Mathematical analysis: Bolzano-Weierstrass theorem For  $x \in \Bbb{R}$ with $x > 0$ let
$f(x) = 1/x^2 + 1/x + x^2$ and let $ m $= inf {$f(x) : x > 0$}
Prove there exists a sequence $x_n > 0$ with $f(x_n)$ goes to $m$ as $n$ goes to $1$ and show
that $m ≤ 3$.
Could someone explain how do I find such a sequence using Bolzano-Weierstrass theorem?
 A: By characterisation of the infimum, there exists a sequence $(x_n)$ such that $f(x_n)$ converges to $m$ as $n$ goes to infinity. Now Bolzano-Weierstrass applies to bounded sequences so you have to prove that $(x_n)$ (or some subsequence of $(x_n)$) is bounded.
Note that
$f(x) \xrightarrow[x \rightarrow 0^+]{} +\infty$
and
$f(x) \xrightarrow[x \rightarrow +\infty]{} +\infty$
so there exists $ 0 < a < 1 $ and $1 < b$ such that $f(x) \geq f(1) + 1$ for all $x$ in
$]0,+\infty[\setminus [a,b]$.
Now as $f(x_n)$ converges to $m$ and $m < f(1) + 1$, there is an integer $n_0$ such that for all $n \geq n_0$,
$f(x_n) < f(1) + 1$ and thus $x_n$ is in $[a,b]$. Therefore by Bolzano-Weierstrass there is a subsequence $(x_{\varphi(n)})$ which converges to $x \in [a,b]$. Since $f$ is continuous, we have
$$
f(x_{\varphi(n)}) \xrightarrow[n \rightarrow +\infty]{} f(x).
$$
However, you don't need this to prove that $m \leq 3$. All you need to note is that $f(1) = 3$ and therefore $m = \inf_{x>0} f(x) \leq 3$.
