$\lim_{(x,y,z)\rightarrow(0,0,0)}((\frac{xyz}{x^2+y^2+z^2})^{(x+y)})$ I'm quite stumped. The answer to this question is it does not exist, but I've tried using $y=x$, $z=x$ and $y=x^2$ and $z=x^2$, or even $y=\sqrt{x}$ and $z=\sqrt{x}$, but they all give me the answer $1$. How can I find two diferent pathways to prove it does not exist?
 A: This may be wrong, but please comment kindly!
$$
\lim_{(x,y,z)\to(0,0,0)}\left(\frac{xyz}{x^2+y^2+z^2}\right)^{(x+y)} \tag{*}
$$
switch to spharical coordinates $x = r\sin\theta\cos\phi,\ y = r\sin\theta\sin\phi,\ z = r\cos\theta$
$$
\lim_{r\to0}\left(\frac{r^3\sin\theta\cos\phi\sin\theta\sin\phi\cos\theta}{r^2}\right)^{r\sin\theta(\cos\phi + \sin\phi)}  \\
= \lim_{r\to0}\left(\frac{r}{2}\sin^2\theta\cos\theta\sin2\phi\right)^{r\sin\theta(\cos\phi + \sin\phi)} \\
$$
take the logarithm of the expression (disregarding that the argument of the logarithm should be positive)
$$
r\sin\theta(\cos\phi + \sin\phi)\ln\left(\frac{r}{2}\sin^2\theta\cos\theta\sin2\phi\right)
$$
make the following abreviations
$$
A = \sin\theta(\cos\phi + \sin\phi),\qquad B = \frac{1}{2}\sin^2\theta\cos\theta\sin2\phi
$$
so we get
$$
\lim_{r\to0} Ar \ln(Br) = 0
$$
but!...this holds if (at least) $B \ne 0$, that is, for $(\theta,\phi) \notin \{0,\frac{\pi}{2},\pi\}\times\{0 ,\frac{\pi}{2},\pi,\frac{3\pi}{4}\}$. The last limit implies that the limit $(*)$ is $1$. What happens for $B = 0$ depends on what $0^0$ should mean.
A: First take limit along the line x=y=z and we get the limit to be 1. Next take limit along the y-z plane , that is when x=0. The limit is 0. Since along two different paths we get two different limits, we conclude that the limit doesn't exist.
