Finding the Limit of a Multi-Variable Function

  • Context: Undergrad 
  • Thread starter Thread starter hazellaw
  • Start date Start date
  • Tags Tags
    Limit Multi-variable
Click For Summary
SUMMARY

The limit of the multi-variable function lim(x,y)->(0,0) x²sin²y/(x²+2y²) approaches zero as (x,y) approaches (0,0). To confirm this, one can utilize polar coordinates or evaluate the limit along different paths to demonstrate convergence. The discussion emphasizes the importance of testing multiple approaches to determine the existence of the limit, particularly when initial attempts yield zero. The expression x²sin²y/(x²+y²) is suggested as a simpler variant for adaptation.

PREREQUISITES
  • Understanding of multi-variable calculus
  • Familiarity with limits in calculus
  • Knowledge of polar coordinates
  • Experience with evaluating limits along different paths
NEXT STEPS
  • Study the application of polar coordinates in limit evaluation
  • Learn techniques for proving limits do not exist through path analysis
  • Explore the behavior of functions near singular points
  • Investigate the limit of x²sin²y/(x²+y²) for further insights
USEFUL FOR

Students and educators in calculus, mathematicians focusing on multi-variable functions, and anyone interested in advanced limit evaluation techniques.

hazellaw
Messages
2
Reaction score
0
i need some help with this question

Find the limit, if it exists, or show that the limit does not exist

lim(x,y)->(0,0) x2sin2y/(x2+2y2)

i've tried to x=y x=0 or x=y2 but i still got 0...
 
Physics news on Phys.org
With these kind of exercises, there are two alternatives:

It could be that the limit does not exist, and then you can prove it by finding two different paths to the origo, such that the limit is different along them.

Or then, it could be that the limit does exist,and then a one good idea is to use polar coordinates, or in some other way obtain some bounds (upper or lower, whatever you need) as a function of [itex]r[/itex] (distance from origo), and then prove that the bounds converge when [itex]r\to 0[/itex].

If you don't know in advance what the correct solution is, then you must try both to see what works.

I believe this expression approaches zero when [itex](x,y)\to 0[/itex], but it could be I made mistake.
 
Hi hazellaw! :smile:

Can you see an easy way of doing it for x2sin2y/(x2 + y2) ?

Then adapt that. :wink:
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
11
Views
2K
  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 3 ·
Replies
3
Views
5K
  • · Replies 1 ·
Replies
1
Views
3K