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How can I prove that $\displaystyle{\lim_{(x, y) \to (0, 0)} \frac{x^3 - y^3}{x^2 + y^2}}=0$? The method I've been taught is the pinching one, where you compare the absolute value of the function to greater limits that are known to equal zero, but I haven't managed to find a comparison that works. Could someone point me in the right direction?

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2 Answers 2

up vote 7 down vote accepted

$$\left|\frac{x^3 - y^3}{x^2 + y^2}\right|\leq \left|\frac{x^3}{x^2 + y^2}\right|+\left|\frac{y^3}{x^2 + y^2}\right|\leq |x| + |y|.$$

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Hint: If $(x,y)$ is such that both $|x|\le\varepsilon$ and $|y|\le\varepsilon$, then your function $f$ is such that $|f(x,y)|\le2\varepsilon$.

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