Continuity of piecewise function of two variables The question looks like this. 
Let $f(x, y)$ = 0 if $y\leq 0$ or $y\geq x^4$, and $f(x, y)$ = 1 if $0 < y < x^4 $. 
(a) show that $f(x, y) \rightarrow 0$ as $(x, y) \rightarrow (0, 0)$ along any path through (0, 0) of the form $ y = mx^a $ with $a < 4$. 
(b) Despite part (a), show that $f$ is discontinuous at (0, 0)
(c) Show that $f$ is discontinuous on two entire curves. 
What I've came to conclusion is that when $ x<0, m>0 $, and $a$ being an odd number, $y$ becomes smaller then zero, so $f(x, y)$ can't be any larger than zero. But I don't think that's not enough. I think I need to find a way to generalize that $ mx^a (a<4) $is larger than $x^4$ or smaller than 0 when $x$ and $y$ is close enough to zero, where I cant' quite get to. 
In regarding (b), I know $f(x, y)$ is discontinuous on certain directions, but can't elaborate it in decent form. 
In regarding (C), How can I show it?

Substitute $e^{-t}$ for $x$. In case $m=|m|$ then substitute $e^{-at+b}$ for $y$ where $b=ln(|m|)$, in case $m=-|m|$ then $y=-e^{-at+b}$. In the first case we have  $t>t_0=-b/(4-a) \Rightarrow y>x^4$ so $f(e^{-t},e^{-at+b})=0,\; \forall t>t_0$. In the second case we have $y<0 \;\forall t\in \Bbb{R}$ so that $f(e^{-t},-e^{-at+b})=0,\; \forall t\in \Bbb{R}$.
