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.

We have that $\forall x>0$
$$\frac{\sqrt{x}}{x+1}>0$$
and 
$$\lim_{x\to \infty}\frac{\sqrt{x}}{x+1}=0$$
then $\inf{S}=0$.
