Analysing continuity of piecewise function.

Question

Question

Given a piecewise function (See below) determine the points of discontinuity.

My Attempt

Looking at the function I can see that the points of discontinuity will be when the denominator = 0 and possibly at (0,0). To check this, I must find the limit of the function at that point. If the limit = 1, then f is continuous there, otherwise no. By simply replacing the point in the function, the limit is 0. However, I have to prove this by using the formal definition of a limit (involving inequalities and norms). I'm lost as where to start so any pointers would be appreciated.

Answer 1

Let $\varphi : \mathbb R^2 \to \mathbb R$ be the function $(x,y)\mapsto 4x^2+y^2-1$, and $E=\varphi^{-1}(0)\;$ (ellipse in $\mathbb R^2$). Given any point $(a,b)\in\mathbb R^2$, we distinguish three cases:

• $\bf (a,b)\notin E, \;(a,b)\neq(0,0).\;\;$ We have $\varphi(a,b)\neq 0$, and as $\varphi$ is continue, by the Theorem of permanence of the sign we deduce that there exists a neighborhood $U$ of $(a,b)$ such that $\varphi(x,y)\neq0$ in $U$; moreover we can suppose $U\not\ni (0,0)$, given the separation properties of $\mathbb R^2$. This means that $U\cap (E\cup\{(0,0)\})=\emptyset$, and therefore: $$ \lim_{(x,y)\to(a,b)}f(x,y)=\lim_{(x,y)\to(a,b)}f_{|U}(x,y)=\lim_{(x,y)\to(a,b)}\;\frac{x}{\varphi(x,y)}=\frac{a}{\varphi(a,b)}=f(a,b), $$ i.e. $f\;$ is continuous in $(a,b)$.
• $\bf (a,b)\in E.\;\;$ Consider the segment $r$ of $\mathbb R^2$ parameterized by $\;x=at, \;y=bt$, with $\;0<t<1$. We have $$ \lim_{(x,y)\to(a,b)}f_{|r}(x,y)=\lim_{t\to1^-}\frac{at}{4a^2t^2+b^2t^2-1}. \tag{*}\label{limite}$$ If $a\neq0$, the numerator in \eqref{limite} goes to $a$ ,while the denominator goes to $4a^2+b^2-1=0$, so the limit is $\infty$. If $a=0$, the limit is zero because the denominator is non-zero: in fact, being $\;0<t<1$, we have $4a^2t^2+b^2t^2-1=t^2(4a^2+b^2)-1<4a^2+b^2-1=0$. In both cases, the limit is different from the value of the function in $(a,b)$, i.e. $f(a,b)=1$, so we conclude that the function is discontinuous in $(a,b)$.
• $\bf (a,b)=(0,0).\;\;$ Similarly to the preceeding case, but the limit \eqref{limite} now must be $$ \lim_{(x,y)\to(0,0)}f_{|r}(x,y)=\lim_{t\to0^+}\frac{at}{4a^2t^2+b^2t^2-1}=0,$$ while $f(0,0)=1$, and here too the function is discontinuous.