I have this limit to calculate:


I solve it by going to the polar coordinates. Since $(x,y)\to 0$ means the same as $\sqrt{x^2+y^2}\to 0$, I get (after dealing with the sine in a standard way),

$$l=\,\lim_{r\to0}\frac{r^3\cos^2\theta\sin\theta+r^5\cos^2\theta\sin^3\theta}{r^2} =\lim_{r\to0}\,r(\cos^2\theta\sin\theta+r\cos^2\theta\sin^3\theta)=0. $$

I'm quite sure this actually works, but the free variable $\theta$ bothers me. It seems to me that some explanation for it is needed since it's not "for every fixed $\theta$", which I think would be the standard interpretation of this formula. How should I phrase this solution so it's rigorous?

  • $\begingroup$ Try something like "the limit is zero as the radius diminishes no matter what the variable (polar) position of a point is." You can also phrase this "as the radius $\,r\,$ goes to zero, all the points on each circle of radius $\,r\,$ equally go to zero" $\endgroup$ – DonAntonio Jan 9 '13 at 3:30

Note that $$\vert \cos^2(\theta) \sin(\theta) + r \cos^2(\theta) \sin^3(\theta) \vert \leq \vert \cos^2(\theta) \sin(\theta) \vert + r \vert \cos^2(\theta) \sin^3(\theta) \vert \leq 1 + r$$ Hence, we have that $$\left \vert r \left( \cos^2(\theta) \sin(\theta) + r \cos^2(\theta) \sin^3(\theta) \right) \right \vert \leq r(1+r)$$ Now let $r \to 0$ and conclude the limit is $0$ using squeeze theorem.


I have one caveat. There is nothing wrong with the accepted answer by @Marvis, but in the OP's question, it states, somewhat loosely:

Since $(x,y)\rightarrow0$ means the same as $\sqrt{x^2+y^2}\rightarrow0$....

It is true that $(x,y)\rightarrow0$ implies $\sqrt{x^2+y^2}\rightarrow0$, but the (implicitly asserted) equivalence $$\lim_{(x,y)\rightarrow(0,0)} f(x,y) = \lim_{r \rightarrow 0} f(r\cos\theta,r\sin\theta)$$ does not hold (in all cases).

For instance, in the standard example $f(x,y) = x\,y^2/(x^2+y^4)$, the limit as $r \rightarrow 0$ for any fixed $\theta$ is $0$. However, in any neighborhood of the origin, the $f(x,y)$ attains the value $1/2$ (at any point of the form $x=t^2$, $y=t)$) So $\lim_{(x,y)\rightarrow(0,0)} f(x,y)$ does not exist in this case.

You have to get $|\,f(x,y)-L\,| \le g(r)$, where $g(r)\rightarrow0$, in order to conclude the limit of $f$ is $L$, by the squeeze theorem (allowing for adaptations according to your version of the squeeze theorem). Note that Marvis did exactly that in the solution.

I'm not sure why the downvote: The above is correct and was meant as a complement to the answer by @user17762 (formerly Marvis). The above points out that the OP was justified in being bothered by "the free variable $\theta$."

  • $\begingroup$ Can you please explain why "in the standard example $f(x,y) = x\,y^2/(x^2+y^4)$, the limit as $r \rightarrow 0$ for any fixed $\theta$ is $0$." I tried substituting to polar coordinates and I don't see how the limit is 0. $\endgroup$ – Cauthon Jul 5 '14 at 13:06
  • $\begingroup$ @Crumbs Given: In polar coordinates, $f(x,y) = r \cdot [\cos\theta \sin^2\theta /(\cos^2\theta + r^2\sin^4\theta)]$, $\theta$ fixed, $r \ne 0$, $r \rightarrow 0$. Hence: If $\cos\theta = 0$, then the second factor equals $0$ for all $r$ (given $r \ne 0$); if $\cos\theta \ne 0$, then the limit of the second factor is $\sin^2\theta/\cos\theta$. In both cases the limit of the second factor exists. The first factor approaches $0$, so the limit is $0$. $\endgroup$ – Michael E2 Jul 5 '14 at 15:05
  • $\begingroup$ @MichaelE2, I think you meant $\;x=t^2\,,\,y=t\;$ , otherwise the limit's again zero. $\endgroup$ – Timbuc Sep 20 '14 at 14:15
  • $\begingroup$ @Timbuc Yes, thanks! $\endgroup$ – Michael E2 Sep 20 '14 at 19:14

First use $|\sin w| \le |w|$. Then go to polar coordinates.


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