Is the following true ? A proof or counter-example or reference would be nice.
A function $f:\mathbb{R}^2 \rightarrow \mathbb{R}$ is continous at $(0,0)$ if and only if if for all $a, b$, the limits $$\lim\limits_{t \to 0} f(at,bt)$$ exist are all are equal.
Can the condition of approach in all directions be weakened to the equallty of approach along two distinct lines ? One direction is of course trivial.
I think it is implicitly assumed that $f$ is continuous on $\mathbb{R}^2-\{(0,0)\}$
The answer is no. For example, $f(0,0)=0$ and $$ f(x,y) = \frac{2x^2y}{x^4+y^2} $$ You can see $t \mapsto (t,mt)$ gives $f(t,mt) = \frac{2mt^3}{t^4+m^2t^2} =\frac{2mt}{t^2+m^2}$ hence all linear paths to $(0,0)$ give trivial path limits for $f$. However, $f$ is not continuous as is easily seen by studying quadratic paths. For those, differing quadratics will yield different path limits. Consider $t \mapsto (t,mt^2)$ note $f(t,mt^2) = \frac{2mt^4}{t^4+m^2t^4} = \frac{2m}{1+m^2}$. Clearly different $m$ give different quadratic path limits for $f$. But, if $f$ was continuous at $(0,0)$ then all path limits must converge to the same value. Hence, $f$ is not continuous at $(0,0)$ despite the temptation to ascertain such from the linear data.
Naturally, examples exist for other classes of paths. This is a subtle issue.
