This function isn't uniformly continuous, but how?

Question

I have $f: B \to \mathbb{R}^n$, such that $f(x) = \displaystyle \frac{x}{1-||x||}$ and $B= \{x \in \mathbb{R}^n: ||x|| < 1 \}$. I need to show that this function is not uniformly continuous.

My idea was take the sequences $x_n$ and $y_n$ $\in B$, then for some $n_0$ when $n>n_0$ we have $||x_n - y_n|| < \delta$ but $||f(x_n) - f(y_n)|| \geq \epsilon$.

I have tried

$x_n= (0, \dots, \frac{1}{n+1})$ and $y_n= (0, \dots,- \frac{1}{n+1})$

$x_n= (0, \dots, \frac{1}{2} + \frac{1}{2^{n+1}})$ and $y_n = (0, \dots, \frac{1}{2} - \frac{1}{2^{n+1}}), $

and a lot of another cases. Everytime I get $\displaystyle \lim_{n \to \infty} ||f(x_n)-f(y_n)|| = 0$.

So, I need some tips. I dunno, who knows the sequences $x_n$ and $y_n$ and the reason to choice it.

Answer 1

Graphing $f(x)$, it seems we want to investigate behaviour near the boundary of the ball.

I might try $x_n = (1-1/n^2, 0, 0, \dots)$ and $y_n = x_{n+1}$, so $||x_n - y_n|| < 1/n$.

Replacing $1/n^2$ with $2^{-n}$ gives a perhaps easier computation.

Answer 2

Thank you guys, I have found the solution. Now I took $x_n=(0,...,\frac{n}{n+1})$ and $y_n= (0,...,\frac{n-1}{n})$. So $||f(x_n) - f(y_n)|| = 1. $