Problems with limits of functions of two variables

Question

I have the following function:

$$f(x,y):=\begin{cases}\frac{x^3y}{x^6+y^2}&,\;\;(x,y)\neq (0,0)\\{}\\0&,\;\;(x,y)=(0,0)\end{cases}$$

I'm asked about continuity at the origin and the limit of function there. Now, the limit doesn't exist since

$$\begin{align*}y=x^3&\implies f(x,x^3)=\frac{x^6}{x^6+x^6}=\frac12\xrightarrow [x\to 0]{}\frac12\;,\;\;\text{whereas}\\y=x&\implies f(x,x)=\frac{x^4}{x^6+x^2}=\frac{x^2}{x^4+1}\xrightarrow[x\to 0]{}0\end{align*}$$

My problem is: if I try to apply what's been shown in several questions in this site, namely polar coordinates, I get

$$\begin{cases}x=r\cos t\\y=r\sin t\end{cases}\implies f(r,t)=\frac{r^2\cos^3t\sin t}{r^4\cos^6t+\sin^2t}$$

and now I argue: if $\;\sin t=0\;$ then $\;x=0\;$ and clearly $\;f(0,y)=0\;$ , otherwise

$$\lim_{r\to 0}\frac{r^2\cos^3t\sin t}{r^4\cos^6t+\sin^2t}=\frac 0{0+\sin t}=0$$

and thus the limit is zero...where am I going wrong?! Thanks.

Answer 1

Parametrizing the function in polar coordinates doesn't change the fact that in order for the limit to exist, for any $\epsilon > 0$, there must exist a $\delta > 0$ such that $|f(x,y) - L| < \epsilon$ whenever $|(x,y)| < \delta$. That is to say, $f$ can be made arbitrarily close to some fixed $L$ for any sufficiently small neighborhood of $(0,0)$. If you parametrize the function in polar coordinates, you can still have trajectories with $r \to 0^+$ for which the angle $t$ varies (and it may even vary as a function of $r$). If you fix $t$ to any particular value, then you are only looking at trajectories that proceed along a ray to the origin. As you observed with the trajectory $y = x^3$, the limit is not zero along this curved path.

Here is a plot of $f$. As you can see, curves of the form $y = cx^3$ for any $c \ne 0$ will give a nonzero limit. We can also show this with direct computation, with the parametrization $y = ct^3$, $x = t$. Then $$\lim_{(x,y) \to (0,0)} f(x,y) = \lim_{t \to 0^+} f(ct^3, t^3) = \frac{c}{1+c^2}.$$ The range of this function of $c$ is clearly $-1/2 \le c \le 1/2$.

Answer 2

If the limit exists all of the paths you use should give the same result but you cannot use this to prove the existence of the limit. We usually use it to show the non existence of the limit.

If you want to actually get the limit you have to consider the general case as you do using the above substitution.

One more thing is that when $x$ and $y$ goe to zero $r$ goes to zero for every value of $t$, which does not happen in your case.

So the limit does not exist.