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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.

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  • $\begingroup$ Do you have to use an $\epsilon$-$\delta$ proof for this? $\endgroup$ – JimmyK4542 Jan 19 '16 at 4:56
  • $\begingroup$ I tried to cook up a slick proof using the fact that $\arctan$ is a contractive map (i.e. $\arctan(x+\varepsilon)-\arctan x<\varepsilon$ for positive $\varepsilon$), but my efforts rapidly became unslick. $\endgroup$ – Lubin Jan 19 '16 at 5:55
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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!

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