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?

It is enough to observe that, if $y\geq 0$,
$$
x^2 + y \geq \frac{1}{2} (|x| +\sqrt{y})^2,
$$
so that
$$
\left|\frac{x+\sqrt{y}}{x^2+y}\right|
\leq\frac{|x|+\sqrt{y}}{x^2+y}\leq \frac{2}{|x|+\sqrt{y}}.
$$
