Infimum of a continuous function on an open set

Question

I'm trying to solve the following question:

For $x\in\mathbb{R}$, $x>0$ let $f(x) = 1/x+x$, and let $m=\inf\{f(x):x>0\}$. Show that there exist a sequence $x_n>0$ such that $f(x_n)\to m$ as $n\to\infty$.

Using a graphing calculator I can see that the function attains its infimum value (so there exist $\xi\in\mathbb{R}$ s.t. $f(\xi) =m$, and then its easy to show that $x_n = \xi + 1/n$ satisfies the requirement), but I can't figure out how to prove it.

Answer 1

Since the infimum in this case is finite, for each $\varepsilon_0 > 0$ there is $x_0>0$ such that $m \leq f(x_0) < m+\varepsilon_0$. For a strictly decreasing sequence of $\varepsilon_n > 0$ converging to $0$ (e.g. $\varepsilon_n = 1/n$) you can find a sequence of $x_n>0$ such that $m \leq f(x_n) < m+\varepsilon_n$. Taking the limit $n \to +\infty$ you find $m \leq \lim_n f(x_n) \leq m$, so $\lim_n f(x_n) = m$.