Find limits for x to infinity with L'Hopital

Click For Summary
SUMMARY

The limit of the function f(x) = [xsinx]/[x^2 + 1] as x approaches infinity does not exist due to the oscillatory nature of sin(x). While applying L'Hôpital's rule initially seems plausible, it leads to complications since the numerator does not tend to infinity. Instead, the correct approach involves using the Squeeze Theorem, which allows for bounding the function by simpler expressions. Specifically, dividing both the numerator and denominator by x reveals that the limits of the bounding expressions converge to zero.

PREREQUISITES
  • Understanding of limits in calculus
  • Familiarity with L'Hôpital's Rule
  • Knowledge of the Squeeze Theorem
  • Basic trigonometric functions and their properties
NEXT STEPS
  • Study the Squeeze Theorem in detail and its applications in limit evaluation
  • Review L'Hôpital's Rule and its limitations in certain scenarios
  • Explore oscillatory functions and their behavior at infinity
  • Practice limit problems involving trigonometric functions and rational expressions
USEFUL FOR

Students studying calculus, particularly those focusing on limits and the application of the Squeeze Theorem, as well as educators seeking to clarify common misconceptions about L'Hôpital's Rule.

Fairy111
Messages
72
Reaction score
0

Homework Statement



For the following function decide whether f(x) tends to a limit as x tends to infinity. If the limit exists find it.

Homework Equations



f(x)=[xsinx]/[x^2 +1]

The Attempt at a Solution



I thought about using l'hospital's rule, so i got:

[sinx + xcosx]/[2x]

So, sinx will keep between 0 and 1, and the xcosx and the 2x would both tend to infinity maybe?

Im not really sure though.
 
Physics news on Phys.org


l'Hopital's rule is going to get you into trouble here. x^2+1 goes to infinity, but x*sin(x) does not. It's oscillates. So l'Hopital doesn't apply. You'll have to approach the limit in a more basic way. How about dividing numerator and denominator by x?
 


A different approach is to use the fact that for x>=0, -x <= xsinx <= x, which means that (again for x >= 0)
[tex]\frac{-x}{x^2 + 1} \leq \frac{x sin(x)}{x^2 + 1} \leq \frac{x}{x^2 + 1}[/tex]

The outermost expressions both have limits as x approaches infinity, and you can exploit this fact to say something about the expression in the middle. This idea is referred to in some texts as the "squeeze" theorem or "squeeze-play" theorem.
 

Similar threads

  • · Replies 11 ·
Replies
11
Views
3K
Replies
1
Views
1K
Replies
9
Views
2K
  • · Replies 12 ·
Replies
12
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
17
Views
3K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 5 ·
Replies
5
Views
1K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 23 ·
Replies
23
Views
2K