What is $\lim_{(x,y)\to(0,0)} \frac{(x^3+y^3)}{(x^2-y^2)}$?

Solution 1:

If you define $$\lim_{\langle x,y\rangle\to\langle a,b\rangle}f(x,y)\tag{1}$$ in such a way that it exists only when the function is defined in some open ball centred at $\langle a,b\rangle$, then what you wrote is correct. This is analogous to defining $$\lim_{x\to a}f(x)$$ only when $f(x)$ is defined in some open interval centred at $a$. However, just as we can talk about one-sided limits on the real line, it makes perfectly good sense to talk about $(1)$ whenever $f(x,y)$ is defined at points arbitrarily close to $\langle a,b\rangle$. In that case it’s understood that we look only at the limit along ‘paths’ within the domain of $f$. On that understanding $$\lim_{\langle x,y\rangle\to\langle 0,0\rangle}\frac{x^3+y^3}{x^2-y^2}$$ still does not exist, but for a more fundamental reason.

Suppose that you approach the origin along the curve $y=\sin x$. Then by l’Hospital’s rule you have

$$\begin{align*} \lim_{\langle x,y\rangle\to\langle 0,0\rangle}\frac{x^3+\sin^3x}{x^2-\sin^2x}&=\lim_{\langle x,y\rangle\to\langle 0,0\rangle}\frac{3x^2+3\sin^2x\cos x}{2x-2\sin x\cos x}\\\\ &=\lim_{\langle x,y\rangle\to\langle 0,0\rangle}\frac{3x^2+\frac32\sin2x\cos x}{2x-\sin2x}\\\\ &=\lim_{\langle x,y\rangle\to\langle 0,0\rangle}\frac{6x-\frac32\sin2x\sin x+3\cos2x\cos x}{2-2\cos2x} \end{align*}$$

which does not exist: the numerator approaches $3$ and the denominator, $0$. The problem is that this path, although it stays within the domain of the function, approaches the line $y=x$ so quickly as it approaches the origin that the denominator approaches $0$ much faster than the numerator, and therefore the function blows up as we approach the origin along this path.

Solution 2:

Brian's answer is nice, so it got me thinking, what about approaching the point along other functions, $y=f(x)$, with $f(0)=0$. Let us plug that into the expression to get $\frac{x^3+f^3}{x^2-f^2}$. This will give $$\lim_{x \to 0} \frac{x^2-xf+f^2}{x-f}\;.$$

We will now perform a l’Hospital’s rule to get $$\lim_{x \to 0} \frac{2x-xf'-f+2ff'}{1-f'}\;.$$ If at this point we assume that $f'(0)=1$, we have another $0/0$. If $f'\neq 1$, we get that the limit is zero. So letus now hit this with another l’Hospital. This will give us $$\lim_{x \to 0} \frac{2-2f'-xf''+2f''f'+2(f')^2}{-f''}\;. \qquad (*)$$

When evaluate at $x=0, f'=1$, we get $$\lim_{x \to 0} -\frac{2f''+2}{f''}\;.$$ And you can decide what you want the limit to be by picking a value for $f''$.

Solution 3:

It doesn't matter that along the lines $y = \pm x$, the function is undefined ; you can still stay inside the domain of $f(x,y) = \frac{x^3 + y^3}{x^2-y^2}$ and try to compute the limit, under the constraint that $y \neq \pm x$. The limit could still be defined at $(0,0)$.

But there are cases where your trick doesn't work : think of $\frac{x^4-y^4}{x^2-y^2} = x^2 + y^2$, which is continuous at $(0,0)$.

Hope that helps,

Solution 4:

For your function, in the domain of $f$ (so $x\ne \pm y)$, to compute the limit you can set $x=r\cos\theta, y=r\sin\theta$, and plug it it. You get $\lim\limits_{r\to 0} \frac{r^3(cos^3\theta+sin^3\theta)}{r^2(cos^2\theta-sin^2\theta)} =\lim\limits_{r\to 0} \frac{r(cos^3\theta+sin^3\theta)}{(cos^2\theta-sin^2\theta)}$, and you can easily see that this is $0$ for any $\theta$ in the domain of $f$ (you need to avoid $\theta = \frac{n\pi}{2}-\pi/4$). Of course, if you are considering the whole plane, then the limit does not exist, because the function isn't even defined at $y=x$, so you can't compute the limit along that path.