b π n+2 c − 2 = |2 − 2| = 0 < ε for n > n0 = n1 . So,
we proved that the limit is 2.
(c) Since | sin n1 − 0| = sin n1 < n1 < ε ⇐⇒ n > 1ε we find that n0 = b 1ε c
and that the limit is zero.
√
√
1 √
1 √
(d) Since | n + 1 − n − 0| = √n+1+
− 0 = √n+1+
< 2√1 n <
n
n
ε ⇐⇒ n > 4ε12 we find that n0 = 4ε12 and that the limit is zero.
10
2
+n+1
>
(e) Let M be a real, positive constant. We have that n n−2
n > M for any n > n0 = bM c + 1. So, the limit is +∞.
n2 −2n
n−2
=
(f) Let M be a real, positive constant and let x be greater than 2. If
+2
4
x+2
> M ⇐⇒ x < 2M
M > 1 we have x−2
M −1 = 2 + M −1 . Otherwise
x+2
if M ≤ 1 then x−2
> 1 ≥ M for any x > 2, specially for 2 < x <
2 + 1 = 3. So, we choose δ to be M4−1 for M > 1 and 1 otherwise.
So, the limit is +∞.
2
x+1
2
2
= −x+1
(g) We have x−1
− 1 = x−1
for x < −1. Now −x+1
< ε ⇐⇒
2
2
x < 1 − ε . So, we choose M < 0 to be min{−1, 1 − ε }. So the limit
is 1.
(h) Since sin (x + 2008π) = sin x and | sin x − 0| = | sin x| ≤ |x| < ε ⇐⇒
−ε < x < ε we conclude that δ = ε is desired δ and that the limit is
zero.
3. (a)
lim
n→+∞
n3 − n2 + 1
n3 + 1
= lim
n→+∞
1+
2n2 +1
= lim
n→+∞
−n2
n3 + 1
2n2 +1
n→+∞
=
lim
1+
n→+∞
=
lim
t→−∞
1
1+
t
= lim
n→+∞
= lim
2n2 +1
n3 − n2 + 1
−1
=
1+
n3 + 1
1+
1+
2
n3 +1
−n
2
3
1
−n
n +1
−n2
(2n2 +1)
n3 +1
=
n3 +1
−n2
t limn→+∞
=
n3 +1
−n2
=
n3 +1
limn→+∞
2
−n
2n2 +1
(2n2 +1)
n3 +1
−n2
1
1
−n2
(2n2 +1)
n3 +1
= e−∞ = 0.
(b) We have that limx→0+ arctan x1 = limt→+∞ arctan t = π2 . On the
other side limx→0− arctan x1 = limt→−∞ arctan t = − π2 . We conclude
that the limit doesn’t exist.
(c)
i. If x < −1 then the sequence bxcn has two accumulation points:
1
−∞ and +∞. Thus, limn→+∞ 1+bxc
n = 0.
ii. If x ∈ [−1, 0) then bxc = −1 which give us next expression in
1
the limit: 1+(−1)
n . This expression is not well defined for odd
naturals, so we conclude that the limit doesn’t exist.
1
1
iii. If x ∈ [0, 1) then bxc = 0. Thus, limn→+∞ 1+bxc
n = limn→+∞ 1+0 =
1.
1
1
iv. If x ∈ [1, 2) then bxc = 1. Thus, limn→+∞ 1+bxc
n = limn→+∞ 1+1 =
1
2.
11
v. If x ≥ 2 then bxc > 1 which means that limn→+∞ bxcn = +∞ .
1
Thus, limn→+∞ 1+bxc
n = 0.
xn−1 (nx2 +n−2x2 )
d
xn
> 0 for all n and x ∈ [0, 1), we
(d) Since dx
1+x2 =
(x2 +1)2
n
x
conclude that the function 1+x
2 is strictly increasing on [0, 1) for all
xn
1n
1
n. This means that supx∈[0,1) 1+x
2 = 1+12 = 2 . Finally, we have
n
x
= limn→+∞ 12 = 12 .
limn→+∞ supx∈[0,1) 1+x
2
(e) Let aj = max{a1 , a2 , . . . , a10 }. Then
v
u 10
uX
n
n
lim t
bk · ak = lim
n→+∞
n→+∞
k=1
v
u 10
n
uX
ak
n
t
=
bk anj ·
aj
k=1
v
u 10
n
uX
ak
n
= lim aj t
bk ·
= aj · lim
n→+∞
n→+∞
aj
k=1
v
u
n
10
X
u
ak
u
n bj +
b
·
.
k
t
a
k=1
k6=j
j
n
< 1. This means that limn→+∞ aakj
=
P10
ak n
= 0 i.e.
0 for all k 6= j. Consequently, limn→+∞ k=1 bk · aj
k6=j
P10
ak n
= o(1), n → +∞. Now,
k=1 bk · aj
For k 6= j we have
ak
aj
k6=j
v
u
n
10
q
X
u
ak
n
n bj +
aj · lim u
b
·
=
a
·
lim
bj + o(1)
k
j
n→+∞ t
n→+∞
a
j
k=1
k6=j
q
p
p
b
and since 1 ← n 2j < n bj + o(1) < n 2bj → 1 when n towards
+∞, using the three limits theorem we obtain that the limit is aj .
(f) limn→+∞ (n + a)α − (n + b)α = limn→+∞ nα (1 + na )α − nα (1 + nb )α =
limn→+∞ nα (1 + na )α − (1 + nb )α .
i. If a 6= b then limn→+∞ nα (1+ na )α −(1+ nb )α = limn→+∞ nα 1+
αb
1
αa
= limn→+∞ (a − b)α · nα−1 + o(nα−1 ).
n −1− n +o n
α−1
+ o(nα−1 ) = (a − b)α ·
A. If α 6= 0 then limn→+∞ (a − b)α
· n
1
α−1
limn→+∞ n
1 + o α(a−b) = (a − b)α · limn→+∞ nα−1 ·
1
limn→+∞ 1 + o α(a−b)
= (a − b)α · limn→+∞ nα−1 =

0,
α<1



(a − b), α = 1
.

+∞,
α > 1, a > b



−∞,
α > 1, a < b
B. If α = 0 then limn→+∞ (a−b)α·nα−1 +o(nα−1 ) = limn→+∞ o(n−1 ) =
0.
12
ii. If a = b then the limit is obviously zero.
(g) We will first develop all expressions in the limit: arcsin n1 = n1 + 6n1 3 +
3
1
1
1
1
1
1
1 sin n12
= 1− 2n1 2 +
40n5 +o n5 , sin
120n5 +o n5 , (cos
n)
n = n − 6n3 +
1
1
1
1
1
1
1
1
1
1
+o
− 2n
2 +o n3
o n13 n2 n3 = e n2 +o n3 ln 1− 2n2 +o n3 = e n2 +o n3
=
1
1
β
e− 2n4 +o n4 = 1− 2n1 4 +o n14 and finally 1+ n13 = 1+ nβ3 +o nβ3 .
Now, let’s plug in this in the limit
1
3
3 1
1
3
1
1
1
1
1
n2 + o n2
n + 6n3 + 40n5 + o n5 − n + 6n3 − 120n5 + o n5 − 3n3
lim
=
1
n→+∞
1 − 2n1 4 + o n14 − 1 − nβ3 + o nβ3 n 2
1
1
n + o(n)
15n5 + o n5
= lim
.
n→+∞ − 1 + o 1 − β + o β
2n4
n4
n3
n3
i. If β 6= 0 then
1
1
15n5 + o n5
lim
n→+∞ − 1 + o 1
2n4
n4
lim
1
15n5
n→+∞
1
15n4
n→+∞ − β
n3
= lim
n + o(n)
=
− nβ3 + o nβ3
1
1
+ o n15
n + o(n)
15n4 + o n4
=
lim
=
n→+∞ − β + o 1
− nβ3 + o n13
n3
n3
1 + o(1)
1 + o(1)
n3
lim
= 0.
= lim
4
n→+∞
n→+∞
−15βn
1 + (1)
1 + o(1)
ii. If β = 0 then
lim
1
15n5
n→+∞
+ o n15
n + o(n)
2
=− .
1
1
15
− 2n4 + o n4
√
√
n
n
4. (a) Let’s calculate ln limn→+∞ nn! = limn→+∞
ln nn! . We have limn→+∞ ln
P
n
Pn
ln k−n ln n
. Now, using
limn→+∞ n1 k=1 ln k − ln n = limn→+∞ k=1 n
Pn+1
ln k−(n+1) ln (n+1)−
Pn
n!
n
=
ln k+n ln n
=
k=1
Stolz–Cesàro theorem we obtain limn→+∞ k=1
n+1−n
limn→+∞ ln (n + 1)−(n+1) ln (n + 1)+n ln n = limn→+∞ −n ln (n + 1)+
1
1
n ln n = limn→+∞ −n ln n+1
n = limn→+∞ −n ln (1 + √n ) = limn→+∞ −n n +
n
o n1 = −1. So, we have next result ln limn→+∞ nn! = −1 which
is equivalent to limn→+∞
√
n
n!
n
= 1e .
(b)
√
n+1
p
n+1
(n + 1)!
√
= lim
n
n→+∞
n!
p
(n + 1)!
n
√
= lim
lim
n
n→+∞
n
+
1 n→+∞
n!
n+1
√
n+1
=
(n+1)!
n+1
√
n
limn→+∞ nn!
limn→+∞
13
=
1
e
1
e
= 1.
(n+1)!
n+1
√
n
n!
n
√
n
=
Let ln an = t. Then
an − 1
et − 1
= lim
= 1.
n→+∞ ln an
t→0
t
lim
5. (a) Let ε be any constant greater than zero. Since f is uniformly continuous on [a, b] it follows that there exists some δ > 0 such that for all
x, y ∈ [a, b] which satisfy |x − y| < δ it follows that |f (x) − f (y)| < ε.
Since xn is a Cauchy sequence we conclude that xn is convergent,
and its limit x is an element of the segment [a, b]. Since xn is convergent, there exists some n0 such that for all n > n0 it follows that
|xn − x| < 2δ . If n is greater than n0 and p is any natural we obtain
next estimation: |xn − xn+p | = |xn − x + x − xn+p | < |x − xn | + |x −
xn+p | < 2δ + 2δ = δ. Finally, this means that |f (xn ) − f (xn+p )| < ε
for n > n0 , so f (xn ) is Cauchy sequence.
(b) It is not always. As counterexample we have restriction of f (x) =
1
1
x on (0, 1) and sequence xn = n . This sequence is Cauchy, so it
is convergent but f (xn ) = n isn’t convergent, so it isn’t Cauchy
sequence either.
(c) Sequence xn is convergent (converges to x),so it’s Cauchy sequence,
too. Since f (x) = arctan ex is continuous on [bxc, bxc + 1] and xn ∈
[bxc, bxc + 1] applying a) we obtain that bn = f (an ) is also a Cauchy
sequence, so it is convergent.
6. Let’s suppose that f (x) = f (y). From inequality |f (x) − f (y)| ≥ |x − y|
it directly follows that x = y, so f is injection. This also means that
f is strictly monotone function. Since |f (x) − f (0)| ≥ |x − 0| → +∞
when x towards +∞, we conclude that f (x) → ±∞ when x → +∞.
Without losing generality let’s suppose that limx→+∞ f (x) = +∞. Since
f is strictly monotone we conclude that f has to be strictly increasing.
Also, limx→−∞ |f (x)| = limx→−∞ |f (x) − f (0)| ≥ limx→−∞ |x − 0| =
+∞ ⇒ limx→−∞ |f (x)| = +∞. Since f is increasing we conclude that
limx→−∞ f (x) = −∞. Now, let’s suppose that there exists some C which
doesn’t belong to the image of f . Since limx→−∞ f (x) = −∞ there exists
some s such that f (s) < C and since limx→+∞ f (x) = +∞ there exists
some t such that f (t) > C. But, f is continuous, so by intermediate value
theorem there exists some c ∈ [s, t] such that f (c) = C. Contradiction.
7. Since f is concave upward and twice differentiable it follows that f 00 (x) ≥ 0
which means that f 0 is increasing on R. Let’s suppose that there exists
some ξ such that f (ξ + f 0 (ξ)) < f (ξ). If f 0 (ξ) = 0 we have obvious
contradiction. If f 0 (ξ) > 0 then ξ < ξ + f 0 (ξ) and since f 0 is increasing it
follows that f 0 (ξ +f 0 (ξ)) is also greater then zero. Thus, f is increasing on
[ξ, ξ + f 0 (ξ)] which give us inequality f (ξ) ≤ f (ξ + f 0 (ξ)). Contradiction.
The proof is similar in case f 0 (ξ) < 0.
14
8. (a) If we suppose that an is convergent that also means an is Cauchy
sequence. Thus, there exists some n0 such that |an − an+p | < ε for
all n > n0 and natural p. This is contradiction. So, an can’t be
convergent.
(b) If we suppose that an is bounded then, by Bolzano–Weierstrass theorem, it exists some convergent subsequence of xn . Let’s note it with
xnk . Since |xn − xm | ≥ ε for all m, n this inequality also stands for
members of sequence xnk . However, using the result from (a), this
leads us to conclusion that xnk is divergent which is contradiction.
So, xn can’t be bounded.
3
9. (a) Since f 0 (x) = 3x2 + 3x2 sin x1 − xx2 cos x, we have that f 0 is differentiable on (−∞, 0) ∪ (0, +∞). However, f is not even defined at x = 0.
Now, from the estimation 0 ≤ |2x3 + x3 sin x1 | ≤ 2|x|3 + |x|3 | sin x1 | ≤
2|x|3 + |x|3 → 0 when x → 0 we conclude that f can be extended to
continuous function
(
f (x), x 6= 0
f (x) =
.
0,
x=0
Now, let’s examine differentiability of f at x = 0:
h3 + h3 sin h1
f (h) − f (0)
1
= lim
= lim h2 +h2 sin
h→0−
h→0−
h→0−
h
h
h
0
f − (0) = lim
Similar estimation as the first 0 ≤ |h2 +h2 sin h1 | ≤ |h|2 +|h2 || sin h1 | ≤
0
2|h|2 → 0 when h → 0 give us the result f − (0) = 0. Analogously,
0
f + (0) = 0, so f can be differentably extended on R.
(b) Let suppose there exists such neighbourhood and let’s note it with
N . Let ε be a constant greater than zero such that (−ε, ε) ⊂ N .
1
1
1
Also, 2nπ−
π ,
2nπ ∈ (−ε, ε) for some natural n. Now, f ( 2nπ− π ) =
2
2
1
1
π
1
1 3
3
3
( 2nπ−
+ ( 2nπ−
π )
π ) sin (2nπ −
2 ) = 0 and f ( 2nπ ) = ( 2nπ ) +
2
2
1 3
1 3
( 2nπ
) sin (2nπ) = ( 2nπ
) > 0 which means that f is decreasing on
1
1
N , since 2nπ− π > 2nπ . This give us inequality 0 = limx→0+ f (x) ≥
2
1
f ( 2nπ
) > 0 which can’t be true. We conclude there isn’t any neighbourhood in which f is monotone.
10. Let’s observe the function F (x) = f (x)−x. We have that F (0) = f (0)−0 >
0 and F 0 (x) = f 0 (x) − 1 < 0 so F is strictly decreasing on [0, +∞). Let’s
suppose that F (x) 6= x for all x ∈ [0, +∞). From F (0) > 0 and the
fact that F is continuous we conclude that F (x) > 0 for x ∈ [0, +∞).
Considering all this it follows that F has horizontal asymptote y = c ≥ 0.
This means that limx→+∞ F 0 (x) = 0. On the other side F 0 (x) = f 0 (x) −
1 = f 0 (x + T ) − 1 = F 0 (x + T ). Since F 0 is periodic with period T > 0 it
follows that limx→+∞ F 0 (x) doesn’t exist. Contradiction.
15
11.
12. (a)
n
n
1
1
nan
lim
1 + + an
= lim
1+ +
=
n→+∞
n→+∞
n
n
n
n n
1
o(1)
1
1+ +
= lim
= 1+ n
=
n→+∞
n
n
1+o(1)
= 1+
1
n(1+o(1))
n
1+o(1)
·
n
n
1+o(1)
= elimn→+∞
n+o(n)
n
= e.
(b) First we have limn→+∞ an = limn→+∞ nann = limn→+∞ o(1)
n = 0 and
1+a 6= 0. Thus, for ε > 0 there exists some n0 such that an ∈ (−ε, ε)
3(1+a)
1+a
for n > n0 . For ε = 1+a
2 we have that 2 < 1 + a + an <
2
3(1+a)
for all n greater than some n0 . Since sgn 1+a
=
sgn
it
follows
2
2
that 1 + a + an 6= 0 for n > n0 . So, the expression (1 + a + an )−n is
well defined for n > n0 . This is enough to say that bn is well defined
starting from n0 . Now, let’s find accumulation points of bn . First,
let’s calculate accumulation points of (1 + a + an )−n .
i. If a < −2 then 1 + a + an < −1 starting from some n0 which
1
means that −1 < 1+a+a
< 0 starting from the same n0 . Thus,
n n
1
(1 + a + an )−n = 1+a+a
has one accumulation point x = 0.
n
ii. If −2 < a < −1 then < −1 < 1 + a + an < 0 starting from
1
some n0 which means that 1+a+a
< −1 starting from the same
n
n
1
−n
n0 . Thus, (1 + a + an )
= 1+a+an
has two accumulation
points: x1 = −∞ for n = 2k − 1 and x2 = +∞ for n = 2k.
iii. If −1 < a < 0 then 0 < 1 + a + an < 1 starting from some
1
n0 which means that 1+a+a
> 1 starting from the same n0 .
n n
1
Thus, (1 + a + an )−n = 1+a+a
has one accumulation point
n
x = +∞.
n
−no(1)
· n
o(1)
iv. If a = 0 then limn→+∞ (1+an )−n = limn→+∞ (1+ o(1)
=
n )
0
e = 1. So, in this case we have one accumulation point x = 1.
v. If a > 0 then 1 + a + an > 1 starting from some n0 which
1
means that 0 < 1+a+a
< 1 starting from the same n0 . Thus,
n
n
1
has one accumulation point x = 0.
(1 + a + an )−n = 1+a+a
n
n
Now, let’s find the accumulation points of np cos nπ = np(−1) .
n
i. If p < 0 then the accumulation points of np(−1) are: +∞ for
n = 2k − 1 and 0 for n = 2k.
16
ii. If p = 0 then we have one accumulation point x = 1.
n
iii. If p > 0 then the accumulation points of np(−1) are: 0 for
n = 2k − 1 and +∞ for n = 2k.
Now, let’s gather this together:
i. If p < 0 and a < −2 then for n = 2k −1 accumulation point can’t
be obtained directly (product 0 · (+∞) isn’t defined expression).
This means that this accumulation point has to be calculated in
some other way which we will do later. On the other side for
n = 2k accumulation point is 0 · 0 = 0.
ii. If p < 0 and −2 < a < −1 then for n = 2k − 1 accumulation
point is (+∞) · (−∞) = −∞ and for n = 2k accumulation point
can’t be obtained directly (0 · (+∞)).
iii. If p < 0 and −1 < a < 0 then for n = 2k − 1 accumulation point
is (+∞) · (+∞) = +∞ and for n = 2k accumulation point can’t
be obtained directly (0 · (+∞)).
iv. If p < 0 and a = 0 then for n = 2k − 1 accumulation point is
+∞ · 1 = +∞ and for n = 2k accumulation point is 0 · 1 = 0.
v. If p < 0 and a > 0 then for n = 2k − 1 accumulation point can’t
be obtained directly (+∞ · 0) and for n = 2k accumulation point
is 0 · 0 = 0.
π
and g(x) = x cos πx + 1. Since
13. Observe functions f (x) = (x + 1) cos x+1
π
π sin x+1
π sin π
x
π
0
0
f (x) =
+ cos x+1 and g (x) =
+ cos πx next logx+1
x
ical move will be to examine monotonicity of g 0 (x). We have g 00 (x) =
π 2 cos( π )
− x3 x . It is obvious that g 00 is negative for x > 2. Using Lagrange theorem we conclude that difference f 0 (x) − g 0 (x) = g 0 (x + 1) −
g 0 (x) = g 0 (ξ), ξ ∈ (x, x + 1) is negative when x towards +∞. This means
that expression f (x) − g(x) is decreasing when x towards +∞. Since
π
π
limx→+∞ ((x + 1) cos x+1
− x cos πx − 1 = limx→+∞ x(cos x+1
− cos πx ) +
π
π2
π2
1
limx→+∞ cos x+1
− 1 = limx→+∞ x 1 − 2(x+1)
=
2 − 1 + 2x2 + o( x2 )
2
3
−x
π
limx→+∞ π 2 x(x+1)
x2 (x+1)2 = 0. Thus f (x) − g(x) > 0 ⇐⇒ (x + 1) cos x+1 −
x cos πx > 1 for x > 2.
14. Since F and G are continuous it follows that inf x∈[0,1] F (x) = minx∈[0,1] F (x) =
F (c) for some c ∈ [0, 1]. Let’s suppose that F (x) 6= G(x) for all x ∈ [0, 1].
Having on mind continuity of F (x) − G(x) we have F (x) − G(x) > 0 or
F (x) − G(x) < 0. Without loosing generality let’s suppose that F (x) −
G(x) > 0. But this means that 0 < F (c)−G(c) = minx∈[0,1] F (x)−G(c) =
inf x∈[0,1] F (x) − G(c) = inf x∈[0,1] G(x) − G(c). Contradiction.
(ξ) 15. Let’s prove that there exists some ξ ∈ [0, 1] such that ln 1 + g(ξ)−f
=
1+f (ξ)
g(ξ)−f (ξ) 1+g(ξ)
1+g(ξ)
0. We have ln 1 + 1+f (ξ)
= ln 1+f (ξ) = 0 ⇐⇒ 1+f (ξ) = 1 ⇐⇒
17
f (ξ) = g(ξ). Now, using the statement in the problem 14. we have that
there exists some ξ ∈ [0, 1] such that f (ξ) = g(ξ) which means the equality
(ξ) ln 1 + g(ξ)−f
= 0 stands for that ξ. On the other side for the same
1+f (ξ)
ξ we have 0 = f (ξ) − g(ξ) = f (ξ) − g(ξ) f (ξ)2 + f (ξ)g(ξ) + g(ξ)2 =
f (ξ)3 − g(ξ)3 .
16. First, let’s prove that xn > 0 using mathematical induction.
B. x1 > 0
S. Since xn > 0 we have that xn+1 =
denominator are both positive.
3x2n
(1+xn )3 −1
> 0 since numerator and
3x2n
(1+xn )3 −1
x3n + 3x2n +
Now let’s examine monotonicity: xn+1 =
3xn
x3n +3x2n +3xn
=
3x2n
x3n +3x2n +3xn
<
3xn . The last inxn ⇐⇒
< 1 ⇐⇒ 3xn <
equality in this sequence of equivalences is true so we conclude that xn is
decreasing. Since x1 ≥ xn > 0 we have that xn is also bounded so it converges. Let’s note its limit with A and let’s calculate it: limn→+∞ xn+1 =
limn→+∞ 3x2n
3A2
3A2
⇐⇒ A = (1+A)
The equality A = (1+A)
3 −1 .
3 −1
(1+limn→+∞ xn )3 −1
3A
stands for A = 0. If A 6= 0 then we have A3 +3A
⇐⇒ 0 =
2 +3A = 1
2
3
3A + A ⇐⇒ A = −3 which is not possible since xn > 0. We conclude
that limn→+∞ xn = 0. The limit limn→+∞ nxn can be calculated using
Stolz–Cesàro theorem:
n
lim nxn = lim
n→+∞ 1
xn
n→+∞
lim
n
n→+∞ 1
xn
1
= lim
)3 −1
n→+∞ (1+xn
3x2n
−
1
xn
= lim
n→+∞
n+1−n
1
1 =
xn+1 − xn
3x2n
=
n→+∞ (1 + xn )3 − 1 − 3xn
= lim
3x2n
= 1.
3
n→+∞ 3x2
n + xn
lim
x (2+x2 )
17. (a) We can easy obtain that xn+2 = 2xn4 +9x2n+8 . Let’s prove that odd
n
n
members are greater then zero:
B. x1 > 0
x
(2+x2
)
S. x2k+1 = x42k−1+9x2 2k−1+8 > 0 since numerator and denominator are
2k−1
2k−1
both positive.
x2k−1 (2+x22k−1 )
x42k−1 +9x22k−1 +8
x42k−1 + 9x22k−1 + 8
Also x2k+1 <
2 + x22k−1
⇐⇒
2+x22k−1
x42k−1 +9x22k−1 +8
< 1
⇐⇒
<
which is true. Thus, x2k−1 is decreasing. On the other side let’s prove that x2k is negative and increasing:
x1
B. x2 = − 2+x
2 < 0 since x1 > 0.
1
x
(2+x2 )
S. x2k+2 = x42k+9x2 2k+8 > 0 since numerator is negative and denomi2k
2k
nator is positive.
x (2+x2 )
2+x2
Also, x2k+2 > x42k+9x2 2k+8 ⇐⇒ x4 +9x2k
⇐⇒ 2 + x22k <
2 +8 < 1
2k
2k
2k
18
2k
x42k + 9x22k + 8 (pay attention that x2k is negative, so it inverts inequality) which is true. Thus, x2k is increasing. We also can conclude
that sequences of odd and even elements both converges.
(b) We see that x2k and x2k−1 have the same limit and that is the so2
)
lution of equation A = AA(2+A
4 +9A2 +8 . One of the solutions is zero and
actually this is the only possible solution since limk→+∞ x2k ≤ 0 and
limk→+∞ x2k−1 ≥ 0. So, xn converges and limn→+∞ xn = 0.
1
x+b 0
= 1+a
18. (a) Since 1+a
we have that f is differentable for x < 1. Since
0
x2
2
x ln (a + x) = a+x + 2x ln (x + a) and since x + a > 0 for x ≥ 0
we have that f is differentable for x > 1. Now, let’s see what is
happening at x = 1:
f (1 + h) − f (1)
= lim
h→0−
h→0−
h
0
f−
(1) = lim
1+h+b
1+a
h
−0
= lim
h→0−
1+h+b
.
(1 + a)h
0
(1) has to be finite we have 1+b = 0, i.e. b = −1. Having this
Since f−
1
0
(1) = 1+a
on mind we obtain that f−
. Now, let’s give the estimation
of right derivative:
0
f+
(1) =
lim
h→0+
f (1 + h) − f (1)
=
h
(1 + h)2 ln (a + 1 + h) − 0
< +∞
h→0+
h
if and only if a = 0. In that case, we have
lim
=
ln (1 + h)
(1 + h)2 ln (1 + h)
= lim (1+h)2 · lim
= 1.
h→0+
h→0+
h→0+
h
h
0
f+
(1) = lim
Also, for a = 0 left derivative at the point x = 1 is 1 which means
that for b = −1 and a = 0 function f is differentiable.
(b) Let’s observe sequences xn = n and yn = n + n1 . It is obvious that
limn→+∞ |xn −yn | = 0. On the other side |f (xn )−f (yn )| = n2 ln n−
(n + n1 )2 ln (n + n1 ) = n2 ln n − n2 ln (n + n1 ) + n2 ln (n + n1 ) − (n +
1
1
1 2
1
2
2
2
n ) ln (n + n ) ≤ n ln n − n ln (n + n ) + n ln (n + n ) − (n +
1 2
1
1
1
1
2
2
2
n ) ln (n + n ) = n ln (1 + n2 ) + (n + 2 + n2 − n ) ln (n + n ) =
2 ln(n) + o(ln n) → +∞ when n towards +∞. Thus, f is not uniformly continuous.
19.
√
3
x2
x2
= q
x3 + 8
x3 1+
8
x3
− 31
8
8
128
1
= x 1+ 3
= x 1− 3 + 6 +o 6
=
x
3x
9x
x
=x−
8
1
+
o
,
2
3x
x2
so we obtain that a = 1, b = 0, c = 0 and d = − 38 .
19
(x−1)
= f (x) − f (x − 1) = f 0 (ξ) for some ξ ∈ (x − 1, x).
20. (a) Since f (x)−f
x−(x−1)
From the inequalities x − 1 < ξ and ξ < x we obtain x → +∞ ⇐⇒
ξ → +∞. Having on mind all that, we have: limx→+∞ f (x) − f (x −
1) = limξ→+∞ f 0 (ξ) = +∞.
Pbxc
(b) Let’s observe f (bxc) − f (cac + 1) =
k=bac+1 f (k + 1) − f (k) =
Pbxc
0
0
0
k=bac+1 f (ξk ). Since limbxc→+∞ f (ξk ) = limk→+∞ f (ξk ) = +∞,
P+∞
0
we conclude that
k=bac+1 f (ξk ) = +∞. From the equivalence
bxc → +∞ ⇐⇒ x → +∞ we obtain limx→+∞ f (x) − f (0) =
Pbxc
P+∞
0
limx→+∞ k=bac+1 f 0 (ξk ) =
k=bac+1 f (ξk ) = +∞. Having on
mind that f (0) is a constant, its removal will not change value of
limit, so limx→+∞ f (x) = +∞.
21. (a) We have f (k + 1) − f (k) ≤ f (k + 2) − f (k + 1) ⇐⇒ 2f (k + 1) ≤
f (k +2)+f (k) ⇐⇒ f (k +1) ≤ 21 f (k +2)+ 12 f (k). If x = k, y = k +2
and t = 21 then inequality f (k + 1) ≤ 12 f (k + 2) + 12 f (k) becomes
f ( 12 x + 12 y) ≤ 12 f (x) + 12 f (y) which is true since f is upward concave.
(b) Let’s assume that an is increasing i.e. f (n) + f (−n) ≤ f (n + 1) +
f (−(n + 1)). Rearranging the last inequality we get f (−n) − f (−(n +
1)) ≤ f (n + 1) − f (n). Using (a) we have the sequence of inequalities
f (−n) − f (−(n + 1)) ≤ f (−(n − 1)) − f (−n) ≤ . . . ≤
≤ f (1) − f (0) ≤ . . . ≤ f (n + 1) − f (n),
which proofs our assumption. Now, let’s assume that an+1 − an is
increasing i.e. f (n + 1) + f (−(n + 1)) − f (n) − f (−n) ≤ f (n + 2) +
f (−(n + 2)) − f (−(n
+ 1)) − f (−(n + 1)). Rearranging this we have
f (n + 1) − f (n) − f (−n)
−
f
(−(n
+
1))
≤
f
(n
+
2)
−
f
(n
+
1)
−
f (−(n+1))−f (−(n+2)) . Since f (n+1)−f (n) ≤ f (n+2)−f (n+1)
and f (−n) − f (−(n + 1)) ≥ f (−(n + 1)) − f (−(n + 2)) which means
that our assumption was true.
22.
n
lim mm
n→+∞ n
n·(n−1)·...·(n−m+1)
m!
n→+∞
nm
= lim
=
nm (1 − n1 )(1 − n2 ) . . . (1 − m−1
1
n )
lim
=
m
n→+∞
m!
n
1
1
2
m−1
1
=
lim
1−
lim
1−
. . . lim
1−
=
.
n→+∞
m! n→+∞
n n→+∞
n
n
m!
=
23. B. If n = 1 then a31 = (a + d)3 . On the other side
2
2
(a + 2d)(a + d) − a(a + d)
(a2 a1 )2 − (a1 a0 )2
=
=
4d
4d
20
=
a4 + 6a3 d + 13a2 d2 + 12ad3 + 4d4 − a4 − 2a3 d − a2 d2
=
4d
= a3 + 3a2 d + 3ad2 + d3 = (a + d)3 .
S.
n+1
X
a3k =
k=1
X
n
a3k
+ a3n+1 =
k=1
(an an+1 )2 − (a0 a1 )2
+ a3n+1 =
4d
a2n+1 a2n + 4dan+1 − (a0 a1 )2
=
=
4d
a2n+1 a2 + 2and + n2 d2 + 4da + 4(n + 1)d2 − (a0 a1 )2
=
=
4d
a2n+1 a2 + 2a(n + 2)d + (n2 + 4n + 4)d2 − (a0 a1 )2
=
4d
2
a2n+1 a + (n + 2)d − (a0 a1 )2
(an+1 an+2 )2 − (a0 a1 )2
=
=
.
4d
4d
24. We have that ab = ef (a)−f (b) ⇐⇒ ln a − ln b = f (a) − f (b). Since a 6= b
(b)
we can divide this equality with ln a − ln b, so we obtain fln(a)−f
a−ln b = 1.
Using Cauchy mean value theorem we have
Rearranging this, the get the proof.
f 0 (c)
1
c
= 1 for some c ∈ (a, b).
25. Let suppose that F (α) = f (α) − f (α − 31 ) 6= 13 f (1) − f (0) . Since F is
continuous that means F (α) < 13 f (1) − f (0) or F (α) > 31 f (1) − f (0)
.
Without loosing generality let’s suppose that F (α) < 13 f (1) − f (0) . It
follows that F ( 13 )+F ( 23 )+F (1) = f (1)−f (0) < f (1)−f (0). Contradiction.
26. Since
lim a2k = lim
k→+∞
k→+∞
p
k
3k − 2 ln k + k =
q
q
= lim k 3k + o 3k = lim 3 k 1 + o 1) = 3,
k→+∞
k→+∞
one of accumulation points is 3. On theqother side, if n = 2k − 1 and
2s−1
1
k = 2s − 1 we have a4s−3 = − (4s − 2)( 1 + 2s−1
− 1)
. Now, let’s
calculate the limit of this sequence:
r
2s−1
1
1+
lim − 4s − 2
−1
=
s→+∞
2s − 1
2s−1
1
1
1
−
+
o
−
1
=
s→+∞
4s − 2 8(2s − 1)2
s2
2s−1
1
1
1
= lim − 4s − 2
−
+o 2
=
s→+∞
4s − 2 8(2s − 1)2
s
= lim −
4s − 2 1 +
21
2s−1
1
1
1
= lim − 1 −
= −e− 4 .
+o
s→+∞
4(2s − 1)
s
1
Thus, the second accumulation point is −e− 4 .Similarly, if k = 2s we have
1
1
lims→+∞ a4s−1 = e− 4 , so the last accumulation point is e− 4 .
27. (a) First an+1 − an = n12 + o( n12 ) = n12 (1 + o(1)) when n towards +∞.
Now, o(1) is some sequence αn which towards 0 when n towards +∞.
Thus, − 12 < αn < 12 starting from some n0 . Using this estimation we
conclude 2n1 2 < n12 (1 + αn ) = an+1 − an < 2n3 2 . Thus, an is increasing
starting from n0 .
2
3
. This is equivalent to 3n
(b) Let’s solve inequation 2n3 2 < n(n+1)
2 + 2 <
2n ⇐⇒ 32 < n2 ⇐⇒ n > 3. So, for n > n1 = max{n0 , 3} we have
2
an+1 − an < 2n3 2 < n(n+1)
.
(c)
am −an = am −am−1 +am−1 −am−2 +am−2 −am−3 +. . .−an+1 +an+1 −an <
2
2
2
+
+ ... +
<
(m − 1)m (m − 1)(m − 2)
n(n + 1)
1
1
1
1
1
1
1
2 2
2
<2
− +
−
+. . .−
+ −
= − < ,
m−1 m m−2 m−1
n+1 n n+1
n m
n
<
for n greater then n1 from (b).
(d) Since inequality 0 < an+p − an < n2 stands starting from some n0 ,
using the three limits theorem we have limn→+∞ an+p −an = 0 which
means that an is Cauchy sequence so it converges.
28. Without loosing generality let’s suppose that a > 0 and let’s observe
the function F (x) = f (x) − ax. Since f is bounded we can give next
estimation −M − ax < f (x) − ax < M − ax for some real M > 0. Using
the three limits theorem we conclude that limx→−∞ F (x) = +∞ and
limx→+∞ F (x) − ax = −∞. Using the intermediate theorem we get that
for some ξ equality F (ξ) = 0, i.e. f (ξ) = aξ stands.
29. (a) Let’s estimate the derivative of f :
a
a
f 0 (x) = ln 1 +
−
.
x
a+x
2
a
0
Since f 00 (x) = − x(a+x)
2 < 0 for x > 0. Thus, inf x>0 f (x) = limx→+∞
0
0
f (x) = 0. Also, this infimum is not reached so f (x) > 0 for x > 0
which means that f is strictly increasing.
x
(b) Since f is strictly increasing, it follows that ef (x) = 1+ xa is strictly
x
increasing so supx>0 ef (x) = limx→+∞ ef (x) = limx→+∞ (1 + xa =
ea . The wanted inequality now follows directly.
22
30. From 28. we get the existence of a solution. Let’s prove its uniqueness.
Since |f 0 (x)| < 1 we have F 0 (x) = (f (x) − x)0 = f 0 (x) − 1 < 0 so F is
strictly decreasing. Thus, F is ”1-1” which means F has only one zero,
i.e. f (ξ) = ξ for a unique ξ.
31.
32. Let u(x) = x2 + 3x and v = e−2x . Using the general Leibniz rule we have
n X
n (k)
f (n) (x) =
u (x)v (n−k) (x).
k
k=0
Let’s calculate these derivatives: u0 (x) = 2x + 3, u00 (x) = 2 and u(k) (x) =
0 for k > 2. On the other side we have v (n−k) (x) = (−2)n−k e−2x =
(−1)n−k 2n−k e−2x . Plugging in these results in the formula we get
f (n) (x) = (−1)n 2n (x2 + 3x)e−2x + (−1)n−1 2n−1 n(2x + 3)e−2x +
(−1)n−2 2n−2 n(n − 1)e−2x .
Thus,
f (n) (0) = (−1)n−1 2n−1 3n+(−1)n−2 2n−2 n(n−1) = (−1)n−2 2n−2 n(n−7).
33. We have 3x + 6 = tan x ⇐⇒ arctan (3x + 6) = x − kπ for x ∈ (kπ −
π
π
2 , kπ + 2 ) where k is a whole number. Let’s observe the function Fk (x) =
arctan (3x + 6)−(x−kπ) defined on [kπ − π2 , kπ + π2 ]. Since F (kπ − π2 ) > 0
and F (kπ + π2 ) < 0 using intermediate theorem we conclude that for any
k ∈ Z function Fk has a zero ξk ∈ (kπ − π2 , kπ + π2 ). Since (kπ − π2 , kπ +
π
π
π
2 ) ∩ (jπ − 2 , jπ + 2 ) = Ø, we conclude that ξk = ξj ⇐⇒ k = j. Thus,
the set {ξk |k ∈ Z} has infinite number of elements.
34. Let’s suppose that limn→+∞ cos n = A. Using this we have limn→+∞
sin2 n = limn→+∞ (1 − cos2 n) = 1 − A2 . Also,
A = lim cos 2n = lim (cos2 n − sin2 n) = A2 − (1 − A2 ) = 2A2 − 1.
n→+∞
n→+∞
Solving equation 2A2 − A − 1 = 0 we obtain that A1 = 1 and A2 = − 21 .
If A = 1 then
2 = lim cos n + cos (n + 2) = 2 cos 1 lim cos (n + 1) = 2 cos 1
n→+∞
n→+∞
⇒ cos 1 = 1
which is false. If A2 =
−1 = lim
n→+∞
− 21
then
cos n + cos (n + 2) = 2 cos 1 lim cos (n + 1) = 2 cos 1
n→+∞
1
2
which is false. Thus, limn→+∞ cos n doesn’t exist, so cos n diverges.
⇒ cos 1 = −
23
3
3
35. We have tan x > x + x3 ⇐⇒ x > arctan (x + x3 ) for x ∈ (0, π2 ). Plugging
x = 0 in the right side of equivalence we obtain 0 = 0. Thus, it suffices
3
1+x2 to prove that x0 = 1 > (arctan (x + x3 )0 =
for x ∈ (0, π2 ).
3 2
1+ x+ x3
3 2
> 1 + x2 which is true for x > 0.
Rearranging this, we have 1 + x + x3
36. Let’s start with little rearrangement of initial expression
lim
n→+∞
ln (n!)
nHn
Hn
= lim
n→+∞
1+
ln (n!) − nHn
nHn
Hn
.
Now, using two times Stolz–Cesàro theorem we have
lim
n→+∞
= lim
n→+∞
ln (n!)
ln (n + 1)
=
= lim
n→+∞ Hn+1 + n
nHn
n+1
ln (n + 1)
ln (n + 1)
= lim
=
Hn+1 + o(Hn+1 ) n→+∞ Hn+1
ln (1 + n1 )
ln (n + 1) − ln n
= lim
= 1.
1
n→+∞ Hn+2 − Hn+1
n→+∞
n+2
= lim
From the last calculation we get
lim
n→+∞
ln (n!) − nHn
ln (n!)
= lim =
− 1 = 0.
n→+∞
nHn
nHn
It is, also known that limn→+∞ Hn = +∞. Having on mind previous
results we have
Hn
Hn (ln n!−nHn )
ln (n!) − nHn
nHn
lim
1+
= elimn→+∞
=
n→+∞
nHn
= elimn→+∞
ln n!−nHn
n
.
Finally, using Stolz–Cesàro theorem again we get
elimn→+∞
ln n!−nHn
n
= elimn→+∞
ln n!−nHn
n
n
= eln (n+1)−Hn+1 − n+1 = e−γ−1 ,
where γ is Euler–Mascheroni constant.
37. Since f is onto [0, 1] there exists some ξ1 , ξ2 ∈ (0, 1) such that f (ξ1 ) = 0
and f (ξ2 ) = 1. Since ξ1 and ξ2 are global extrema it follows f 0 (ξ1 ) =
f 0 (ξ2 ) = 0. Since ξ1 6= ξ2 , using Rolle’s theorem we conclude that there
exists some ξ ∈ (ξ1 , ξ2 ) such that f 00 (ξ) = 0.
38. Since f 00 (x) < 0 it follows that f 0 strictly decreases. By Lagrange theorem
we have that f (b) − f (a) = (b − a)f 0 (c) for some c ∈ (a, b). Since c < b it
follows that f 0 (c) > f 0 (b). Thus, f (b) = f (a) + (b − a)f 0 (c) > f (a) + (b −
a)f 0 (b).
24
39. (a) Using Lagrange theorem we have f (n + c) − f (n) = cf (ξ) for some
ξ ∈ (n, n + c). Since f 0 (ξ) > ξ > n, the desired inequality follows
directly.
(b) Let’s observe sequences xn = n + n1 and yn = n. It is obvious that
limn→+∞ |xn − yn | = 0. On the other side, from (a) we conclude that
f (n + n1 ) − f (n) > 1, so limn→+∞ |f (xn ) − f (yn )| ≥ 1 which means
that f is not uniformly continuous.
> 1 it follows that aan+1
> 1 starting from some n0 .
40. Since limn→+∞ aan+1
n
n
Thus, sequence an strictly increases starting from that n0 which means
that its limit surely exists. Let’s suppose that it is some finite constant c.
=
Since an is positive we conclude that c > 0. Now we have limn→+∞ aan+1
n
limn→+∞ an+1
limn→+∞ an
= cc = 1 > 1. From this contradiction and having on mind
that an is increasing we conclude that limn→+∞ an = +∞.
41. Let’s observe the function F (x) = f (x)e−λx . Since F (a) = F (b) = 0,
using Rolle’s theorem we conclude that there exists some c ∈ (a, b) such
that 0 = F 0 (c) = f 0 (c)e−λc − λf (c)e−λc . Since e−λc > 0 it follows that
f 0 (c) = λf (c).
42. We will start with next estimation: 0 ≤ |f (x0 + h) − f (x0 )| ≤ c|h|p → 0
when h towards zero. This means that f is continuous on (a, b). Let’s
(x0 )
(x0 )|
estimate the derivative of f : 0 ≤ | f (x0 +h)−f
| = |f (x0 +h)−f
≤
h
|h|
p−1
0
|h|
→ 0 when h towards zero. Thus, f vanishes on (a, b). Having on
mind all that we conclude that f is constant.
43. All constant functions satisfies the condition. Now, let’s put x = a, y =
z = 0. Then, 2f (a) + f (0) ≥ 3f (a) ⇒ f (0) ≥ f (a). Now let’s put x =
y = a2 , z = − a2 . We obtain f (a) + 2f (0) ≥ 3f (0) ⇒ f (a) ≥ f (0). This
means that f (a) = f (0). Since a is arbitrary we conclude that f must be
constant function.
44. Suppose there is such a function. Let’s define C = inf{y|(∀x ∈ D(f ))(y ≥
f (x))}. This C exists since f is bounded. We have f (2) = f (1 + 112 ) =
f (1) + f (1)2 = 2, so C ≥ 2. By characterization of infimum there exists
2
some a such that f (a) > C − 41 . So, c ≥ f a + a12 = f (a) + f a1 >
2
2
c − 14 + f a1 ⇒ f a1 < 41 . This also means that f a1 > − 12 . Now,
2
c ≥ f (a + a12 ) = f (a) + f a1
> − 21 + (c − 14 )2 . Rearranging this we
3
7
2
get c − 2 c − 16 > 0 ⇐⇒ (c − 34 )2 < 1 which is false. From this
contradiction we conclude there isn’t any function that satisfies conditions
of the problem.
45. Let’s observe an arbitrary real number d ∈ [−1, 1]. There exists some c ∈ R
such that cos c = d. Since Z+2πZ is dense in R, there exists some sequence
xn + 2πyn where xn , yn ∈ Z such that limn→+∞ (xn + 2πyn ) = c. Since
cosine is continuous and even we have d = limn→+∞ cos (xn + 2πyn ) =
limn→+∞ cos xn = cos |xn |. Since |xn | ∈ N ∪ {0} our prove is done.
25
46. Let a and b be zeros of the function, i.e. f (a) = f (b) = 0 and let’s suppose
that a < b. We know that function f can’t change the sign on (a, b).
However, f is continuous so there exists some extremum on (a, b), let’s
note it c. Also, there exists some d ∈ [a, b] such that f (c) = f (d). Without
loosing generality let’s suppose that c and d are maxima and c < d which
also means that f will be positive on (a, b). Since c and d are maxima
on (a, b) we have 0 < f (x) < f (c) = f (d) for all x ∈ (c, d). Let’s choose
some x0 ∈ (c, d) and observe f (x0 ). Using intermediate value therem there
exist some ξ1 ∈ (a, c) and ξ2 ∈ (d, b) such that f (ξ1 ) = f (ξ2 ) = f (x0 ).
Contradiction.
47. Let’s suppose that f (x) 6= g(x) for all x ∈ [0, 1]. Since f − g is continuous
function this means that f (x) − g(x) < 0 for all x ∈ [0, 1] or f (x) −
g(x) > 0 for all x ∈ [0, 1]. Without loosing generality let’s suppose that
f (x) − g(x) < 0. Let’s observe the set A = {x ∈ [0, 1]|f (x) = x}. Since
f (0) − 0 ≤ 0 and f (1) − 1 ≥ 0 if the equality doesn’t stand in any of
these two expressions using intermediate value theorem we have that there
exists some ξ ∈ (0, 1) so f (ξ) − ξ = 0. Thus, A is not empty set. Also,
A ⊂ [0, 1] so it is bounded which lead us to conclusion that there exists
some c = sup A. Let’s observe the sequence of intervals (c − n1 , c]. For
each n there exists some xn ∈ A such that xn ∈ (c − n1 , c]. Now, it is
obvious that limn→+∞ xn = c and since f is continuous it follows c =
limn→+∞ xn = limn→+∞ f (xn ) = f (c) which means that c ∈ A. Since
g(f (c)) = g(c) = f (g(c)), we conclude g(c) ∈ A which lead us to inequality
g(c) ≤ c = f (c), i.e. f (c) − g(c) ≥ 0. Contradiction.
48. Let’s prove that f is ”1 − 1”. Since
f (x) = f (y) ⇐⇒ f (f (x)) = f (f (y)) ⇐⇒ . . . ⇐⇒
⇐⇒ x = f ◦ · · · ◦ f (x) = f ◦ · · · ◦ f (y) = y,
| {z }
| {z }
n times
n times
it follows that f is injection. Since f is injection it follows that f is strictly
monotone and since f (0) = 0 < 1 = f (1) it follows that f is strictly
increasing. Let’s suppose that there exists some ξ so f (ξ) 6= ξ. Let’s
suppose that f (ξ) < ξ. Using that f is strictly increasing we have
ξ = f ◦ · · · ◦ f (ξ) < . . . < f (f (ξ)) < f (ξ) < ξ.
| {z }
n times
Contradiction. Similar conclusion would be obtained if we suppose that
f (ξ) > ξ.
Pn−1
49. Let’s observe the function g(x) = f (x + n1 ) − f (x). We have i=0 g ni =
Pn−1 i+1 − f ni = f (1) − f (0) = 0. If we suppose that g ni ≥ 0
i=0 f
n
it follows that g ni = 0 for all i and so f ( ni + n1 ) = f ( ni ) for all i. If
we suppose the opposite, then there exists some i0 such that g in0 < 0.
26
Pn−1
Also since i=0 g ni = 0 there must be some i1 such that g( in1 ) > 0.
Using intermediate value theorem we obtain that there exists some ξ so
0 = g(ξ) = f (ξ + n1 ) − f (ξ).
50. The condition is equivalent to saying
ln f (0) +
n
X
ln f (jx) ≤ ln f (0) + ln a + k ln n.
j=1
Taking x =
α
n,α
> 0 we get
n
X
α
j=0
n
ln f (
αj
α ln f (0) α ln a αk ln n
)≤
+
+
.
n
n
n
n
The left side of inequality is the Riemann sum for the function ln f (x) on
[0, α]. Letting n tend to infinity we get
Z α
α ln f (0) α ln a αk ln n
+
+
= 0.
ln f (x)dx ≤ lim
n→+∞
n
n
n
0
Rα
Since f (x) ≥ 1, we have ln f (x) ≥ 0 which means that 0 ln f (x)dx ≥ 0.
Rα
Having on mind these two inequalities we get 0 ln f (x)dx = 0. Since
ln f (x) is continuous and non-negative, it follows that ln f (x) = 0, i.e.
f (x) = 1 for all x ∈ [0, α]. Finally, since α is a random constant we
conclude that f (x) = 1 for x ≥ 0. A similar argument yields f (x) = 1
for x < 0. So, f (x) = 1 is the only candidate for solution. Plugging in
f (x) = 1 into the starting condition we get 1 ≤ ank . Choosing a ≥ 1 and
k ≥ 1 we conclude that f (x) = 1 is the only solution.
51. Using L’Hôpital’s rule we have
2x + f 0 (x)
x2 + f (x)
= lim
=
lim = √
√ x
x→±∞
x→±∞
1 + x2
1+x2
p
p
f 0 (x)
f 0 (x)
= lim
1 + x2 2 +
= lim
1 + x2 lim
2+
.
x→±∞
x→±∞
x→±∞
x
x
0
Since f 0 is bounded it follows that limx→±∞ f x(x) = 0. Thus,
p
f 0 (x)
2
lim
1 + x lim
2+
= +∞.
x→±∞
x→±∞
x
52. Let’s calculate net limit limn→+∞
orem we have
Pn
lim
n→+∞
Pn
arctan
ln n
k=1
1
k
. Using Stolz–Cesàro the-
1
arctan n+1
arctan k1
= lim
= lim
n→+∞ ln (1 + 1 )
n→+∞
ln n
n
k=1
This means that
Pn
k=1
1
n+1
+o
1
n
+o
1
n+1
= 1.
1
n
arctan k1 = ln n + o(ln n) → +∞ when n → +∞.
27
53. First, let’s prove that an is bounded. We have next estimation an =
a1+1+1+...+1 ≤ na1 , i.e. ann ≤ a1 . Since an ≥ 0, we have that 0 ≤ ann ≤ a1 .
Now, let a be the infimum of A = { ann |n ∈ N}. Let ε > 0. Since a is
infimum of A there exists some amm ∈ A such that a + 2ε > amm . Now, let’s
give next estimation: for n > m we have
an
aqm+r
aqm
ar
qam ar
am ar
ε ar
=
≤
+
≤
+
=
+
< a+ + ,
n
qm + r
qm + r qm + r
qm
n
m
n
2
n
for some natural q and r < m. If M = max{a1 , a2 , . . . , am−1 } we have that
M
ε
n < 2 for all n greater then some n1 . Finally, for n > n0 = max{n1 , m}
ε
ε
we have ann < a + 2ε + anr ≤ a + 2ε + M
n < a + 2 + 2 = a + ε which means
an
that n converges.
54. Let m = inf x∈R f (x) ∈ R and M = supx∈R f (x) ∈ R and let’s suppose
that m 6∈ Im(f ) and M 6∈ Im(f ). Now, let’s observe some m < c < M .
Since m is infimum of Im(f ), for ε = c − m there exists some a ∈ Im(f )
such that m + ε = m + c − m = c > a > m. Similarly, there exists
some b ∈ Im(f ) so c < b < M. Using intermediate therem it follows
that c belongs to Im(f ). Since c is chosen randomly we conclude that
Im(f ) = (−m, M ). Since f is not constant it follows that m is less then
M , so (m, M ) is not an non-empty set. Since the cardinality of non-empty
interval is the same as cardinality of R this special case of the problem is
solved. The proof is similar if m or M (or both) belongs to Im(f ).
28