Find $\lim\limits_{(x,y)\to(0,0)}\frac{x^6+y^6}{x^3+y^3}$ using the $\epsilon-\delta$ definition?

Question

My textbook asks the question

$$f(x,y) = \frac{x^6+y^6}{x^3+y^3}$$ Does $f(x,y)$ have a limit as $(x,y) \rightarrow (0,0)$?

I used polar coordinates instead of solving explicitly in $\mathbb R^2 $, and it went as the following:

$$ x = r \cos \theta, \qquad y = r\sin\theta $$

Hence,

$$\lim_{(x,y) \to (0,0)} \frac{x^6+y^6}{x^3+y^3} = \lim_{r \to 0}\frac{{r^6\cos^6\theta + r^6\sin^6\theta}}{r^3\cos^3\theta + r^3\sin^3\theta}$$

This simplifies to,

$$ \lim_{r \to 0} \frac{r^3({\cos^6\theta + \sin^6\theta})}{\cos^3\theta + \sin^3\theta}$$

Now from the above, we find that as r→$0$ the limit is $0$.

So now i have got $0$ as a possible limit.I have tried to procceed further using $\epsilon-\delta$ definition but I cannot get anywhere.

Any thoughts on how to prove that $0$ is in fact the limit?

Answer 1

Top tip: don't trust wolfram! It gets the answer wrong sometimes!

Suppose we set $y = x$ and take $x \to 0$. Then \begin{align*} \lim_{(x,y(x)) \to (0,0)}\frac{x^6+y^6}{x^3+y^3} &= \lim_{x \to 0}\frac{2x^6}{2x^3}\\ &= 0 \end{align*} Now let's suppose we take $y = -x$ and take $x \to 0$. Then \begin{align*} \lim_{(x,y(x)) \to (0,0)}\frac{x^6+y^6}{x^3+y^3} &= \lim_{x \to 0}\frac{2x^6}{x^3 - x^3}\\ &= \lim_{x \to 0}\frac{2x^6}{0} \end{align*} which is undefined. Since we have two approaches of $(x,y) \to (0,0)$ that give different results, the overall limit must not exist.