How to Prove $\lim_{(x, y) \rightarrow (0, 0)}\frac{x^3-y^3}{x^2+y^2}=0$

Question

I wanted to prove how $$ \lim_{(x, y) \to (0, 0)} \frac{x^3-y^3}{x^2+y^2} = 0. $$ Specifically, I want to use the Squeeze Theorem for multivariable calculus. Then I know that I should pick an arbitrary function $g(x, y)$ where $|f(x, y)-L| \leq g(x, y)$ and the limit of $g(x, y)$ is $0$. Since the limit of $f(x, y) =0$, I just have to make $$ \Biggl\lvert \frac{x^3-y^3}{x^2+y^2} \Biggr\rvert \leq g(x, y), $$ where the limit of $g(x,y)=0$. I was given a hint that $|x^3-y^3| \leq |x|^3+|y|^3$. So I divided both sides by $|x^2+y^2|$, $$ \frac{\bigl\lvert x^3-y^3 \bigr\rvert}{\bigl\lvert x^2+y^2 \bigr\rvert} \leq \frac{\lvert x\rvert^3+ \lvert y\rvert^3}{\bigl\lvert x^2+y^2 \bigr\rvert}, $$ meaning that $$ \frac{\bigl\lvert x^3-y^3 \bigr\rvert}{\bigl\lvert x^2+y^2 \bigr\rvert} \leq \lvert x\rvert + \lvert y\rvert. $$ What do I do next?

Answer 1

Here it is a similar way to approach it:

\begin{align*} 0\leq x^{2} \leq x^{2} + y^{2} & \Rightarrow 0\leq \frac{x^{2}}{x^{2} + y^{2}}\leq 1\\\\ & \Rightarrow \left|\frac{x^{2}}{x^{2} + y^{2}}\right| \leq 1\\\\ & \Rightarrow \left|\frac{x^{3}}{x^{2} + y^{2}}\right|\leq |x| \end{align*}

Then you can apply the squeeze theorem as $(x,y)\to(0,0)$.

Similarly, we do also have that \begin{align*} \left|\frac{y^{3}}{x^{2} + y^{2}}\right| \leq |y| \end{align*} where we can apply the squeeze theorem as well.

Along with the triangle inequality, the desired result holds.

Answer 2

Now, $$ 0\leqq\left|\frac{x^3-y^3 }{x^2+y^2} \right| \leq \lvert x\rvert + \lvert y\rvert$$

Letting $(x,y)\to(0,0)$, you get $$ \lim_{(x,y)\to(0,0)} \ \ \ 0\leq\lim_{(x,y)\to(0,0)}\ \ \left|\frac{x^3-y^3 }{x^2+y^2} \right| \leq \lim_{(x,y)\to(0,0)}\lvert x\rvert + \lvert y\rvert$$

The LHS and RHS are $0$ so due to the squeeze theorem, you get $$\lim_{(x,y)\to(0,0)}\ \ \left|\frac{x^3-y^3 }{x^2+y^2} \right| =0.$$

This means $$\lim_{(x,y)\to(0,0)}\ \ \ \frac{x^3-y^3 }{x^2+y^2} =0.$$

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But I think that you should show why $\frac{\bigl\lvert x^3-y^3 \bigr\rvert}{\bigl\lvert x^2+y^2 \bigr\rvert} \leq \frac{\lvert x\rvert^3+ \lvert y\rvert^3}{\bigl\lvert x^2+y^2 \bigr\rvert} $ means $ \frac{\bigl\lvert x^3-y^3 \bigr\rvert}{\bigl\lvert x^2+y^2 \bigr\rvert} \leq \lvert x\rvert + \lvert y\rvert. $