Why $x^2$ is not uniformly continuous The definition from the book I am using state: $f(x):D\rightarrow\mathbb{R}$ is said to be uniformly continuous provided that whenever $\{u_n\}$ and $\{v_n\}$ are sequences in $D$ such that $\lim_{n\rightarrow\infty}(u_n-v_n)=0$, then $\lim_{n\rightarrow\infty}(f(u_n)-f(v_n))=0$. 
To show $f(x)=x^2$ is not uniformly continuous, I know I can pick $\{u_n\}=\{n\}$ and $\{v_n\} = \{n+1/n\}$. 
What if I don't know those two sequences or any sequences that I can apply to show $x^2$ is not uniformly continuous.  

Assume that $f(x)=x^2$ is uniformly continuous from $D$ to $\mathbb{R}$, then let $\{u_n\}$ and $\{v_n\}$ in $D$ such that $\lim_{n\rightarrow\infty}|u_n-v_n|=0$; by the definition, we have $$\lim_{n\rightarrow\infty}|f(u_n)-f(v_n)|=\lim_{n\rightarrow\infty} |u_n^2-v_n^2|=\lim_{n\rightarrow\infty} |(u_n-v_n)(u_n+v_n)|=\lim_{n\rightarrow\infty} |(u_n-v_n)||(u_n+v_n)|=0$$

I don't quite see a contradiction, can someone tell me where I did wrong? Thanks in advanced.  

Under the definition, to prove that $x \mapsto x^{2}$ is not uniformly continuous it suffices to find some sequences $(u_{n}), (v_{n})$ such that $u_{n} - v_{n} \to 0$ and $(u_{n}-v_{n})(u_{n} + v_{n})$ does not; consider the sequences $(u_{n}), (v_{n})$ defined by
$$
u_{n} := n^{2} + \frac{1}{n}, v_{n} := n^{2}
$$
for all $n \geq 1$.
