Please check my proof on "$f(x)=\frac{1}{x}$ is not uniformly continuous on $(0,\infty)$" 
$f(x)=\frac{1}{x}$. Prove that $f$ is not uniformly continuous on $(0,\infty)$.

We want to find $\epsilon_0 >0$ such that $\forall \delta>0$, there are some $x,y \in (0,\infty)$ such that $|x-y|<\delta$, yet $|\frac{1}{x} - \frac{1}{y}|\ge\epsilon_0$.
Take $\delta_n=\frac{2}{n}.$ Take $x_n=n, y_n=n+\frac{1}{n}$. Then apparently we have $|x_n - y_n|=|\frac{1}{n}|<\delta_n$. However, $|f(x_n)-f(y_n)|=\frac{n^2 +1}{n}=n + \frac{1}{n} \ge2.$ So $\epsilon_0=2$, and indeed this function is not uniformly continuous on the interval. 
 A: what Robearz pointed out is not a big problem because you've shown a sequence of $\delta_n$ converging to $0$. what is a problem however is your formula for $f(x_n) - f(y_n)$. actually
$$ |f(x_n) - f(y_n)| = |\frac{1}{n} - \frac{1}{n + 1/n}| = \frac{1/n}{n^2+1}$$
which is pretty small. you might want to consider $x_n = 1/n$ and $y_n = 1/(n+1)$ for example
A: Alternately you can see $f$ being not uniformly continuous on $(0,\infty)$ because for $x = \epsilon$, and $y = 2\epsilon$, then $\displaystyle \text{Sup}_{\epsilon > 0} |\dfrac{1}{\epsilon} - \dfrac{1}{2\epsilon}| = +\infty$.
A: Your proof strategy is correct, but your details are not. Notice: $$|f(x_n) - f(y_n)| = \left|\frac{1}{n} - \frac{n}{n^2 + 1}\right| = \frac{1}{n(n^2+1)}$$
Clearly this is getting small as $n$ gets big: not what you want. Rather, take $x_n$ and $y_n$ near $0$, e.g. $x_n = \frac{1}{2n}$, $y_n = \frac{1}{n}$.
A: Moreover you can take $\{x_n\}=\frac{1}{n}$ and $\{y_n\}=\frac{1}{n+1}$.
A: If one takes $(x_n) = (1/n)$ and $(y_n) = (1/2n)$, then we find that $$\lim_{n \to \infty} |f(x_n) - f(y_n)| = \lim_{n \to \infty} |n| = \infty .$$
Meaning the image of the function is not "eventually close" so to speak even though $(x_n)$ and $(y_n)$ are.
I think your idea could have worked, but as others have pointed out you only made a minor algebraic mistake which can be easily fixed.
