Find the Limit of Multivariable Function

Click For Summary
SUMMARY

The limit of the multivariable function lim (x,y) ->(0,0) (x^2*y)/(x^2 + y^4) approaches 0 as demonstrated through polar coordinates. By substituting x = rcosθ and y = rsinθ, the expression simplifies to L = r (cos^2θ * sinθ) / (cos^2θ + r^2sin^4θ). Multiple paths confirm that the limit consistently evaluates to 0, although proving the limit exists requires further analysis beyond the current methods discussed.

PREREQUISITES
  • Understanding of multivariable calculus concepts, specifically limits.
  • Familiarity with polar coordinate transformations.
  • Knowledge of trigonometric identities, particularly sin²θ + cos²θ.
  • Experience with limit evaluation techniques in calculus.
NEXT STEPS
  • Study the epsilon-delta definition of limits in multivariable calculus.
  • Learn about the Squeeze Theorem and its application in proving limits.
  • Explore alternative paths for limit evaluation in polar coordinates.
  • Investigate the behavior of functions near singular points in multivariable contexts.
USEFUL FOR

Students and educators in calculus, particularly those focusing on multivariable functions and limit evaluation techniques. This discussion is beneficial for anyone seeking to deepen their understanding of limits in a multivariable context.

Rapier
Messages
83
Reaction score
0

Homework Statement


Find the limit of:
lim (x,y) ->(0,0) (x^2*y)/(x^2 + y^4)

Homework Equations


x=rcosθ
y=rsinθ

The Attempt at a Solution


lim r->0 for all steps
L = (rcosθ)^2*(rsinθ)/[(rcosθ)^2 + (rsinθ)^4]
L = r^3 (cosθ)^2 (sinθ) / [ r^2 * (cosθ)^2 + r^4 * (sinθ)^4]
L = r (cos^2 * sin ) / (cos^2 + r^2*sin^4)

That's as far as I can get. I thought about trying to use r^2 in the denominator to work back around to a sin^2 + cos^2 or trying to convert the terms in the denominators into r and either cos or sin so I could get rid of one of the terms and nothing seems to be working.

Oh, I should mention that I tried a couple of different paths and my limit seemed to always equal 0. I know that I cannot prove a limit exists with this method, I can only prove that the limit does not exist. I'm fairly certain the limit = 0, but I can't figure out a way to determine it.

This is a problem that I'm supposed to be thinking about for discussion on Thursday. So I don't want the answer, but guidance. Thanks!

-r
 
Physics news on Phys.org
try using the fact x^2+y^4>=x^2
 

Similar threads

Limits in multivariable calculus
  • · Replies 6 ·
Replies
6
Views
2K
##\lim_{x \to0} \left(\dfrac{1}{\sin^2 x}-\dfrac{1}{x^2}\right)##
  • · Replies 5 ·
Replies
5
Views
2K
Multivariable Limit Problem: Find Values of k That Make Limit Exist
  • · Replies 9 ·
Replies
9
Views
2K
Find limit of multi variable function
  • · Replies 10 ·
Replies
10
Views
2K
Setting the limits of an integral
  • · Replies 3 ·
Replies
3
Views
2K
Deriving the formula for arc length of a polar function
  • · Replies 4 ·
Replies
4
Views
2K
Can this function still be a constant function?
  • · Replies 11 ·
Replies
11
Views
2K
How should I use the averaging approximation to find this?
  • · Replies 3 ·
Replies
3
Views
2K
Uniform Continuity of functions
  • · Replies 40 ·
2
Replies
40
Views
5K
Center of mass of spherical shell inside of cone (Apostol Problem).
  • · Replies 3 ·
Replies
3
Views
2K