Is the following function radially unbounded or not? $$V(x) = \frac{x_{1}^2}{1 + x_{1}^2} + x_{2}^2$$ I know that if $x_{2} \to \infty$ in which case $||x|| \to \infty$ and $V(x) \to \infty$ but if $x_{1} \to \infty$ which makes $||x|| \to \infty$ but then $V(x)$ does not go to $\infty$. What I would like to know is that is a function radially unbounded if only one or more of its variables makes it go to infinity or should it be true for every variable that that function depends on?
Edit: I am not a student of math. In a engineering subject, this popped up. I have seen other answers but can't get my head around it. So can you kindly answer the question in simpler words? instead of down voting the question.
You can think of it geometrically. The condition that $V(x)$ is radially unbounded means that all level sets are closed. Now, take $x_2 = 0$ and determine the level set $V(x) = 1$, you get
$$ V(x_1, 0) = \frac{x_1^2}{1 + x_1^2} = 1. \tag{1} $$
Obviously, this cannot hold for finite $x_1$, so $V(x)$ is radially bounded.
Edit: To adress your second question, I will give an example. In general it means that any combination of $x_1$, $x_2$ that makes $||x|| \rightarrow \infty$ also has to make $V(x)\rightarrow \infty$.
This holds for any norm, but lets assume Euclidean norm for simplicity. Since for $x_1 \rightarrow \infty, x_2 = 0$ you get $||x||_2 = \sqrt{x_1^2 + x_2^2} = \sqrt{\infty^2 + 0^2} \rightarrow \infty$. However, for $V(x)$, you get in this case
$$ \lim_{x_1 \rightarrow \infty} V(x_1, 0) = 1 \neq \infty. $$
So your function is not radially unbounded.