Trying to do Delta-Epsilon proof of a multivariable inverse tri-function

Question

I was given $$\lim_{{(x, y) \to (0,0)}} \arctan(x - y)(x + y)$$ I noticed the limit is just zero. $$\lim_{{(x, y) \to (0,0)}} \arctan(x - y)(x + y) = L = 0$$ I set up the proof $$WTS:\forall \epsilon > 0, \exists \delta > 0 \space s.t. if 0<\sqrt{x^2+y^2}<\delta, then |\arctan(x - y)(x + y) - L| < \epsilon$$ So far I noticed a change in the coordinate system can be handy. $$x=r\cos{\theta} \space \space y=r\sin{\theta}$$ $$\therefore 0<\sqrt{x^2+y^2}<\delta \implies 0<\sqrt{r^2(\cos^2{\theta}+\sin^2{\theta})}<\delta \implies 0<|r|<\delta$$ and $$|\arctan(x - y)(x + y) - L| < \epsilon \implies |\arctan{r^2(\cos^2{\theta}-\sin^2{\theta})}|< \epsilon$$ becomes $$|\arctan{(r^2cos(2\theta))}| < \epsilon$$ But what should I do now? How do I find δ in terms of ϵ?(edit: fixed a tri identity mistake)

Answer 1

Take $\varepsilon>0$. Now, take $\delta>0$ such that $\delta<\min\left\{\frac\varepsilon2,\frac{\tan1}2\right\}$. Then, if $\sqrt{x^2+y^2}<\delta$, both $|x|$ and $|y|$ are smaller than $\frac\varepsilon2$, and therefore $|x+y|\leqslant|x|+|y|<\varepsilon$. On the other hand, $|x-y|<\tan1$, and therefore $\left|\arctan(x-y)\right|<1$. So, $\left|\arctan(x-y)(x+y)\right|<\varepsilon$.

Answer 2

Rather than the Euclidean metric, I would use the taxicab metric. If the proof is valid under the taxicab metric it will be valid under the euclidean metric.

$d((x,y),(0,0)) = |x| + |y|$

$|x+y| \le |x|+|y| = \delta$ and $|x-y| \le |x|+|y| =\delta$

For all $x,y$

$|\arctan (x-y)| < |x-y|$

$|(x+y) \arctan (x-y)| \le \delta^2$

Set $\delta < \sqrt \epsilon$ to complete the proof.