Proving a sequence has the given limit.

  • #1

Homework Statement


Prove that the following sequence has the following limit:

[tex]\lim_{n\rightarrow \infty} {{{\frac{n}{n^2+3n+1}}} = 0 [/tex]


Homework Equations


N/A


The Attempt at a Solution



First I did the following:

[tex]{\frac{n}{n^2+3n+1} < {\frac{n}{n^2+3n} = {\frac{n}{n(n+3)} = {\frac{1}{n+3}[/tex]

Then:
[tex]{\frac{1}{n+3} < \epsilon[/tex]

[tex]{{n+3} > \frac{1}{\epsilon}[/tex]

[tex]{{n} > \frac{1}{\epsilon}-3[/tex]


Putting it all together:

Let [tex]\epsilon > 0[/tex] be given

Let [tex]{{n} > \frac{1}{\epsilon}-3[/tex]

Then: [tex]{\left|\frac{n}{n^2+3n+1}\right| - 0 = \frac{n}{n^2+3n+1} < \frac{n}{n^2+3n} = \frac{1}{n+3} < {\epsilon }[/tex]

Therefore [tex]\left|\frac{n}{n^2+3n+1}-0\right| < \epsilon [/tex]
QED

Am I going about this the right way?
 
Last edited:

Answers and Replies

  • #2
Any help would be appreciated. Thanks. :)
 
  • #3
HallsofIvy
Science Advisor
Homework Helper
41,847
967

Homework Statement


Prove that the following sequence has the following limit:

[tex]\lim_{n\rightarrow \infty} {{{\frac{n}{n^2+3n+1}}} = 0 [/tex]


Homework Equations


N/A


The Attempt at a Solution



First I did the following:

[tex]{\frac{n}{n^2+3n+1} < {\frac{n}{n^2+3n} = {\frac{n}{n(n+3)} = {\frac{1}{n+3}[/tex]

Then:
[tex]{\frac{1}{n+3} < \epsilon[/tex]

[tex]{{n+3} > \frac{1}{\epsilon}[/tex]

[tex]{{n} > \frac{1}{\epsilon}-3[/tex]
Since [itex]\epsilon-3< 1/\epsilon[/itex] it would be sufficient to take [itex]n> 1/\epsilon[/itex].

Other than that, it looks good to me.

Putting it all together:

Let [tex]\epsilon > 0[/tex] be given

Let [tex]{{n} > \frac{1}{\epsilon}-3[/tex]

Then: [tex]{\left|\frac{n}{n^2+3n+1}\right| - 0 = \frac{n}{n^2+3n+1} < \frac{n}{n^2+3n} = \frac{1}{n+3} < {\epsilon }[/tex]

Therefore [tex]\left|\frac{n}{n^2+3n+1}-0\right| < \epsilon [/tex]
QED

Am I going about this the right way?
 
  • #4
Since [itex]\epsilon-3< 1/\epsilon[/itex] it would be sufficient to take [itex]n> 1/\epsilon[/itex].

I don't follow that. How did you get [itex]\epsilon-3< 1/\epsilon[/itex] and why can you take [itex]n> 1/\epsilon[/itex] from it?
 
  • #5
35,646
7,520
I think HallsOfIvy meant [tex]1/\epsilon - 3 < 1/\epsilon[/tex]
 

Related Threads on Proving a sequence has the given limit.

  • Last Post
Replies
5
Views
4K
  • Last Post
Replies
24
Views
2K
Replies
1
Views
5K
  • Last Post
Replies
4
Views
825
  • Last Post
Replies
3
Views
910
  • Last Post
Replies
11
Views
2K
Replies
33
Views
21K
Replies
17
Views
2K
Replies
1
Views
963
Top