Is $h(x_1,...,x_n)=\sqrt{x_1^2+...+x_n^2}$ continuous?

Question

How would I go about showing whether or not $h(x_1,...,x_n)=\sqrt{x_1^2+...+x_n^2}$ is continuous?

I have shown that the partial derivatives exist everywhere except $(0,..,0)$.

Answer 1

Note that $h$ simply represents the euclidean norm of the vector $(x_1, \ldots,x_n)^T$. It therefore satisfies all axioms of a norm and I would like to refer you to this article Why are norms continuous? to find your answer.

Answer 2

$h$ is just the $l_2$-norm, it is a continuous function in virtue of the Cauchy-Schwarz inequality. Or: $h^2$ is a non-negative polynomial function, hence its square root is well-defined and continuous.

Answer 3

for partial derivate we have $\frac{\partial}{\partial x}\sqrt{u}=\frac{1}{2\sqrt{u}}\frac{\partial}{\partial x}u$, wich gives for $a\in\{1,2,\cdots,n\}$:

$$h_a(x_1,x_2,\cdots,x_n)=\\ \frac{\partial}{\partial x_a}\sqrt{x_1^2+x_2^2+\cdots+x_n^2}=\frac{\frac{\partial}{\partial x_a}(x_1^2+x_2^2+\cdots+x_n^2)}{2\sqrt{x_1^2+x_2^2+\cdots+x_n^2}}=\frac{x_a}{\sqrt{x_1^2+x_2^2+\cdots+x_n^2}}$$

then if $h_a(x_1,x_2,\cdots,x_n)$ is continuos at $(x_1,x_2,\cdots,x_n)=(0,0,\cdots,0)$ then the limit $(1)$ exists:

$$\lim_{(x_1,\cdots,x_n)\to(0,\cdots,0)}h_a(x_1,x_2,\cdots,x_n)\tag{1}$$

lets proof that the limite $(1)$ does not exists, taking the path $x_a=x$ and $x_i=0,i\ne a$, then $(x_1,\cdots,x_n)\mapsto(0,\cdots,0)=x\mapsto0$

$$\lim_{(x_1,\cdots,x_n)\to(0,\cdots,0)}h_a(x_1,x_2,\cdots,x_n)=\lim_{x\to0}\frac{x}{\sqrt{x^2}}=\lim_{x\to0}\frac{x}{|x|}$$

since

$$\lim_{x\to0^+}\frac{x}{|x|}=1$$

and

$$\lim_{x\to0^-}\frac{x}{|x|}=-1$$

and

$$\lim_{x\to0^+}\frac{x}{|x|}\ne\lim_{x\to0^-}\frac{x}{|x|}$$

then the limit $\displaystyle\lim_{x\to0}\frac{x}{|x|}=\nexists$ wich implies that the limit $(1)$ does not exists and then $h_a(x_1,\cdots,x_n)$ is not continuos at $(0,\cdots,0)$

to proof the continuity on other points, since $h(x_1,x_2,\cdots,x_n)$ are continuos and $x_a$ is continuos and for all $(x_1,\cdots,x_n)\ne(0,0,\cdots,0)\Rightarrow h(x_1,\cdots,x_n)\ne0$, then $h_a(x_1,\cdots,x_n)=\frac{x_a}{h(x_1,\cdots,x_n)}$ is continuos for all $(x_1,\cdots,x_n)\ne(0,0,\cdots,0)$