$\epsilon - \delta$ proof of limit of $\arctan \left(\frac {x+z}{y}\right)$ as $(x, y, z) \to (1, 2, -3)$ Find the limit and prove the limit for 
$$\lim_{(x,y,z)\rightarrow(1,2,-3)}\arctan \left(\frac{x+z}{y}\right).$$
This is a homework problem. While I am comfortable with general epsilon-delta proof techniques, I am at a loss for inverse trigonometric inequalities. I was trying to use the trigonometric identity that 
$$\arctan(\alpha)+\arctan(\beta)= \arctan\left(\frac{\alpha +\beta}{1-\alpha \beta}\right),$$
but trying to select the correct $\alpha$ and $\beta$, the argeument of the arctan became ugly. I could use some suggestions on arctan inequalities and identities.

We have
$$\begin{align}
\left|\arctan\left(\frac{x+z}{y}\right)-\arctan\left(-1\right) \right|&=\left|\arctan\left(\frac{x+y+z}{x-y+z}\right)\right|\\\\
&=\left|\arctan\left(\frac{(x-1)+(y-2)+(z+3)}{x-y+z}\right)\right|\\\\
&\le\left|\frac{(x-1)+(y-2)+(z+3)}{x-y+z}\right|\\\\
&\le \frac{3\sqrt{(x-1)^2+(y-2)^2+(z+3)^2}}{|x-y+z|} \tag1
\end{align}$$
First we choose $\delta =1$ so that if $\sqrt{(x-1)^2+(y-2)^2+(z+3)^2}<\delta =1$, then $0<x<2$, $-3<-y<-1$, and $-4<z<-2$.  
Therefore, with $\delta=1$, we see that $\frac{1}{|x-y+z|}<1$.  Then, from $(1)$ we have
$$\begin{align}
\left|\arctan\left(\frac{x+z}{y}\right)-\arctan\left(-1\right) \right|&\le 
 3\sqrt{(x-1)^2+(y-2)^2+(z+3)^2} \\\\
&<\epsilon
\end{align}$$
whenever $\sqrt{(x-1)^2+(y-2)^2+(z+3)^2}<\min \left(1,\epsilon/3\right)$.  And we are done!
