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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?

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  • $\begingroup$ Can you link to the WA calculation, or provide the exact formula you typed in there? $\endgroup$ – Barry Cipra Dec 15 '18 at 13:53
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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

$$\frac{u+v}{u^2+v^2}=\frac1r(\cos \theta +\sin \theta)\to 0$$

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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.

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

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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 $0.$

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