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?

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 $0.$
