Limit $\lim_{(x, y) \to (\infty, \infty)} \frac{x+\sqrt{y}}{x^2+y}$ Show whether the limit exists and find it, or prove that it does not.
$$\lim_{(x, y) \to(\infty,\infty)}\frac{x+\sqrt{y}}{x^2+y}$$
WolframAlpha shows that limit does not exist, however, I do fail to conclude so. 
$$\lim_{(x, y) \to(\infty,\infty)}\frac{x+\sqrt{y}}{x^2+y} = [x=r\cos\theta, y = r\sin\theta] = \lim_{r\to\infty}\frac{r\cos\theta+\sqrt{r\sin\theta}}{r^2\cos^2\theta+r\sin\theta} = \lim_{r\to\infty}\frac{\cos\theta\frac{\sqrt{\sin\theta}}{\sqrt{r}}}{r\cos^2\theta+\sin\theta} = 0.$$
Having gotten the exact results for whatever the substitution is made (such as $y = x, y = x^2, [x = t^2, y = t])$, my conclusion is that limit does exist and equals $0.$ 
Did I miss something?
 A: For $x,y\gt0$, we have
$$\begin{align}
{x+\sqrt y\over x^2+y}
&={x\over x^2+y}+{\sqrt y\over x^2+y}\\
&\le{x\over x^2}+{\sqrt y\over y}\\
&={1\over x}+{1\over\sqrt y}\\
&\to0+0
\end{align}$$
The key step here uses the fact that a smaller denominator makes for a larger fraction.
A: It is enough to observe that, if $y\geq 0$,
$$
x^2 + y \geq \frac{1}{2} (|x| +\sqrt{y})^2,
$$
so that
$$
\left|\frac{x+\sqrt{y}}{x^2+y}\right|
\leq\frac{|x|+\sqrt{y}}{x^2+y}\leq \frac{2}{|x|+\sqrt{y}}.
$$
A: By $x=u$ and $y=v^2$ the limit becomes 
$$\lim_{(x, y) \to (\infty, \infty)} \frac{x+\sqrt{y}}{x^2+y}=\lim_{(u,v) \to (\infty, \infty)} \frac{u+v}{u^2+v^2}=0$$
indeed for example by polar coordinates 
$$\frac{u+v}{u^2+v^2}=\frac1r(\cos \theta +\sin \theta)\to 0$$
A: Since $(x,y)\rightarrow (\infty,\infty),$ assume $x,y>0.$ $$0<\frac{x+\sqrt{y}}{x^2+y}=\frac{x\sqrt {y}\left(\frac{1}{\sqrt y}+\frac 1x\right)}{x\sqrt {y}\left(\frac{x}{\sqrt y}+\frac {\sqrt y}{x}\right)}\leq 
\frac{\frac{1}{\sqrt y}+\frac 1x}{2} \rightarrow 0$$
as $\frac ab + \frac ba \geq 2$ for $a,b>0.$ Thus by the squeeze theorem is our limit $0.$
