Math 142A
(P. Fitzsimmons)
First Midterm Exam
Solutions
2.
Use the
definition
of limit (of a sequence) to show that
lim
n
→∞
n
n
+
√
n
= 1
.
Solution.
Let us begin by estimating the distance from the sequence to the limit:
(2
.
1)
vextendsingle
vextendsingle
vextendsingle
vextendsingle
n
n
+
√
n

1
vextendsingle
vextendsingle
vextendsingle
vextendsingle
=
vextendsingle
vextendsingle
vextendsingle
vextendsingle
n

n

√
n
n
+
√
n
vextendsingle
vextendsingle
vextendsingle
vextendsingle
=
vextendsingle
vextendsingle
vextendsingle
vextendsingle

√
n
n
+
√
n
vextendsingle
vextendsingle
vextendsingle
vextendsingle
=
√
n
n
+
√
n
≤
√
n
n
=
1
√
n
.
Given
ǫ >
0 (and with the above calculation in mind) we choose the cutoff
n
*
to be any integer
larger that 1
/ǫ
2
. If
n
≥
n
*
, then
n >
1
/e
2
, so
√
n >
1
/ǫ
, so 1
/
√
n < ǫ
. That is
(2
.
2)
n
≥
n
*
=
⇒
1
√
n
< ǫ.
Consequently, by (2.1) and (2.2),
n
≥
n
*
=
⇒
vextendsingle
vextendsingle
vextendsingle
vextendsingle
n
n
+
√
n

1
vextendsingle
vextendsingle
vextendsingle
vextendsingle
< ǫ,
as required.
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 Winter '01
 Fitzsimmons
 Math, Calculus, Limit, Monotone convergence theorem, P. Fitzsimmons

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