# 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?

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

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

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

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.

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

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 $$0.$$