1549901369-Elements_of_Real_Analysis__Denlinger_

2.1 Basic Concepts: Convergence and Limits 59

The latter inequality will be true if

n^2 + 5

4 n + 15 > 2, 000; i.e.,

n^2 + 5 > 8000n + 30, 000

n^2 - 8000n > 29, 995

n(n - 8000) > 29, 995.

Note that 8004(8004 - 8000) = 8004 · 4 = 32, 016 > 29, 995. In fact

n 2: 8004 =? n(n - 8000) 2: 8004(4) > 29, 995.

Thus, we take n 0 = 8004. The above analysis shows that

I

3n

2

n 2: 8004 =? - 4n I

n^2 +5 -^3 < .0005..

(c) Let c > 0 be a fixed but arbitrary positive number. We want to find an

1

3n

2

no EN 3 n 2: no=? n -4n I

2 + 5 -^3 < t=:. As shown above,

I

3n

2

• 4n _ 31 = 4n + 15.
  n^2 + 5 n^2 + 5

Thus, our objective is to find an no E N 3

4n+ 15

n 2: no =? n2 + 5 < c.

Notice that

4n + 15 4n + 15

<

n^2 + 5 n^2

<

4n+n

n2

5

n

if n > 15

4n + 15 5

Thus, 2 < c if n > 15 and - < c. But the last inequality will be

n +5 n

guarantee d i .f -n > - ;^1. i.e., n > -.^5
5 c c
By the Archimedean property, 3 n 0 E N 3 no > ~ + 15.

• ----c--
  ( We want to be sure that both n > 15 ~nd n > ~·)