Does $\lim_{(x,y) \to (0,0)} \frac{x^4+y^4}{x^3+y^3}$ exist? 
Does the following limit exist?
  $$ \lim_{(x,y) \to (0,0)} \frac{x^4+y^4}{x^3+y^3}$$

I am not looking for any work, just a quick yes or no answer. I have already done the work on this problem and I just want to know if I my answer is consistent with a general consensus.
 A: No. It depends on the path.
To see why, pick the path $y = mx,$ and substitute. You will get 
$$
x\frac{1 + m^4}{1 + m^3}
 $$
which depends on the path.
A: If you let $x=r\cos\theta$ and $y=r\sin\theta$ (which is good for checking how the limit looks as you approach the origin from different directions), then the limit becomes
$$
\lim_{r\to0}\left(r\frac{\cos^4\theta+\sin^4\theta}{\cos^3\theta+\sin^3\theta}
\right)=\left\{
\matrix{
\pm\infty&\quad&\theta\to\left(k-\frac14\right)\pi\text{ for }k\in\mathbb{Z}\\\\
\text{undefined}&\quad&\theta=\left(k-\frac14\right)\pi\text{ for }k\in\mathbb{Z}\\\\
0&\quad&\text{otherwise}
}
\right.
$$
and since this depends on $\theta$, the limit does not exist. This is because, on the unit circle, the trigonometric ratio is well-defined except on the two points where $\cos\theta=-\sin\theta$, and at these points, the denominator vanishes while the numerator is positive, producing a ratio that blows up for all $r\ne0$ (and, in our limit, $r$ is never $0$). For other values of $\theta$, the ratio, which can also be represented as $\frac{1+\tan^4\theta}{1+\tan^3\theta}$, is a finite number, so that multiplying it by $r$ scales the result down to $0$ as $r\to0$.
A: Putting $f(x,y)= \frac{x^4+y^4}{x^3+y^3}.$
Using a special curve, $y=\sqrt[3]{k x^4-x^3}$, we obtain
$$\lim_{x\to 0} f\left(x, \sqrt[3]{k x^4-x^3}\right)=\frac{2}{k}$$
depends on $k$. Therefore, the limit does not exist. 
A: This one is easier if you don't worry about precise paths.  Obviously, you have many easy paths (such as $x=y$) where the pathwise limit is $0$.  On the other hand, on the circle of radius $\epsilon$ about $0$, you obviously have the numerator larger than $\epsilon^4/16$ and you can make the denominator as small as you like by taking $x$ very close to $-y$, so the function is unbounded on any disk around $(0,0)$, no matter how small, so the limit cannot exist.  It's much easier to see this, which is all that is needed, than to find precise paths where the pathwise limit is something other than $0$ or does not exist.
