Finding xn as n Tends to Infinity: Q1 & Q2

  • Thread starter Thread starter Mattofix
  • Start date Start date
  • Tags Tags
    Infinity
Click For Summary
SUMMARY

The discussion focuses on finding the limit of sequences as n approaches infinity, specifically addressing two questions involving the expressions (1/n)log(n^2) and sin(exp(n)). It is established that (1/n)log(n^2) simplifies to (2/n)log(n), which definitively tends to 0 as n increases. For the second question, it is confirmed that the limit of the sequence \sqrt{n}/(n + e^{-n}) converges to 0, leveraging the bounded nature of the sine function and the behavior of e^{-n}.

PREREQUISITES
  • Understanding of limits in calculus
  • Familiarity with L'Hôpital's rule
  • Knowledge of logarithmic functions and their properties
  • Basic concepts of oscillatory functions, particularly sine
NEXT STEPS
  • Study L'Hôpital's rule in depth for limit evaluation
  • Explore properties of logarithmic functions in calculus
  • Learn about the behavior of exponential functions as n approaches infinity
  • Investigate oscillatory behavior of trigonometric functions in limits
USEFUL FOR

Students studying calculus, particularly those tackling limits and sequences, as well as educators looking for examples of limit evaluation techniques.

Mattofix
Messages
137
Reaction score
0

Homework Statement



2 Questions, both find xn as n tends to infinity.

http://img229.imageshack.us/img229/5154/scan0002un5.th.jpg

Homework Equations


The Attempt at a Solution



Have attempted question one but am unsure if (1/n)log(n^2) tends to 0, and if it does do i need to prove it? I don't know how to do the second q, i know that sin(expn) oscillates between -1 and 1 and exp(-n) tends to 0 as n tends to infinity
 
Last edited by a moderator:
Physics news on Phys.org
Yes, (1/n) log(n^2) = (2/n)log(n) goes to 0. You might prove that by looking at 2ln(x)/x^2 and using L'Hopital's rule.

As for the second one, since sin is always between -1 and 1, you really just need to show that [itex]\sqrt{n}/(n+ e^{-n})< \sqrt{n}/n[/itex] (since [itex]e^{-n}[/itex] is always positive) converges to 0.
 
thanks :smile:
 

Similar threads

  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 5 ·
Replies
5
Views
1K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 13 ·
Replies
13
Views
3K
Replies
3
Views
2K
  • · Replies 24 ·
Replies
24
Views
2K
  • · Replies 10 ·
Replies
10
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 23 ·
Replies
23
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K