Using polar coordinates to determine the limit

In summary, the limit in question is not -1, but rather 0 as x and y approach 0. The function can be simplified by factoring out a common factor of r squared from the numerator and denominator, resulting in a denominator of 1 and a numerator of (Rcos^3(x)+rsin^3(x)). To prove that the limit is 0, one can choose a specific value for ε and then find a corresponding δ value that satisfies the delta-epsilon definition of a limit.
  • #1
rmiller70015
110
1
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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  • #2
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}##.
 
  • #3
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
 

1. What are polar coordinates?

Polar coordinates are a coordinate system commonly used in mathematics and physics to describe the position of a point in a two-dimensional space. They use a distance (radius) and an angle to define the location of a point, rather than using the traditional x and y coordinates.

2. How do polar coordinates work?

Polar coordinates use a reference point, typically the origin, and a reference line called the polar axis. The distance from the reference point to the point of interest is represented by the radius, while the angle formed between the polar axis and a line connecting the reference point to the point of interest is represented by the angle.

3. How are polar coordinates used to determine limits?

In mathematics, limits are used to describe the behavior of a function as the input approaches a certain value. To determine the limit using polar coordinates, we can convert the function from Cartesian coordinates to polar coordinates and then evaluate the limit by approaching the point of interest along various angles.

4. Can polar coordinates be used to determine limits for all functions?

Yes, polar coordinates can be used to determine limits for all continuous functions. However, for discontinuous functions, the limit may not exist or may be different depending on the approach angle.

5. What are the advantages of using polar coordinates to determine limits?

Polar coordinates can be useful in determining limits for functions that are difficult to evaluate using Cartesian coordinates. They also allow for a more intuitive understanding of the behavior of a function as the input approaches a certain value, as the radius and angle directly correspond to the distance and direction from the point of interest.

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