If given the vector field $\mathbf{F}(x,y)=\langle\frac{1}{x}+2xy,\frac{1}{y}+x^2-\cos{y}\rangle$, it is clear that the component functions are continuous everywhere expect at the origin.
Now, when asked to sketch the largest $\mathit{simply \, connected}$ region, let's call it $D$, which contained the point $P(-1,-2)$ on which $\mathbf{F}$ is continuous, I am getting slightly tripped up.
Because, to my understanding, if the component functions of $\mathbf{F}$ were continuous everywhere, we could conclude that the largest simply connected region containing $P$ on which $\mathbf{F}$ is continuous would be $\mathbb{R}^2$. Yet, with the domain restriction on $\mathbf{F}$, the origin would be, for lack of a better term, a "hole" and therefore would make $\mathbb{R}^2$ not simply connected.
I can sketch a simply connected region containing said point on which $\mathbf{F}$ is continuous, yet I don't seem to understand how one would find and/or sketch the $\mathbf{largest}$ region fulfilling said conditions.
Would the location of $P$ have any affect on the size or location of this region?
I apologize if my notation is poor, so please feel free to correct it at any time because I am big on using proper notation wherever applicable.
Thank you,
Sam