Let $S=\{\frac{\sqrt{x}}{x+1}:x>0\}$. Show that $\inf{S}=0$. Let $S=\{\frac{\sqrt{x}}{x+1}:x>0\}$. Show that $\inf{S}=0$. 
My solution so far:
Since $\frac{\sqrt{x}}{x+1}>0$ for all $x>0$, $0$ is a lower bound of $S$. Next I know I need to show that $0$ is the largest lower bound of the set but I don't know how.

1) $0$ is a lower bound: $0<\dfrac{√x}{x+1}=:f(x)$, $x >0.$
2) Assume there is a greater lower bound $L >0$, with
$L <\dfrac{√x}{x+1}$, $x >0$.
Choose $x=1/n^2$;
$f(1/n)= (1/n)\dfrac{1}{1/n^2+1}<1/n$.
Archimedean principle;
There is a $n_0 \in \mathbb{N}$ s.t.
$n_0 >1/L$.
We have 
$0 < f(n_0) <1/n_0 < L$, for $x=1/n_0^2$, a contradiction.
Hence $\inf$ {$f(x)| x>0$}$=0$.
