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How would you prove or disprove that the function given by $f(x,y) = \begin{cases} \frac{x^3y^2}{x^4 + y^4} & (x,y) \neq (0,0) \\ 0 & (x,y) = (0,0) \end{cases}$ is continuous at $(0,0$). I tried to think of a function where the limit approached zero over which tended to an answer apart from zero, but they all went to zero! So this makes me think that the function is continuous at $(0,0)$. But how can I prove this? Thanks!

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We have, since $(x^2-y^2)^2\geq 0$, that $2x^2y^2\leq x^4+y^4$ hence $0\leq |f(x,y)|\leq \frac 12|x|$.

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    @DonAntonio The previous inequality was true, now it's more accurate.2012-06-02
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You could take advantage of the fact that numerator and denominator are homogeneous polynomials to write the expression, as far as possible as a function of $y/x$ (or $x/y$): Divide both numerator and denominator by $x^4$ to get $\frac{z}{1+z^4}\cdot y,\qquad z=\frac{y}{x}.$ Now use ordinary single variable calculus to show that the fraction is bounded, so the whole expression is bounded by a constant times $\lvert y\rvert$. (For $x=0$, the original expression is $0$, so no problem there.)

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    I offer this solution not because it is an improvement on Davide's solution – it is not – but because the method can be easily used on a number of similar cases.2012-06-02