Math 3210, Fall 2013
Instructor: Thomas Goller
2 October 2013
Learning Celebration #2
Example (10 points)
Give an example of each of the following or state that no example exists:
(1) A sequence that has a limit but does not converge. (2 points)
{n}
(2) A convergent subsequence of the sequence {(−1)n n}. (2 points)
No example exists
(3) A nested sequence of closed and bounded intervals I1 ⊃ I2 ⊃ · · · such that
(2 points)
No example exists
!∞
n=1 In
= ∅.
(4) A sequence that is monotone and bounded. (2 points)
0, 0, 0, . . .
(5) A function with domain R that is not continuous at exactly two points in R. (2 points)


if x < 0;
1
f (x) = 0
if 0 ≤ x ≤ 3;


−1 if x > 3
Computation (4 points)
(6) Use the Main Limit Theorem to compute the limit of the sequence
Show your work! (4 points)
lim
'
n3 + 3
4n3 − 7n
(49
= lim
'
1+
4−
3
n3
7
n2
(49
=
'
lim 1 + 3(lim n1 )3
lim 4 − 7(lim n1 )2
(49
Precision (6 points)
(7) State the Monotone Convergence Theorem. (3 points)
Every bounded monotone sequence converges.
(8) State the Bolzano Weierstrass Theorem. (3 points)
Every bounded sequence has a convergent subsequence.
=
'
&'
3
n +3
4n3 − 7n
1 + 3 · 03
4 − 7 · 02
(49
(49 )
' (49
1
=
4
.
Math 3210, Fall 2013
Instructor: Thomas Goller
2 October 2013
Proof (20 points)
2
n
(9) Guess the limit of the sequence { 2n−1
}, state your guess as a claim, and prove the claim
using the definition of limit. (5 points)
2
n
Claim. lim 2n−1
= ∞.
Proof. Given M ∈ R, let N = 2M . Then for all n > N ,
n2
n2
n
N
>
= >
= M.
2n − 1
2n
2
2
(10) Let {an } be a sequence in R \ {0} and assume that lim an = ∞. Prove that lim 1/an = 0.
(5 points)
Proof. Given ! > 0, since lim an = ∞, there is N ∈ R such that for all n > N , an > 1! .
Thus for all n > N ,
*
*
*1
*
* − 0* = 1 = 1 < !.
* an
* |an |
an
(11) Let λ ∈ R \ {0} and let f be the function with rule f (x) = λx. Prove f is continuous
using the !-δ definition of continuity. (5 points)
Proof. Given a ∈ R and ! > 0, let δ =
!
.
|λ|
Then for all x ∈ R such that |x − a| < δ,
|f (x) − f (a)| = |λx − λa| = |λ| · |x − a| < |λ| · δ = !.
(12) Let g be the function with rule g(x) =
points)
&
x + 1 if x < 0
. Prove g is not continuous. (5
x − 1 if x ≥ 0
Proof. Let xn = − n1 for all n ∈ N. Then xn → 0, but g(xn ) = − n1 + 1 → 1, which is not
equal to g(0) = −1.