1549901369-Elements_of_Real_Analysis__Denlinger_

60 Chapter 2 • Sequences

}-Let no be such a natural number. The above analysis shows that

n ~no=? 1

3

~: ~ ~n - 31 < ~ < €. D

"

Example 2.1.8 Prove that lim (
3
n:-

4

n) = 3.

n->oo n + 5

Note: As in Example 2.1.6, to prove this limit statement, we essentially.
work Part (c) of Example 2.1.7 backwards. But remember, our proof must
stand alone, independently of Example 2.1.7.

. 5
Proof. Let c > 0. By the Arch1medean property,^5 :J no EN 3 n 0 > -+ 15.
€
Then,
5 5
n ~ no =? n > -+ 15 =? n > - and n > 15
€ €

n 1

=? - > -and n > 15

5 €

5

=? - < c and n > 15

n

4n+n

=? -- 2 - < c and n > 15

n

4n+ 15 4n+n

n^2 +^5 <-n - 2- <c

=? I (3n

2

• 4n) -3(n
  2
  
  
  • 
    
    
    
    5. I < €
       n^2 + 5
    
    
    
    
  
  
  
  

=? 13~: ~ ~n - 31 < €.

Thus, n ~no =? I 3n: -

4

n - 31 < c. Therefore, lim (

3

n: -

4

n) = 3. D

n + 5 n->oo n + 5

5. This choice of no came from Example 2.1.7. We know that, but there is no need to say
   that here. It would not help the proof in any way.