Using polar coordinates to determine the limit

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SUMMARY

The limit of the function (x^3 + y^3)/(x^2 + y^2) as (x, y) approaches (0, 0) is 0, not -1. The discussion illustrates the use of polar coordinates, specifically substituting x = r cos(θ) and y = r sin(θ), leading to the expression r(cos^3(θ) + sin^3(θ)). As r approaches 0, this expression converges to 0 for any angle θ, confirming that the limit does not approach -1.

PREREQUISITES
  • Understanding of limits in multivariable calculus
  • Familiarity with polar coordinates and their application
  • Knowledge of trigonometric identities, specifically sin²(θ) + cos²(θ) = 1
  • Basic algebraic manipulation and factoring techniques
NEXT STEPS
  • Study the concept of limits in multivariable calculus
  • Learn about the application of polar coordinates in evaluating limits
  • Explore trigonometric identities and their implications in calculus
  • Practice problems involving limits approaching multiple variables
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Students and educators in calculus, mathematicians focusing on multivariable analysis, and anyone interested in understanding the behavior of functions near critical points.

rmiller70015
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Lim (x, y)->(0,0)(X^3+y^3)/(x^2+y^2)

The answer is -1, but I can't get it there. Here is what I did.
((Rcosx)^3 +(rsinx)^3)/((rcosx)^2+(rsinx)^2)
Then by factoring out a r squared from top and bottom I'm left with a denominator of (sin^2(x ) + cos^2 (x)) which simplifies to 1. And a numerator of:
Rcos^3(x)+rsin^3 (x)

This is about as far as I've gotten and I can't figure out how to get it to a -1.
 
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rmiller70015 said:
Lim (x, y)->(0,0)(X^3+y^3)/(x^2+y^2)

The answer is -1, but I can't get it there. Here is what I did.
((Rcosx)^3 +(rsinx)^3)/((rcosx)^2+(rsinx)^2)
Then by factoring out a r squared from top and bottom I'm left with a denominator of (sin^2(x ) + cos^2 (x)) which simplifies to 1. And a numerator of:
Rcos^3(x)+rsin^3 (x)

This is about as far as I've gotten and I can't figure out how to get it to a -1.

The answer is certainly not -1. The answer is 0 as ##r → 0##.

Conversely, if ##\epsilon > 0## is given, then choose ##\delta = \frac{\epsilon}{2}##.
 
The limit is not -1. If it were, the function should get closer and closer to it, when (x,y) gets closer and closer to (0,0). If you choose x=y, f(x,x)=x →0. If you choose x=rcosθ, y=rsinθ, f=r(cos3θ+sin3θ) which goes to 0 if r→0, for any value of θ.

ehild
 

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