Supremum and Infimum of set I just got the set and Ι tried to find the supremum and infimum and prove it. $$(x-2)\sqrt{\frac{x+1}{x-1}} \quad \text{ for } \quad 2< x\leq 54$$
I succeed to get to this set $\frac{x-2}{x-1}\sqrt{x^{2}-1}$ but I'm stuck.
What can I do now ?
Thanks.

On one hand, since $x>2$ we have 
$$
(x-2)\sqrt{\frac{x+1}{x-1}}>0.
$$
On the other hand
$$
\lim_{x\rightarrow 2^+}(x-2)\sqrt{\frac{x+1}{x-1}}=0.
$$
Hence
$$
\inf_{2<x\leq 54}(x-2)\sqrt{\frac{x+1}{x-1}}=0.
$$
Since $2<x\leq 54$
$$
(x-2)\sqrt{\frac{x+1}{x-1}}=\frac{x-2}{x-1}\sqrt{x^2-1}=\left(1-\frac{1}{x-1}\right)\sqrt{x^2-1}\leq \left(1-\frac{1}{54-1}\right)\sqrt{54^2-1}.
$$
Hence
$$
\sup_{2<x\leq 54}(x-2)\sqrt{\frac{x+1}{x-1}}=\max_{2<x\leq 54}(x-2)\sqrt{\frac{x+1}{x-1}}=\left(1-\frac{1}{54-1}\right)\sqrt{54^2-1}.
$$
