$$\lim_{x,y\to\infty} \frac{x-y}{x^2+y^2}\tag{$\star$}$$

I'm used to do the following substitution when I see $``x^2+y^2"$ and that $x,y\to 0$

$$x^2+y^2 = r^2,\;x=r\cos\theta,\;y=r\sin\theta$$

plug these values in the function and compute the limit as $r\to0$ I know I can do that because the only way for $x$ & $y$ to approach $0$ is $r$ approaching $0.$

I cannot do this substitution everytime because if for example: $(x,y)\to(-1,7)$ there's no value $u$ that guarantee me if $r\to u$ then $(x,y)\to(-1,7).$

but here since $x,y\to\infty$ I think that logically this phenomenon can only happen if $r\to\infty$ as well.

So computing $(\star)$ is the same as computing this :

$$ \lim_{r\to\infty} \frac{r\cos\theta-r\sin\theta}{r^2} =\lim_{r\to\infty} \frac{\cos\theta-\sin\theta}{r}=0. $$

I'm 90% sure that what I've done is correct but I still want a confirmation and if possible show me other ways to compute this limit.

Sorry if this question sounds kinda dumb but I'm still new to multivariable calculus and today is my first time dealing with MVC limits.

Thank you !


2 Answers 2


Others can add and/or correct, but I'm not sure if you can do the polar trick. If $x$ and $y$ tend to infinity, then clearly $r \to \infty$. But if you have $r \to \infty$, then you don't necessarily have $x$ and $y$ to infinity since, for example: $x \to \infty$ and $y \to c$ (constant) will also lead to $r \to \infty$.

and if possible show me other ways to compute this limit.

Rewrite: $$\left| \frac{x-y}{x^2+y^2} \right| =\left| \frac{x}{x^2+y^2}- \frac{y}{x^2+y^2} \right| \le \left| \frac{x}{x^2+y^2}\right|+ \left|\frac{y}{x^2+y^2} \right|$$ Now: $$\left| \frac{x}{x^2+y^2} \right| \le \left| \frac{x}{x^2} \right| = \left| \frac{1}{x} \right| \to 0 \quad\mbox{ and }\quad \left| \frac{y}{x^2+y^2} \right| \le \left| \frac{y}{y^2} \right| = \left| \frac{1}{y} \right| \to 0$$ Alternatively, if you feel more comfortable with limits to $(0,0)$, substitute $\left( x,y \right) \to \left( \tfrac{1}{u},\tfrac{1}{v} \right)$ and take the limit $(u,v) \to (0^+,0^+)$ and you could follow up with your classical polar substitution.

With no other answers so far, I'll add this example: consider $f(x,y)=x+y$, then clearly: $$\lim_{(x,y)\to (+\infty,+\infty)} \bigl( x+y \bigr) = +\infty$$ However, switching to polar coordinates, we get: $$f(r,\theta) = r \left( \cos\theta + \sin\theta \right)$$ Now simply taking $r \to +\infty$, the limit: $$\lim_{r \to +\infty} \bigl( r \left( \cos\theta + \sin\theta \right) \bigr)$$ depends on $\theta$ since $\cos\theta + \sin\theta$ can be positive, negative or zero. This makes sense since fixing $\theta$ to a value outside the interval $(0,\tfrac{\pi}{2})$ would not correspond to $(x,y)\to (+\infty,+\infty)$.

  • 1
    $\begingroup$ @rapidracim Thanks for accepting but you may want to wait for some more input of others, specifically regarding your question on the polar substitution for the limit to $(+\infty,+\infty)$. $\endgroup$
    – StackTD
    Apr 11, 2017 at 12:52
  • $\begingroup$ I think I've found a counterexample $$\lim_{x,y \to \infty} \frac{x}{y^2}+x^2$$ here If I use the polar substitution I'll get $+\infty$. but wolfram-alpha says it doesn't exist because it is path dependent do you think I can trust wolframalpha ? if it gave me a "value" maybe but since it told me it doesn't exist I presume other than computing the limit along two different paths and finding two different results (which actually proves that the limit D.N.E) a computer can't make such a conclusion (that the limit is path-dependant) using a different algorithm . $\endgroup$ Apr 11, 2017 at 16:15

"but here since $x,y\to \infty$ I think that logically this phenomenon can only happen if $r\to \infty.$" The definition of $ x,y\to \infty,$ which you haven't given us, is probably exactly the same as $r\to \infty.$ Let's assume this. You have correctly arrived at

$$f(r\cos t, r \sin t) = \frac{\cos t - \sin t}{r}.$$

This leads to

$$0\le |f(r\cos t, r \sin t)| = \frac{|\cos t - \sin t |}{r} \le \frac{|\cos t| + |\sin t|}{r} \le \frac{1 +1}{r} = \frac{2}{r}\to 0.$$

Thus $|f(r\cos t, r \sin t)|\to 0,$ which is the same as saying $f(r\cos t, r \sin t)\to 0.$


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