Q:

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

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. 

A:

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.