# $f(x,y)=\arcsin \frac{x}{y}$ is continuous but not uniformly continuous in its domain

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 $$|\arcsin \frac{x}{y} - \arcsin \frac{x'}{y'}|<\epsilon$$.

• Have you tried $\sin(\arcsin(x)+\arcsin(y))$ and developing the $\sin$ and carry on your idea? – EDX May 20 at 22:36
• @EDX I don't understand how can I relate $\arcsin \frac{x}{y}$ with $\sin (\arcsin x+ \arcsin y)$? – Emo May 21 at 9:26
• I was meaning $\sin(\arcsin(u)+\arcsin(v))$ where you can choose $u=\dfrac{u}{v}$ and $v=\dfrac{u'}{v'}$. I could work. – EDX May 21 at 10:51
• 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? – Emo May 22 at 10:20
• Why can't you just use the fact that both arcsin and $x/y$ are continuous? – Sam May 22 at 22:52

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 $$\left|\left(\tfrac 1n, \tfrac 1n\right)- \left(\tfrac 1n, \tfrac 2n\right)\right|=\tfrac 1n$$.