Consider the function $f\colon \mathbb{R^2} \rightarrow \mathbb{R}$ defined on all of $\mathbb{R^2}$ by
$f(x,y) = \left\{ \begin{array}{lr} 1, & \text{if } (x,y) \in A\\ 0, & \text{if } (x,y) \notin A \end{array} \right.$
Where $A \subset \mathbb{R^2}$ is the set $A=\lbrace(x,y):x>0\text{ and }x^2<y<2x^2 \rbrace$
Show that $ \lim_{t \to 0}f(tv)=0$ for every non-zero vector $v=(a,b) \in \mathbb{R^2}$
I find the notation a little confusing, but I suppose it makes sense; $tv=(ta,tb)$, and of course vectors from the origin correspond to points. I'm just more used to physics or vector calculus notation.
Anyway, I checked all cases except when $v$ is in the first quadrant. I need to find $\delta >0$ such that $f(tv)=0$ for $t \in (-\delta,\delta)$. This ensures that $f(tv)<\epsilon$ for any positive $\epsilon$, proving that the limit is indeed $0$.
I feel like this should be simple, but I seem to be going in circles. $f(tv)=0 \implies y<x^2$ or $2x^2<y$, so I found that the parametric line
$x=ta, y=tb$ intersects $y=x^2$ at $t=\frac{b}{a^2}$, but I don't see how this can give me a radius $\delta$ that I need.
Intuitively, we want to show that if we draw any line through than the origin, then this line will be outside of $A$ provided that we zoom in to the origin close enough to some circle with radius $\delta > 0$ that is sufficiently small.
To this end, consider a line with direction vector $v$ that intersects $A$. If we start at the origin and walk along the line until we hit $A$, then the first border that we hit is the curve $y = 2x^2$. Since $x = ta$ and $y = tb$, this occurs when $t = \frac{b}{2a^2}$. We claim that this is our desired radius $\delta$.
Proof: Fix any non-zero vector $v = (a,b)$ in the first quadrant so that $a,b > 0$. Now consider $tv = (ta, tb)$, where $t < \delta = \frac{b}{2a^2}$. Then since: $$ 2(ta)^2 = t(2a^2t) < t(2 a^2 \cdot \tfrac{b}{2a^2}) = tb $$ it follows that $(ta, tb) \notin A$ so that $f(tv) = 0$ for all $t < \delta = \frac{b}{2a^2}$, as desired.