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EXERCISES FOR CHAPTER 1: Sequences and Limits
1. (a) 1
2
n
(b) 2
3
n
(c) 1 + n
1 2n
Solution
(a) limn
1
2
n
= limn
( 1)n
2n= 0
(b) limn
2
3
n
= limn
2n
3n= 0
(c) limn
1+ n
1 2n= lim
n
1
2=
1
2
2. (a) 1 + ( 1)n
2 (b)
1 + n2
1+ n (c)
n + 2
n2 + 3
Solution (a) Limit does not exist.
(b) limn
1+ n2
1+ n= lim
n
2n
1=
(c) limn
n + 2
n2 + 3= lim
n
1
2n= 0
3. (a) 6n5 + n
3n5 + 1 (b)
sin2n
n (c)
ln2n
lnn
Solution
(a) limn
6n5 + n
3n5 +1= lim
n
6n5 (1+16n4
)
3n5 (1+13n5
)= 2
(b)
1
n
sin2n
n
1
n
limn
sin2n
n= 0
since limn
1
n= 0
(c) limn
ln2n
lnn= lim
n
ln2
lnn+ lim
n
lnn
lnn= 1
4. (a) 73 / n (b) 5n3n (c) 2n
n2
Solution
K. A. Tsokos: Series and D.E.
2
(a) y = limn
73/n ln y =3
nlimn
ln 7 = 0 y = 1
(b) limn
5n3n= lim
n51/n lim
nn3/n = 1
(c) limn
2n
n2= lim
n
2n ln2
2n= lim
n
2n (ln2)2
2=
5. (a) n3
3n (b)
n!+ n2
2n!+ n (c)
n!+ 3n
1+ n
Solution
(a) limn
n3
en= lim
n
3n2
3n ln 3= lim
n
6n
3n (ln 3)2= lim
n
6
3n (ln 3)3= 0
(b) limn
n!+ n2
2n!+ n= lim
n
n!(1+n2
n!)
2n!(1+n
2n!)=1
2
(c) limn
n!+ 3n
1+ n= lim
n
n!(1+3n
n!)
n(1+1n)=
6. (a) en
1+ 2en (b)
e2n
(1+ 2en )2 (c) nsin
2n
Solution
(a) limn
en
1+ 2en= lim
n
en
2en=1
2
(b) limn
e2n
(1+ 2en )2= lim
n
e2n
4e2n (1+12en
)2=1
4
(c) limn
nsin2n
= limn
sin2n1n
= limn
2n2cos
2n1n2
=2limn
cos2n
=2
7. (a) lnen +1
en 1
(b)
2n +1
n (c)
3
2
n
Solution
(a) limn
lnen +1
en 1= ln lim
n
en +1
en 1= ln lim
n
en
en= ln1 = 0
(b) limn
2n +1
n= lim
n
n( 2 +1n)
n= 2
(c) limn
3
2
n
= since 3
2> 1
K. A. Tsokos: Series and D.E.
3
8. Show that (a) limn
1+x
n
n
= ex , (b) limn
1+x
n
n+
= ex and (c) limn
1+x
n +
n
= ex
where is a constant. Solution
(a) Let y = limn
1+x
n
n
. Call x
n=1
m so that m as n . Then n = mx and so
y = limm
1+1
m
mx
= limm
1+1
m
m x
= ex
(b)
limn
1+x
n
n+
= limn
1+x
nlimn
1+x
n
n
= 1 limn
1+x
n
n
= 1 ex = ex
(c) Call m = n + . Then as n , m . So
limn
1+x
n +
n
= limm
1+x
m
m
= limm
1+x
mlimm
1+x
m
m
= 1 ex = ex
9. (a) 11
n2
n
(b) 1 +n
n
(c) n 1
n
n
Solution (a)
limn
11
n2
n
= limn
11
n1+
1
n
n
= limn
11
n
n
limn
1+1
n
n
= e 1e1
= 1
(b)
K. A. Tsokos: Series and D.E.
4
limn
1+n
n
= limn
1+n
n
= e( )
= e2
(c)
limn
n 1
n
n
= limn
11
n
n
= e 1
10. (a) n + 3
n + 2
n
(b) n
n + 3
n
(c) 12
n
2n
Solution (a)
limn
n + 3
n + 2
n
= limn
n + 2 +1
n + 2
n
= limn
1+1
n + 2
n
= e
(b)
limn
n
n + 3
n
= limn
n + 3 3
n + 3
n
= limn
13
n + 3
n
= e 3
(c) limn
12
n
2n
= limn
12
n
n 2
= (e 2 )2 = e 4
11. (a) ln n( )7
n2 (b) n3nn (c) n
1+1
n
Solution
(a)
limn
lnn( )7
n2= lim
n
7 lnn( )6
2n2= lim
n
42 lnn( )5
4n2= = 0
(b) limn
n3nn= lim
nnn lim
n3nn
= 1 3 = 3
(c) limn
n1+1
n = limn
n limn
n1
n = 1 =
12. (a) n7nn (b) 1
n
1/ lnn
(c) arctan n
Solution
K. A. Tsokos: Series and D.E.
5
(a) limn
n7nn= lim
nnn lim
n7nn
= 1 7 = 7 .
(b) Let y =1
n
1/ln n
. Then limn
ln y = limn
ln(1n)
lnn= lim
n
lnn
lnn= 1 .
Thus limn
1
n
1/ln n
= e 1 .
(c) limn
arctann =2
.
13. Show that limn
1+ xn( )1/n
= 1 if x < 1 . Hence find limn
5n + 7n( )1/n
.
Solution
y = limn
1+ xn( )1/n
ln y = limn
ln(1+ xn )
n= 0
y = 1
limn
5n + 7n( )1/n
= limn
7 1+5
7
n 1/n
= 7 limn
1+5
7
n 1/n
= 7
14. Find the limit (a) limn
2n + 4n( )1/n
.
Solution
limn
2n + 4n( )1/n
= limn
1
2n + 4n( )1/n
= limn
1
424
n
+1
1/n
=1
4
15. (a) 2n + 4n
3n + 6n (b)
2n + 4n
5n + 6n
1/n
Solution
(a) limn
2n + 4n
3n + 6n= lim
n
4n
6n
12
n
+1
12
n
+1
= limn
4n
6n= 0
(b) limn
2n + 4n
5n + 6n
1/n
=4
6limn
24
n
+1
56
n
+1
1/n
=2
3
K. A. Tsokos: Series and D.E.
6
16. (a) n137
137n (b)
n17
en (c)
32n+1
5n2
Solution
(a) limn
n137
137n= 0 , standard result.
(b) limn
n17
en= 0 , standard result.
(c) limn
32n+1
5n2= lim
n
32n+12 ln 3
10n= lim
n
32n+1(2 ln 3)2
10=
17. (a) arctann
(b) ln(n + 1) ln n (c) ln(n2 + n) lnn2
Solution
(a) limn arctann
=
2
= 2
(b) limn
ln(n +1) lnn( ) = limn
ln(n +1
n) = ln lim
n(n +1
n) = ln1 = 0
(c) limn
ln(n2 + n) lnn2( ) = limn
ln(n2 + n
n2) = ln lim
n(n2 + n
n2) = ln1 = 0
18. (a) 8n!
5nn (b)
n!
10n!+(n 1)! (c)
ln9n
n3
Solution
(a) limn
8n!
5nn= 0 , standard result.
(b) limn
n!
10n!+ (n 1)!= lim
n
n!
10n!(1+110n
)=1
10
(c) limn
ln9n
n3= lim
n
ln9
n3+ lim
n
lnn
n3= 0 + lim
n
1n
13n 2 /3
= 3limn
1
n1/3= 0
19. (a) n2 +1
n (b)
sin n
n (c)
n!
5n2
Solution
(a) limn
n2 +1
n= lim
n
n 1+1n
n=
(b) 1
n
sin n
n
1
nlimn
sin n
n= 0 since lim
n
±1
n= 0
(c) limn
n!
5n2= , standard result.
K. A. Tsokos: Series and D.E.
7
20. (a) n
3n (b) n3e 2n (c)
n!+(n 1)!
(n 1)!+(n 2)!
Solution
(a) limn
n
3n= lim
n
1
3n ln 3= 0
(b) limn
n3e 2n= lim
n
n3
e2n= lim
n
3n2
2e2n= lim
n
6n
4e2n= lim
n
6
8e2n= 0
(c) limn
n!+ (n 1)!
(n 1)!+ (n 2)!= lim
n
n!(1+1n)
(n 1)!(1+2
n 1)=
21. (a) lnn
n (b) n ln(1+
2
n) (c)
32n
7n
Solution
(a) limn
lnn
2n= lim
n
1n
21
2 n
= limn
2
n= 0
(b) limn
n ln(1+2
n) = ln lim
n1+
2
n
n
= lne2 = 2
(c) limn
32n
7n= lim
n
9n
7n= lim
n
9n ln9
7=
22. (a) n!
1+ 5n! (b)
2nn
nn + 3nn 1
Solution
(a) limn
n!
1+ 5n!= lim
n
n!
5n!(1+15n!)=1
5
(b) limn
2nn
nn + 3nn 1= lim
n
2nn
nn (1+3n)= 2
23. (a) 2nn
nn + n! (b)
n!
e2n + 3n! (c)
ln2n
2n
Solution
(a) limn
2nn
nn + n!= lim
n
2nn
nn (1+n!nn)= 2
(b) limn
n!
e2n + 3n!= lim
n
n!
3n!(1+e2n
3n!)=1
3
(c) limn
ln2n
2n= lim
n
ln2
2n+ lim
n
lnn
2n= 0 + lim
n
1n
21
2 n
= limn
2
n= 0
K. A. Tsokos: Series and D.E.
8
24. James Stirling (1692-1770) showed that for large values of n, n!n
e
n
2 n . Use
this approximation for the factorial of n to show that limn
n!
nn= 0 .
Solution
limn
n!
nn= lim
n
(n / e)n 2 n
nn= lim
n
2 n
en= 0 .
(Notice that Stirling’s result leads to the elegant limit limn
n!en
nn n= 2 .)
25. (a) Show that limn
n2
n2 +1= 1 . (b) Establish this limit by an N proof.
Solution
(a) limn
n2
n2 +1= lim
n
n2
n2 (1+1n2)= 1
(b) n2
n2 +11 < <
n2
n2 +11< <
1
n2 +1< . Hence we have that
n2 +1<1
n >11 . Thus
n2
n2 +11 < for all n > N where N =
11 .
26. (a) Show that limn
1
2n= 0 . (b) Establish this limit by an N proof.
Solution
(a) Let y =1
2n. Then lim
nln y = lim
nln
1
2n= lim
nn ln2 = . Hence y e = 0 .
(b) 1
2n< <
1
2n< . Hence
1
2n< 2n >
1n >
ln
ln2. Thus
1
2n<
for all n > N where N =ln
ln2.
27. (a) Show that limn
n!
1+ 2n!=1
2. (b) Establish this limit by an N proof.
Solution
(a) limn
n!
1+ 2n!= lim
n
n!
2n!(1+12n!)=1
2.
(b) n!
1+ 2n!
1
2< <
n!
1+ 2n!
1
2< . I.e. <
2n! 1 2n!
2(1+ 2n!)< i.e.
<1
2(1+ 2n!)< . We must then ensure that
K. A. Tsokos: Series and D.E.
9
1
2(1+ 2n!)>
1
2(1+ 2n!)<
2n!>21
n!>1 1
2
Now n!> n and so we if we demand n >1 1
2 we automatically ensure n!>
1 1
2.
Thus n!
1+ 2n!
1
2< for all n > N where N >
1 1
2. As a check take
= 10 1 N >101
2. With N = 10 we have that
10!
1+ 2 10!
1
2< 6.89 10 8
<10 1 .
This shows that the choice of N is very loose. This is so because the inequality n!> n is
very easily satisfied so a much lower value of N than N >1 1
2 could have done.
28. A sequence of positive numbers is defined recursively through un+1 = 1 un , with
u1 =1
2. Given that the sequence is convergent, find the limit of the sequence,
limn
un .
Solution Let the limit be L. Then lim
nun = L , lim
nun+1 = L and so L = 1 L L2 + L 1 = 0 .
The positive root is L =1+ 5
2.
29. Find limx 0
1 cos x
x2.
Solution
Limit is of the 0/0 type. Hence limx 0
1 cos x
x2= lim
x 0
sin x
2x= lim
x 0
cos x
2=1
2.
30. Find the limit limx 0
ln(1 2x)
x.
Solution
Limit is of the 0/0 type. Hence limx 0
ln(1 2x)
x= lim
x 0
21 2x1
= 2 .
31. Evaluate the limit limx 0
tan x
x.
Solution
Limit is of the 0/0 type. Hence limx 0
tan x
x= lim
x 0
sec2 x
1= 1 .
K. A. Tsokos: Series and D.E.
10
32. Find limx 0
tan 3x
x.
Solution
Limit is of the 0/0 type. Hence limx 0
tan 3x
x= lim
x 0
3sec2 3x
1= 3 .
33. Find limx 0
arcsin(3x) 3x
x3.
Solution Limit is of the 0/0 type. Hence
limx 0
arcsin(3x) 3x
x3= lim
x 0
3
1 9x23
3x2= lim
x 0
3 ( 18x)(12)(1 9x2 ) 3/2
6x
= limx 0
27(1 9x2 ) 3/2 + 27x(32)(1 9x2 ) 3/2 ( 18x)
6=9
2
34. Evaluate limx 0(1
x
1
sin x) .
Solution
limx 0(1
x
1
sin x) = lim
x 0(sin x x
x sin x)
= limx 0(
cos x 1
sin x + x cos x)
= limx 0(
sin x
cos x + cos x x sin x)
= 0
35. Find the limit limx 0
sin(x + 2sin x)
sin x.
Solution
limx 0
sin(x + 2sin x)
sin x= lim
x 0
cos(x + 2sin x)(1+ 2cos x)
cos x= 3
36. Find limx 0(1
x
1
tan x) .
Solution
We may write 1
x
1
tan x=x tan x
x tan x which is a 0/0 limit. Using L’ Hôpital’s rule we
have
limx 0(1
x
1
tan x) = lim
x 0
x tan x
x tan x= lim
x 0
1 sec2 x
tan x + x sec2 x
= limx 0
2sec x(sec x tan x)
sec2 x + sec2 x + x2sec x(sec x tan x)= 0
K. A. Tsokos: Series and D.E.
11
37. Find limx 1(1
ln x
1
x 1) .
Solution 1
ln x
1
x 1=x 1 ln x
(x 1)ln x which results in a 0/0 limit. Using L’ Hôpital’s rule we have
limx 1
x 1 ln x
(x 1)ln x= lim
x 1
11x
ln x +x 1x
= limx 1
1x2
1x+1x2
=1
2, using L’ Hôpital’s rule again.
38. Find limx 0
1+ x 1 x
x.
Solution The limit is of the 0/0 type. Using L’ Hôpital’s rule we have that
limx 0
1+ x 1 x
x= lim
x 0
1
2 1+ x+
1
2 1 x1
= 1 .
39. Consider the function f (x) = 1+ ex + e2x + + enx . (a) Show that f (x) =
e(n+1)x 1
ex 1.
(b) Show that
df
dx= 1ex + 2e2x + + nenx . (c) Hence show that
1+ 2 + + n =df
dx x=0
. (d) Hence evaluate 1+ 2 + + n . (e) Using this method find
an expression for (but do not attempt to evaluate) 1k+ 2k + + nk where k is a
positive integer. Solution
(a)
1+ ex + e2x + + enx =e(n+1)x 1
ex 1, by summing a geometric progression.
(b)
df
dx= 1ex + 2e2x + + nenx
(c) Setting x=0 in the result above gives the answer. (d) df (x)
dx=(n +1)e(n+1)x (ex 1) (e(n+1)x 1)ex
(ex 1)2
=ex e(n+1)x ne(n+1)x + ne(n+2)x
(ex 1)2
This gives a 0/0 limit if we put x=0. So by L’ Hôpital’s rule
K. A. Tsokos: Series and D.E.
12
df (0)
dx= lim
x 0
ex (n +1)e(n+1)x n(n +1)e(n+1)x + n(n + 2)e(n+2)x
2(ex 1)ex
= limx 0
ex (n +1)2e(n+1)x n(n +1)2e(n+1)x + n(n + 2)2e(n+2)x
4e2x 2ex
=1 (n +1)2 n(n +1)2 + n(n + 2)2
2
=1 n2 2n 1 n3 2n2 n + n3 + 4n2 + 4n
2
=n2 + n
2
=n(n +1)
2
(e) We saw that
df
dx= 1ex + 2e2x + + nenx . Differentiating again gives
d 2 f
dx2= 12ex + 22e2x + + n2enx
and so differentiating k times gives
dk f
dxk= 1k ex + 2k e2x + + nkenx .
Thus at x = 0 we have that
1k + 2k + + nk =dk f
dxkx=0
=dk
dxke(n+1)x 1
ex 1x=0
.
Applying L’ Hôpital’s rule to this expression is hopeless. To make progress requires more advanced work.
