Find the limit $\lim_{(x,y,z)\to(0,0,0)}\frac{xy+xz+yz}{x^2+y^2+z^2}$ Find the limit if it exists
$$\lim_{(x,y,z)\to(0,0,0)}\frac{xy+xz+yz}{x^2+y^2+z^2}$$
 A: Hint: Consider the limit along the paths $t \mapsto (t,t,t)$ and $t\mapsto (t,0,0)$.
A: It is an overkill for sure, but since the quadratic form associated with $xy+xz+yz$ has zero trace, there is a positive eigenvalue and a negative eigenvalue too, so the ratio is positive along a direction and negative along a different direction, and the limit cannot exist.
A: Once consider x=t and y=t and z=t and then x=t and y=t and z=2t.
In first case limit is 1 and in second case 5/6.So limit does not exist.
A: Let's see the limit in the plane $z=0$. So,
\begin{align*}
     \lim_{(x, y, z)\rightarrow (0,0,0)} \frac{xy + yz + xz}{x^2 + y^2 + z^2} &= \lim_{(x, y) \rightarrow (0,0)} \frac{xy} {x^2 + y^2}
\end{align*}
Now analyzing by paths, we have to take the path $y=mx$, with $m>0$.
\begin{align*}
\require{cancel}
     \lim_{(x, y)\rightarrow (0,0)} \frac{xy}{x^2 + y^2} &= \lim_{x\rightarrow 0} \frac{mx^2}{x^2 + m^2x^2}\\
     & = \lim_{x\rightarrow 0} \frac{mx^2}{x^2(1 + m^2)}\\
     & = \lim_{x\rightarrow 0} \frac{m\cancel{x^2}}{\cancel{x^2}(1 + m^2)}\\
     & = \frac{m}{1 + m^2}.
\end{align*}
The above tells us that the limit is not unique, because it varies according to the value that $m$ takes. Therefore the limit $\displaystyle \lim_{(x, y, z) \rightarrow (0,0,0)}\frac{xy + yz + xz}{x^2 + y^2 + z^2}$ does not exist.
A: The $x^2+y^2+z^2$ in the denominator is a dead giveaway to switch to spherical coordinates. The $r$'s will cancel leaving you with a trigonometric expression you can use identities to simplify. 

 You will find you won't be able to get rid of the dependency on the polar and azimuth angles, which means that the limit doesn't exist, as its value depends on the trajectory you approach the origin on.

