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?
Best Answer
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.