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I need to prove that $f(x,y)=\arcsin \frac{x}{y}$ is continuous, but not uniformly continuous on its domain. I noticed that the domain of the function is $D_f=\{(x,y)|-y\leq x \leq y$ if $y>0$, and $y\leq x \leq -y$ if $y<0\}.$ I started to prove the continuity by the epsilon-delta deffinition, but I'm stuck at proving that $|\arcsin \frac{x}{y} - \arcsin \frac{x'}{y'}|<\epsilon$.

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    $\begingroup$ Have you tried $\sin(\arcsin(x)+\arcsin(y))$ and developing the $\sin$ and carry on your idea? $\endgroup$ – EDX May 20 at 22:36
  • $\begingroup$ @EDX I don't understand how can I relate $\arcsin \frac{x}{y}$ with $\sin (\arcsin x+ \arcsin y)$? $\endgroup$ – Emo May 21 at 9:26
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    $\begingroup$ I was meaning $\sin(\arcsin(u)+\arcsin(v))$ where you can choose $u=\dfrac{u}{v}$ and $v=\dfrac{u'}{v'}$. I could work. $\endgroup$ – EDX May 21 at 10:51
  • $\begingroup$ On that case we'll have $\arcsin \frac{x}{y} - \arcsin \frac{x'}{y'} = \arcsin \left( \frac{x}{y} \sqrt{1-\frac{x'}{y'} - \frac{x'}{y'} \sqrt{1-\frac{x}{y}} \right)$. Now how should we prove that the RH in absolute values is small enough when points are sufficiently close to each other? $\endgroup$ – Emo May 22 at 10:20
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    $\begingroup$ Why can't you just use the fact that both arcsin and $x/y$ are continuous? $\endgroup$ – Sam May 22 at 22:52
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To show that the function $f$ is not unifromly continous first recall that $\operatorname{arcsin} 1 =\tfrac{\pi}2$ and $\operatorname{arcsin} \tfrac 12 =\tfrac{\pi}6$. Thus for each natural $n$ we have $f\left(\tfrac 1n, \tfrac 1n\right)= \tfrac{\pi}2$ and $f\left(\tfrac 1n, \tfrac 2n\right)= \tfrac{\pi}6$, whereas $\left|\left(\tfrac 1n, \tfrac 1n\right)- \left(\tfrac 1n, \tfrac 2n\right)\right|=\tfrac 1n$.

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