EXERCISES FOR CHAPTER 1: Sequences and Limits
1 n
1. (a) 2
2 n
(b) 3
(c)
1+n
1 2n
(c)
n+2
2
n +3
(c)
ln2n
ln n
(c)
2n
n2
Solution
n
(1)n
1
(a) lim = lim n = 0
n n 2
2
n
2n
2
(b) lim = lim n = 0
n 3 n 3
1
1+ n
1
(c) lim
= lim
=
n 1 2n
n 2
2
n
1 + (1)
1 + n2
2. (a)
(b)
2
1+ n
Solution
(a) Limit does not exist.
1 + n2
2n
= lim
=
(b) lim
n 1 + n
n 1
n+2
1
(c) lim 2
= lim
=0
n n + 3
n 2n
6n 5 + n
3. (a)
5
3n + 1
(b)
sin 2n
n
Solution
1
6n 5 (1 + 4 )
6n 5 + n
6n = 2
= lim
(a) lim 5
n 3n + 1
n
1
5
3n (1 + 5 )
3n
(b)
1
sin 2n
1
1
n
n
n
=0
since lim
n
sin 2n
n
lim
=0
n
n
ln 2n
ln 2
ln n
(c) lim
= lim
+ lim
=1
n ln n
n ln n
n ln n
3 /n
4. (a) 7
Solution
(b) n 5n 3
2
K. A. Tsokos: Series and D.E.
(a) y = lim 7 3/n ln y =
n
3
lim ln 7 = 0 y = 1
n n
(b) lim n 5n 3 = lim 51/n lim n 3/n = 1
n
n
n
n
n
2
2 ln 2
2 n (ln 2)2
= lim
=
(c) lim 2 = lim
n n
n
n
2
2n
n!+ n 2
n3
5. (a) n
(b)
2n!+ n
3
n!+ 3n
(c)
1+ n
Solution
n3
3n 2
6n
6
= lim n
= lim n
=0
(a) lim n = lim n
2
n e
n 3 ln 3
n 3 (ln 3)
n 3 (ln 3)3
n2
n!(1 + )
n!+ n 2
n! = 1
= lim
(b) lim
n 2n!+ n
n
n
2n!(1 +
) 2
2n!
n
3
n!(1 + )
n!+ 3n
n! = = lim
(c) lim
n 1 + n
n
1
n(1 + )
n
n
e 2n
e
(b)
(c) nsin
6. (a)
n 2
n
2n
(1+ 2e )
1+ 2e
Solution
1
en
en
=
lim
=
n
n
n 1 + 2e
n 2e
2
2n
1
e
e2n
=
lim
=
(b) lim
n
2
n (1 + 2e )
n
1
4e2n (1 + n )2 4
2e
sin
2 cos
2n = lim 2n
2n = lim cos = = lim
(c) lim n sin
n
n
1
2n n 1
2 n
2n 2
2
n
n
n
en + 1
3
2n + 1
7. (a) ln n (b)
(c) 2
e 1
n
Solution
en + 1 en + 1
en
(a) lim ln n
=
ln
lim
=
ln
lim
= ln1 = 0
n
n e n 1
n e n
e 1 (a) lim
1
n( 2 + )
2n + 1
n = 2
= lim
(b) lim
n
n
n
n
n
3
3
(c) lim = since > 1
n 2 2
3
K. A. Tsokos: Series and D.E.
n
n +
n
x
x
x x
x
x
8. Show that (a) lim 1+ = e , (b) lim 1+ = e and (c) lim 1+
=e
n n
n
n
n
n +
where is a constant.
Solution
n
x
x 1
(a) Let y = lim 1 + . Call =
so that m as n . Then n = mx and so
n n
n m
1
y = lim 1 + m m
mx
x
m
1 = lim 1 + m m = ex
(b)
x
lim 1 + n n
n +
x
x
= lim 1 + lim 1 + n n n
n
x
= 1 lim 1 + n n
n
n
= 1 ex = ex
(c) Call m = n + . Then as n , m . So
n
x x
lim 1 +
= lim 1 + n m n +
m
x
= lim 1 + m m
m x
lim 1 + m m
= 1 ex = ex
n
(b) 1 + n
n
1
9. (a) 1 2 n Solution
(a)
n
1
1
1 lim 1 2 = lim 1 1 + n n n
n n = lim 1 n = e1e1
=1
(b)
n
1
lim 1 +
n
n
n
1
n
n
n 1 n
(c) n m
4
K. A. Tsokos: Series and D.E.
n
= lim 1 + n n n
lim 1 + n n
( )
= e
= e
2
(c)
1
n 1
lim = lim 1 n n n
n n
n
= e1
n + 3
10. (a) (b)
n + 2 Solution
(a)
n
n
n + 3
n + 2 + 1
lim lim = n
n n + 2 n + 2 n
1 = lim 1 +
n n + 2
n n
n + 3
2 2n
(c) 1 n
n
=e
(b)
n n + 3 3
lim = lim n n + 3 n n+3 n
3 = lim 1 n n + 3
n
n
= e3
2
(c) lim 1 n n
11. (a)
2n
2
n
2 = lim 1 = (e2 )2 = e4
n n (ln n) 7
(b)
n2
n
n3
n
1+
(c) n
1
n
Solution
7
6
5
ln n )
7 ( ln n )
42 ( ln n )
(
= lim
= lim
== 0
(a) lim
n
n
n
n2
2n 2
4n 2
(b) lim n n3n = lim n n lim n 3n = 1 3 = 3
n
n
1+
(c) lim n
1
n
n
n
12. (a) n7
Solution
n
1
n
= lim n lim n = 1 = n
n
n
1 1/ ln n
(b) n
(c) arctan n
5
K. A. Tsokos: Series and D.E.
(a) lim n n7 n = lim n n lim n 7 n = 1 7 = 7 .
n
n
1
(b) Let y = n
1
Thus lim n n 1/ln n
n
1
ln( )
ln n
= 1 .
. Then lim ln y = lim n = lim
n
n ln n
n ln n
1/ln n
= e1 .
.
2
(c) lim arctan n =
n
(
13. Show that lim 1 + x n
n
)
1/n
(
= 1 if x < 1 . Hence find lim 5 n + 7 n
n
)
1/n
.
Solution
(
y = lim 1 + x n
n
(
lim 5 + 7
)
1/n
n
n
=7
(
14. Find the limit (a) lim 2 n + 4 n
n
)
1/n
.
Solution
(
n
)
1/n
= lim
n
= lim
n
=
(2
1
n
+ 4n
)
1/n
1
2n 4 + 1
4
1/n
1
4
2n + 4 n (b) n
5 + 6 n 2n + 4 n
15. (a) n
3 + 6n
Solution
n
1
n
n
n +1
2 +4
4 2
4n
= lim n
= lim n = 0
(a) lim n
n
n 3 + 6 n
n 6
n 6
1
+
1
2
2n + 4 n (b) lim n
n 5 + 6 n 1/n
5n
= lim 7 1 + n
7 1/n
5n
= 7 lim 1 + n
7 ln(1 + x n )
ln y = lim
=0
n
n
y =1
lim 2 n + 4 n
)
n 1/n
2n
+ 1
4
4
= lim 6 n 5 n
6 + 1 1/n
=
2
3
1/n
1/n
6
K. A. Tsokos: Series and D.E.
n1 3 7
n1 7
(b)
137n
en
Solution
n137
= 0 , standard result.
(a) lim
n 137 n
n17
(b) lim n = 0 , standard result.
n e
32n +1
32n +12 ln 3
32n +1 (2 ln 3)2
=
lim
=
lim
=
(c) lim
n 5n 2
n
n
10
10n
16. (a)
arctan n
Solution
= =2
(a) lim
n arctan n
2
(b) ln(n + 1) ln n
17. (a)
(c)
32n +1
5n 2
(c) ln(n 2 + n) ln n 2
n +1
n +1
) = ln lim (
) = ln1 = 0
n
n
n
n2 + n
n2 + n
= lim ln( 2 ) = ln lim ( 2 ) = ln1 = 0
n
n
n
n
(b) lim ( ln(n + 1) ln n ) = lim ln(
n
n
(
(c) lim ln(n 2 + n) ln n 2
n
)
n!
ln9n
8n!
(b)
(c) 3
n
10n!+(n 1)!
5n
n
Solution
8n!
(a) lim n = 0 , standard result.
n 5n
1
n!
n!
= lim
=
(b) lim
n 10n!+ (n 1)!
n
1
) 10
10n!(1 +
10n
1
ln 9n
ln 9
ln n
1
= lim 3 + lim 3 = 0 + lim n = 3lim 1/ 3 = 0
(c) lim 3
n
n
n
n
n
1
n
n
n
n
n 2 / 3
3
2
n +1
n!
sin n
19. (a)
(b)
(c)
5n 2
n
n
Solution
1
n 1+
n2 + 1
n =
= lim
(a) lim
n
n
n
n
1
sin n
1
sin n
±1
lim
= 0 since lim
=0
(b) n
n
n
n
n
n
n
n!
(c) lim 2 = , standard result.
n 5n
18. (a)
7
K. A. Tsokos: Series and D.E.
n!+(n 1)!
n
3 2n
(b) n e
(c)
n
(n 1)!+(n 2)!
3
Solution
n
1
(a) lim n = lim n
=0
n 3
n 3 ln 3
n3
3n 2
6n
6
(b) lim n 3e2n = lim 2n = lim 2n = lim 2n = lim 2n = 0
n
n e
n 2e
n 4e
n 8e
1
n!(1
+
)
n!+ (n 1)!
n
= lim
=
(c) lim
n (n 1)!+ (n 2)!
n
2
(n 1)!(1 +
)
n 1
32n
2
ln n
21. (a)
(b) n ln(1 + )
(c)
7n
n
n
Solution
1
ln n
2
n
(a) lim
= lim
= lim
=0
n
n
n
1
n
2n
2
2 n
n
2
2
(b) lim n ln(1 + ) = ln lim 1 + = ln e2 = 2
n
n n
n
20. (a)
32n
9n
9 n ln 9
= lim
= lim
=
(c) lim
n 7n
n 7n
n
7
n!
2n n
22. (a)
(b) n
1+ 5n!
n + 3n n1
Solution
1
n!
n!
= lim
=
(a) lim
n 1 + 5n!
n
1
) 5
5n!(1 +
5n!
n
2n
2n n
lim
=
lim
=2
(b)
n n n + 3n n 1
n n
3
n (1 + )
n
n
n!
2n
23. (a) n
(b) 2n
e + 3n!
n + n!
Solution
2n n
2n n
= lim
=2
(a) lim n
n n + n!
n n
n!
n (1 + n )
n
1
n!
n!
(b) lim 2n
= lim
2n =
n e
e
3
+ 3n! n
)
3n!(1 +
3n!
(c)
ln2n
2n
1
ln 2n
ln 2
ln n
2
n
(c) lim
= lim
+ lim
= 0 + lim
= lim
=0
n
n
n
1
n
2n n 2n n 2n
2
2 n
8
K. A. Tsokos: Series and D.E.
n n
24. James Stirling (1692-1770) showed that for large values of n, n! 2n . Use
e
n!
this approximation for the factorial of n to show that lim n = 0 .
n n
Solution
n!
(n / e)n 2 n
2 n
lim n = lim
= lim n = 0 .
n
n n
n
n
n
e
n!en
= 2 .)
(Notice that Stirling’s result leads to the elegant limit lim n
n n
n
n2
= 1 . (b) Establish this limit by an N proof.
25. (a) Show that lim 2
n n + 1
Solution
n2
n2
= lim
=1
(a) lim 2
n n + 1
n 2
1
n (1 + 2 )
n
2
n
n2
1
(b) 2
1 < < 2
1< < 2
< . Hence we have that
n +1
n +1
n +1
n2 + 1 <
1
1
1
n2
n>
1 . Thus 2
1 .
1 < for all n > N where N =
n +1
1
= 0 . (b) Establish this limit by an N proof.
n 2 n
26. (a) Show that lim
Solution
1
1
. Then lim ln y = lim ln n = lim n ln 2 = . Hence y e = 0 .
n
n
n
n
2 2
1
1
1
1
1
ln . Thus n < (b) n < < n < . Hence n < 2 n > n > 2
2
2
ln 2
2
ln for all n > N where N = .
ln 2
n!
1
27. (a) Show that lim
= . (b) Establish this limit by an N proof.
n 1 + 2n!
2
(a) Let y =
Solution
n!
n!
1
= lim
= .
n 1 + 2n!
n
1
2n!(1 +
) 2
2n!
n!
1
n!
1
< <
<.
(b)
1 + 2n! 2
1 + 2n! 2
1
<
< . We must then ensure that
2(1 + 2n!)
(a) lim
I.e.
<
2n! 1 2n!
<
2(1 + 2n!)
i.e.
9
K. A. Tsokos: Series and D.E.
1
> 2(1 + 2n!)
1
<
2(1 + 2n!)
2
2n! > 1
1 1
n! > 2
1 1
1 1
Now n! > n and so we if we demand n > we automatically ensure n! > .
2
2
n!
1
1 1
< for all n > N where N > . As a check take
Thus
1 + 2n! 2
2
10!
1
1
< 6.89 10 8 < 10 1 .
= 10 1 N > 10 . With N = 10 we have that
1 + 2 10! 2
2
This shows that the choice of N is very loose. This is so because the inequality n! > n is
1 1
very easily satisfied so a much lower value of N than N > could have done.
2
28. A sequence of positive numbers is defined recursively through un +1 = 1 un , with
1
u1 = . Given that the sequence is convergent, find the limit of the sequence,
2
lim un .
n Solution
Let the limit be L. Then lim un = L , lim un +1 = L and so L = 1 L L2 + L 1 = 0 .
n
The positive root is L =
29. Find lim
x0
1 cos x
.
x2
Solution
n
1 + 5
.
2
1 cos x
sin x
cos x 1
= lim
= lim
= .
2
x0
x0 2x
x0
2
x
2
Limit is of the 0/0 type. Hence lim
30. Find the limit lim
x0
Solution
ln(1 2x)
.
x
2
ln(1 2x)
= lim 1 2x = 2 .
Limit is of the 0/0 type. Hence lim
x0
x0
1
x
tan x
.
31. Evaluate the limit lim
x0
x
Solution
tan x
sec 2 x
= lim
= 1.
Limit is of the 0/0 type. Hence lim
x0
x0
x
1
K. A. Tsokos: Series and D.E.
10
tan 3x
.
x0
x
32. Find lim
Solution
tan 3x
3sec 2 3x
= lim
= 3.
Limit is of the 0/0 type. Hence lim
x0
x0
x
1
arcsin(3x) 3x
.
x0
x3
33. Find lim
Solution
Limit is of the 0/0 type. Hence
3
1
3
3 (18x)( )(1 9x 2 )3/2
2
arcsin(3x) 3x
2
lim
= lim 1 9x2
= lim
x0
x0
x0
6x
x3
3x
3
27(1 9x 2 )3/2 + 27x( )(1 9x 2 )3/2 (18x) 9
2
= lim
=
x0
2
6
1
1
34. Evaluate lim( ).
x0 x
sin x
Solution
1
1
sin x x
lim( ) = lim(
)
x0 x
x0
sin x
x sin x
cos x 1
)
= lim(
x0 sin x + x cos x
sin x
)
= lim(
x0 cos x + cos x x sin x
=0
sin(x + 2 sin x)
35. Find the limit lim
.
x0
sin x
Solution
sin(x + 2 sin x)
cos(x + 2 sin x)(1 + 2 cos x)
lim
= lim
x0
x0
sin x
cos x
=3
1
1
36. Find lim( ).
x0 x
tan x
Solution
1
x tan x
1
We may write which is a 0/0 limit. Using L’ Hôpital’s rule we
=
x tan x
x tan x
have
1
1
x tan x
1 sec 2 x
lim( ) = lim
= lim
x0 x
x0 x tan x
x0 tan x + x sec 2 x
tan x
2 sec x(sec x tan x)
= lim 2
=0
x0 sec x + sec 2 x + x2 sec x(sec x tan x)
11
K. A. Tsokos: Series and D.E.
37. Find lim(
x1
1
1
).
ln x x 1
Solution
1
x 1 ln x
1
=
which results in a 0/0 limit. Using L’ Hôpital’s rule we have
ln x x 1 (x 1)ln x
1
1
1
2
1
x 1 ln x
x
lim
= lim
= lim x
= , using L’ Hôpital’s rule again.
x1 (x 1)ln x
x1
x 1 x1 1 1
ln x +
+ 2 2
x x
x
38. Find lim
x0
1+ x 1 x
.
x
Solution
The limit is of the 0/0 type. Using L’ Hôpital’s rule we have that
1
1
+
1+ x 1 x
lim
= lim 2 1 + x 2 1 x = 1 .
x0
x0
x
1
e(n +1)x 1
39. Consider the function f (x) = 1 + e + e + + e . (a) Show that f (x) =
.
ex 1
df
(b)
Show
that
(c)
Hence
show
that
= 1e x + 2e2 x + + nenx .
dx
df
. (d) Hence evaluate 1 + 2 + + n . (e) Using this method find
1+ 2 ++ n =
dx x = 0
x
2x
nx
an expression for (but do not attempt to evaluate) 1k + 2 k + + n k where k is a
positive integer.
Solution
e(n +1)x 1
(a) 1 + e x + e2 x + + enx =
, by summing a geometric progression.
ex 1
df
(b)
= 1e x + 2e2 x + + nenx
dx
(c) Setting x=0 in the result above gives the answer.
(d)
df (x) (n + 1)e(n +1)x (e x 1) (e(n +1)x 1)e x
=
(e x 1)2
dx
e x e(n +1)x ne(n +1)x + ne(n + 2)x
(e x 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)
e x (n + 1)e(n +1)x n(n + 1)e(n +1)x + n(n + 2)e(n + 2)x
= lim
x0
2(e x 1)e x
dx
e x (n + 1)2 e(n +1)x n(n + 1)2 e(n +1)x + n(n + 2)2 e(n + 2)x
x0
4e2 x 2e x
1 (n + 1)2 n(n + 1)2 + n(n + 2)2
=
2
2
1 n 2n 1 n 3 2n 2 n + n 3 + 4n 2 + 4n
=
2
2
n +n
=
2
n(n + 1)
=
2
= lim
df
= 1e x + 2e2 x + + nenx . Differentiating again gives
dx
d2 f
= 12 e x + 2 2 e2 x + + n 2 enx
2
dx
and so differentiating k times gives
dk f
= 1k e x + 2 k e2 x + + n k enx .
dx k
Thus at x = 0 we have that
dk f
d k e(n +1)x 1 .
1k + 2 k + + n k = k
= k x
dx x = 0 dx e 1 x = 0
(e) We saw that
Applying L’ Hôpital’s rule to this expression is hopeless. To make progress requires
more advanced work.