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While doing revision I stumbled upon this problem:

Is this function continuous at the origin?

$$ f_3(x,y) = \begin{cases} \frac{x^3-y^3}{x^2+y^2}, & \text{if }(x,y)\not= (0,0) \\ 0, & \text{if } (x,y)=(0,0) \end{cases} $$

The answer is yes, but how do I prove it?

Sincere thanks for help.

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2 Answers 2

up vote 1 down vote accepted

For every $u=(x,y) \ne 0=(0,0)$ we have $$ |f_3(0)-f_3(u)|=|f_3(u)| \le \frac{|x|^3+|y|^3}{|u|_2^2} \le \frac{2|u|_2^3}{|u|_2^2}=2|u|_2. $$ For every $\epsilon>0$, if $|u|_2<\epsilon/2$ then $|f_3(0)-f_3(u)|<\epsilon$, i.e. $f_3(u) \to f_3(0)=0$ as $u \to 0$.

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This function is continuous at (0,0). Consider the function in polar form,put $x=rcos\theta$ and $y=rsin\theta$ in the given function, you will get $f(r,\theta) = r(cos\theta-sin\theta)(1+sin\theta.cos\theta)$. As $x \to 0$ and $y \to 0$, limits in polar coordinates becomes $r \to 0$ and no limit on $\theta$ , but as $r \to 0$, your function $f(r,\theta) = r(cos\theta-sin\theta)(1+sin\theta.cos\theta) \to 0$ whatsoever value $\theta$ takes.Therefore,$limit_{(x,y)\to(0,0)} \frac{x^3-y^3}{x^2+y^2} =0$ and the exact value at (0,0) is also $0$, hence the function is continuous at $(0,0)$.

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