Question: Is it a logically correct thing to use spherical coordinates to evaluate limits? For example, consider $\lim \limits \_{(x,y,z)\rightarrow (0,0,0)}\frac{yz}{x^2+4y^2+9yz}$

**Approach 1:** take $y=mx$ and $z=nx$, this gives

$\lim \limits \_{(x)\rightarrow (0)}\frac{mnx^2}{x^2+4m^2x^2+9nmx^2}=\frac{mn}{1+4m^2+9mn}$ So the limit does not exist

**Approach 2:** $x=\rho\sin \phi \cos \theta$, $y=\rho \sin \phi \sin \theta, z= \rho\cos \phi$

$\lim \limits \_{(x,y,z)\rightarrow (0,0,0)}\frac{yz}{x^2+4y^2+9yz}=\lim \limits \_{\rho \rightarrow 0}\frac{\rho ^2\sin \phi \sin \theta \cos \phi}{\rho ^2\sin ^2 \phi \cos ^2 \theta +4\rho ^2\sin ^2\phi \sin ^2 \theta +9\rho ^2\sin \phi \sin \theta \cos \phi }=\frac{\sin \theta \cos \phi}{\sin \phi \cos ^2 \theta +4\sin \phi \sin ^2 \theta +9 \sin \theta \cos \phi }$
so the limit does not exist.

I have doubts about the second approach, personally, I have not seen it being used before, is it valid or is this just a coincidence that I get the same result? i have only seen spherical coordinates in triple integration

Answer: 
Both approaches rely on computing the limit along particular paths leading to $(0,0,0)$, and they are both correct.
Your second approach requires some further explanation but can be made to work. Recall that a function $f:\mathbb{R}^3\rightarrow \mathbb{R}$ has a limit $L$ at a point $p\in\mathbb{R}^3$ if and only if for any sequence of points $\{ p\_n \}\_{n=1}^\infty \subseteq \mathbb{R}^3$ you have $\underset{n\rightarrow \infty}{\lim}f(p\_n)=L$. Translating this to your spherical expression, you can find a sequence of points $p\_n$ with spherical coordinates $(\frac{1}{n},\theta\_n,\phi\_n)$ such that $f(p\_n)$ diverges. For example you need to have two different partial limits.

One more thing that you should recall (or know), is that a sequence of points $\{p\_n \}$ converge to $0\in \mathbb{R}^3$ if and only if $\Vert p\_n\Vert\rightarrow0$, where $\Vert \cdot \Vert$ is the euclidean norm.