Calculus III - Multivariate Continuity

Jonmundsson
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Homework Statement


Let
\begin{equation*}<br /> f(x,y) = \begin{cases} \dfrac{x^3 - y^3}{x^2 + y^2}, \hspace{1.1em} (x, y) \neq (0,0) \\ 0, \hspace{4em} (x,y) = (0,0) \end{cases}<br /> \end{equation*}

Is f continuous at the point (0,0)? Are f_x og f_y continuous at the point (0,0)?

Homework Equations



Polar coords

The Attempt at a Solution



If you convert f to polar it's easy to see that it is continuous (since it doesn't depend on \theta. I'm just wondering if the derivative of the polar function is dependent on \theta then it isn't continuous and therefore neither are f_x and f_y

Steps:

\displaystyle \lim _{(x,y) \to (0,0)} \dfrac{x^3 - y^3}{x^2 + y^2} = \lim _{r \to 0} \hspace{0.3em} r (cos^3 \theta - sin^3 \theta) = 0

Define g(r) = r (cos^3 \theta - sin^3 \theta) then g&#039;(r) = cos^3 \theta - sin^3 \theta and \displaystyle \lim _{r \to 0} cos^3 \theta - sin^3 \theta doesn't exist.

Cheers.
 
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Jonmundsson said:

Homework Statement


Let
\begin{equation*}<br /> f(x,y) = \begin{cases} \dfrac{x^3 - y^3}{x^2 + y^2}, \hspace{1.1em} (x, y) \neq (0,0) \\ 0, \hspace{4em} (x,y) = (0,0) \end{cases}<br /> \end{equation*}

Is f continuous at the point (0,0)? Are f_x og f_y continuous at the point (0,0)?

Homework Equations



Polar coords

The Attempt at a Solution



If you convert f to polar it's easy to see that it is continuous (since it doesn't depend on \theta. I'm just wondering if the derivative of the polar function is dependent on \theta then it isn't continuous and therefore neither are f_x and f_y

Steps:

\displaystyle \lim _{(x,y) \to (0,0)} \dfrac{x^3 - y^3}{x^2 + y^2} = \lim _{r \to 0} \hspace{0.3em} r (cos^3 \theta - sin^3 \theta) = 0

Define g(r) = r (cos^3 \theta - sin^3 \theta) then g&#039;(r) = cos^3 \theta - sin^3 \theta and \displaystyle \lim _{r \to 0} cos^3 \theta - sin^3 \theta doesn't exist.

Cheers.
How about ∂f/∂x and ∂f/∂y? Are the continuous at (0, 0) ?
 
They aren't. I'm asking whether it is a coincident that the derivative of g and the partials of f aren't continuous.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...

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