Is $f(x,y)=\frac{y}{x^2 +1}$ uniformly continuous?

Question

Is $f(x,y)=\frac{y}{x^2 +1}$ uniformly continuous on $\mathbb{R}^2$?

Intuitively it doesn't seem to be uniformly continuous because the denominator grows quadratically and the numerator grows linearly.

I've tried to prove that $f$ is not uniformly continuous by using the sequential characterization of uniform continuity:

Let $A\subset \mathbb{R}^n$ and a function $f:A \rightarrow \mathbb{R}^m$. $f$ is uniformly continuous if and only if for any two sequences $\{x_k\}, \{y_k\}$ of points of $A$ such that $\{x_k - y_k\}\rightarrow 0$ then $\{f(x_k) - f(y_k)\}\rightarrow 0$.

So I've tried to find two sequences such that $\{x_k - y_k\}\rightarrow 0$ but $\{f(x_k) - f(y_k)\}\not \rightarrow 0$.

However none of the pairs of sequences that i have tried have been succesful. These are a couple of them:

• $\{k,k\}$ and $\{k+\frac{1}{\sqrt{k}},k\}$.

• $\{k,\frac{1}{k}\}$ and $\{k+\frac{1}{k},\frac{1}{k}\}$.

• $\{\frac{1}{\sqrt{k-1}},\frac{1}{k}\}$ and $\{\frac{1}{\sqrt{k}},\frac{1}{k}\}$

Any hint or a "good" pair of sequences to try would be appreciated.

Answer 1

The $x$-derivative of $f$ gets large when $x = 1$ and $y$ is large. Thus, I would try a sequence like $v_k = (1,k)$ and $z_k = (1+1/k,k)$. We see $$\|v_k - z_k\| = \frac{1}{k} \to 0$$ but $$f(v_k) - f(z_k) = \frac{k}{2} - \frac{k}{2 + \frac 2 k + \frac 2 {k^2}} = \frac{2k + 2 + \frac 2 k - 2k}{2(2 + \frac 2 k + \frac{2}{k^2})} = \frac{2 + \frac{2}{k}}{4 + \frac{4}{k} + \frac 4{k^2}} \to \frac 1 2 \,\,\,\, \text{as } k \to\infty. $$ Thus $f(v_k)$ and $f(z_k)$ stay at a bounded distance away from each other while $v_k - z_k \to 0$ which means $f$ is not uniformly continuous.

Answer 2

Consider $\mathbf{x}_1=(0,y)$ and $\mathbf{x}_2=(\delta,y)$ where $\delta>0$. Then $|\mathbf{x}_1-\mathbf{x}_2|=\delta$. But as $y\to\infty$, $f(\mathbf{x}_2)-f(\mathbf{x}_1)=y\delta^2/(1+\delta^2)\to\infty$. Therefore $f$ is not uniformly continuous.