Difference between these two propositions

Question

In my multivariable calculus course they ask me to prove the following propositions:

1) $lim_{(x,y)\to(0,0)}f(x,y) = L \Leftrightarrow \forall\epsilon>0, \exists\delta>0$ such that $|g(r,\theta)-L|<\epsilon, \forall r \in(0,\delta), \theta\in[0,2\pi)$ where g is f in polar coordinates

2) $lim_{(x,y)\to(0,0)}f(x,y) = L \Rightarrow \lim_{r\to0^+}g(r,\theta)=L, \forall \theta \in[0,2\pi)$

I'm also asked to prove that the converse of the second one is false, which means the second proposition is somehow different from the first one. Can someone show me how? They look the same to me.

EDIT: As Jonas pointed out in the comments, θ is fixed in the second proposition but not in the first one. That makes intuitive sense to me, but I wouldn't be able to tell that's how I should interpret the propositions just by reading them. I feel as though I'm artifitially interpreting it that way just to make sense of them. How should I read them to naturally interpret them the right way?

Sorry for slightly changing the question.

Answer 1

In the second proposition, $\lim_{r\to0^+}g(r,\theta)=L$ holds for each fixed $\theta$, while in the first proposition $\theta$ can take any value.

Answer 2

First question

If $\lim_{(x,y)\to (0,0)} f(x,y)=L$ then for any $\epsilon>0$ there exist $\delta>0$ such that $$\|(x,y)\|<\delta \implies |f(x,y)-L|<\epsilon.$$ But $\|(x,y)\|<\delta \iff r<\delta.$ So, $$r<\delta \implies |g(r,\theta)-L|<\epsilon, \forall \theta \in [0,2\pi).$$

The converse can be obtained in a similar way. If for any $\epsilon>0$ there exist $\delta>0$ such that $$r<\delta \implies |g(r,\theta)-L|<\epsilon,$$ then (using again that $\|(x,y)\|<\delta \iff r<\delta$) it is

$$\|(x,y)\|<\delta \implies |f(x,y)-L|<\epsilon.$$ That is, $\lim_{(x,y)\to (0,0)} f(x,y)=L$

Second question

We have seen that if $\lim_{(x,y)\to (0,0)} f(x,y)=L$ then for any $\epsilon>0$ there exist $\delta>0$ such that $r<\delta \implies |g(r,\theta)-L|<\epsilon, \forall \theta \in [0,2\pi).$ Thus, $\lim_{r\to 0^+} g(r,\theta)=L.$ Indeed, $\theta$ doesn't need to have a given value. It follows from above that for any curve $\theta(r)$ it is $\lim_{r\to 0^+} g(r,\theta(r))=L.$

Next we will give an example to show that the converse is false. Consider $f(x,y)=\displaystyle \frac{x^2y}{x^4+y^2}.$ Then $$g(r,\theta)=\frac{r^3\cos\theta\sin^2\theta}{r^2(r^2\cos^2\theta+\sin^2 \theta)}=\frac{r\cos\theta\sin^2\theta}{r^2\cos^2\theta+\sin^2 \theta}.$$ Thus, $g(r,\theta)\to 0$ as $r\to 0,$ $\forall \theta.$ But, through the path $y=x$ it is $$\lim_{x\to 0}\frac{x^3}{x^4+x^2}=\lim_{x\to 0}\frac{x}{x^2+1}=0$$ and through the path $y=x^2$ it is $$\lim_{x\to 0}\frac{x^4}{x^4+x^4}=\frac 12.$$ Thus the limit doesn't exist.

The reason because it fails is that $$\frac{\cos\theta\sin^2\theta}{r^2\cos^2\theta+\sin^2 \theta}$$ is not uniformly bounded in $\theta.$ If previous expression were bounded then the converse would hold.

Answer 3

Here is how I would intuitively interpret the statements:

$lim_{(x,y)\to(0,0)}f(x,y) = L \quad \quad$The limit of $f(x,y)$ along any path to $(0,0)$ is $L$. At you travel closer to $(0,0)$, the value of $f(x,y)$ becomes closer to $L$.

$\Leftrightarrow$

$\forall\epsilon>0, \exists\delta>0$ such that $|g(r,\theta)-L|<\epsilon, \forall r \in(0,\delta), \theta\in[0,2\pi)$

Consider a circle of radius $r$ centered at $(0,0)$. As $r$ becomes smaller, the values of $g(r,\theta)$ inside the circle become closer to $L$.

$\Rightarrow$

$ \lim_{r\to0^+}g(r,\theta)=L, \forall \theta \in[0,2\pi)\quad \quad$ Pick some $\theta \in [0,2\pi]$. As you travel along a straight line at this angle towards $(0,0)$, the value of $g(r,\theta)$ becomes closer to $L$. So this is like saying the limit of $f(x,y)$ along linear paths to $(0,0)$ is L.