Let $S=\{\frac{\sqrt{x}}{x+1}:x>0\}$. Show that $\inf{S}=0$.

Question

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.

Answer 1

In order to show that $0$ is the greatest lower bound, you need to show that for any $\epsilon >0$ there exists an $x>0$ such that $$ \frac {\sqrt x}{x+1} <\epsilon$$

Note that $$\frac {\sqrt x}{x+1} <\sqrt x$$ therefore if you pick $x=\epsilon ^2$ then you have $$\frac {\sqrt x}{x+1} <\sqrt x=\epsilon$$

Thus the infimum is $0$

Answer 2

We have that $\forall x>0$

$$\frac{\sqrt{x}}{x+1}>0$$

and

$$\lim_{x\to \infty}\frac{\sqrt{x}}{x+1}=0$$

then $\inf{S}=0$.

Answer 3

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 $\inf$ {$f(x)| x>0$}$=0$.