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Proof about the continuity of a function of norm

ianchenmu

Member
Feb 3, 2013
74
Prove that the function $f : \mathbb{R}^2→\mathbb{R}$ defined by
$f(x)=\left\{\begin{matrix}
\frac{|x|_2}{|x|_1} , if x\neq 0 \\
a, if x = 0\end{matrix}\right.$


is continuous on $\mathbb{R}^2$\{$0$} and there is no value of $a$ that makes $f$ continuous at $x = 0$.
 

Fernando Revilla

Well-known member
MHB Math Helper
Jan 29, 2012
661
Prove that the function $f : \mathbb{R}^2→\mathbb{R}$ defined by
$f(x)=\left\{\begin{matrix}
\frac{|x|_2}{|x|_1} , if x\neq 0 \\
a, if x = 0\end{matrix}\right.$


is continuous on $\mathbb{R}^2$\{$0$} and there is no value of $a$ that makes $f$ continuous at $x = 0$.
Hello ianchenmu. Welcome to MHB.

We have $f : \mathbb{R}^2→\mathbb{R}$ defined by


$f(x_1,x_2)=\left\{\begin{matrix}
\dfrac{\sqrt{x_1^2+x_2^2}}{\left|x_1\right|+\left|x_2\right|} & \mbox{ if} &(x_1,x_2)\neq (0,0) \\(a_1,a_2) &\mbox{ if}& (x_1,x_2)=(0,0) \end{matrix}\right.$

Consider $(\alpha,\beta)\neq (0,0)$ then, there exists a neigbordhood $V$ of $(\alpha,\beta)$ not containing $(0,0)$. In $V$, $f$ is continuous (the square root of a non negative continuous funtion is continuous, the absolute value of a continuous funtion is continuous, and the denominator is $\neq 0$ in $V$). We conclude that $f$ is continuous in $\mathbb{R}^2-\{(0,0)\}$.

Now, prove that $\displaystyle\lim_{(x_1,x_2)\to (0,0)}f(x_1,x_2)=1$ along the line $x_2=0$ and $\displaystyle\lim_{(x_1,x_2)\to (0,0)}f(x_1,x_2)=\sqrt{2}/2$ along the line $x_1=x_2$ which implies that there is no limit at $(0,0)$.
 
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