When I can and when I cannot use polar coordinates for a limit?

Question

I was doin the next exercise: ¿Is this function differential in (0,0)? $$ \text { Let } f(x, y)= \begin{cases}\frac{x(1-\cos y) \sqrt{x^2+y^2}}{x^2+y^4}, & \text { if }(x, y) \neq(0,0) \text {; } \\ 0, & \text { if }(x, y)=(0,0) .\end{cases} $$

So i just took the jacobian on (0,0) as a candidate for the differential:

$$ \begin{aligned} & \frac{\partial f}{\partial x}(0,0)=\lim _{h \rightarrow 0} \frac{f(h, 0)-f(0,0)}{h}=0 \\ & \frac{\partial f}{\partial y}(0,0)=\lim _{h \rightarrow 0} \frac{f(0, h)-f(0,0)}{h}=0 \end{aligned} $$

And the candidate jacobian is: (0,0), then I use the definition to see whether is differential or not:

$$ \begin{aligned} \lim _{(x, y) \rightarrow(0,0)} \frac{x(1-\cos y) \sqrt{x^2+y^2}}{\sqrt{x^2+y^2}\left(x^2+y^4\right)} & =\lim _{(x, y) \rightarrow(0,0)} \frac{(1-\cos y)}{y^2} \frac{x y^2}{x^2+y^4} \\ & =\frac{1}{2} \lim _{(x, y) \rightarrow(0,0)} \frac{x y^2}{x^2+y^4} \end{aligned} $$

Then, we only have to check out what happens when we choose (t,0) and $(t^2,t)$ to see that we have two different limits, then is not differential in (0,0) BUT, if we use polar cordinates. we get:

$$ \frac{1}{2} \lim_{(x, y) \to (0,0)} \frac{r^3(\cos(\theta)\sin^2(\theta))}{r^2\cos^2(\theta)+r^4\sin^4(\theta)} = \frac{0}{2\cos(\theta)} = 0 $$

which tell us is actually differentiable in (0,0). Is this because it could be the case that cos(θ)=0?

thanks!

Answer 1

Note that by polar coordinates

$$\frac{r^3(\cos \theta \sin^2 \theta)}{r^2\cos^2 \theta+r^4\sin^4 \theta}=\frac{r(\cos \theta \sin^2 \theta )}{\cos^2 \theta+r^2\sin^4 \theta}$$

therefore we can't conclude limit is zero for the presence of the $\cos^2(\theta)$ term which goes to zero for $\theta \to \frac \pi 2$ which is consistent with the result you have found for the parametrization $(x,y)=(t^2,t)$ which correspond to a path with "vertical" tangent at $t=0$ (namely $y=\sqrt x$).

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For the general question, after the standard check by direct substitution and continuity, we can always use polar coordinates for this kind of limits and it can be useful in two ways

1. to conclude for the existence of the limit when we obtain an expression in the form $r^af(\theta)$ with $a>0$ and $f(\theta)$ bounded;
2. to guess the problematic paths when we obtain an expression with $f(\theta)$ not clearly bounded, looking at the value of $\theta$ such that $f(\theta)$ diverges at infinity, as in the case in hand.