I have two question concerning continuity and differentiability, because something is unclear to me.
Let’s say I have a function : $f:\mathbb{R} \rightarrow \mathbb{R}$ (this function is not necessarily continuous), and now let’s defined the function $F$ as follow :
$$\forall (r,t) \in \mathbb{R}^+\times \mathbb{R} F(r,t)=(r \cos f(t), r \sin f(t))$$
Now my questions are :
Is $F$ continuous on a neighborhood of $(0,0)$ even-if $f$ is not continuous ?
According to me it’s most likely true because : $$\lim_(r,t \rightarrow (0,0)) F(r,t) = 0$$
If it wasn’t continuous on a neighborhood of $(0,0)$ we can’t talk about the limit of such function at the point $(0,0)$, so the function is continuous at $(0,0)$.
Now I would like to know why I can’t say that $F$ is différentiable at the point $(0,0)$ using the following argument :
If $F$ is differentiable at the point $(0,0)$ then there is a function $u$ such that :
$$F(h,h) = F(0,0) + u(0)(h) + o(||h||)$$
So by taking the limit as $h \rightarrow 0$ we have : $$u(0)(h) = 0 $$
And hence $F$ is differentiable at the point $(0,0)$, and $\mathrm{d}F(0) = x \mapsto 0$.
Yet this argument is actually incorrect, but why ?
Thank you, for taking your time !
We have that $F$ is continuous, because $\sin$ and $\cos$ are bounded and $r\to 0$. The function $f$ has no say in what happens here. But if ${\rm D}F(0,0)$ is the zero map, we would have to check that $$\lim_{(r,t)\to (0,0)} \frac{F(r,t)}{\sqrt{r^2+t^2}} =(0,0),$$since $F(0,0)=(0,0)$. This is equivalent to $$\lim_{(r,t)\to (0,0)} \frac{r}{\sqrt{r^2+t^2}}\cos f(t) =\lim_{(r,t)\to (0,0)} \frac{r}{\sqrt{r^2+t^2}}\sin f(t) =0. $$ Although $r/\sqrt{r^2+t^2}$ is bounded, the terms $\cos f(t)$ and $\sin f(t)$ need not go to zero (take $f=1$ or whatever).