I would like to know the sign the function $f: [1,\infty)\to \Bbb R$ given by $$f(t) = (t-1)^{s/2}-t^{s/2}+1$$ with $0<s<1.$ Differentiating I have $$f(t) = \frac{s}{2}\left[(t-1)^{s/2-1}-t^{s/2-1}\right]\ge 0$$ indeed, $0<t-1<t$ and $-1<\frac{s}{2}-1<-\frac{1}{2}<0.$
Therefore the function $f$ is increasing and we have $ f(1) =0$
Hence I could conclude that $$ f(t)\ge 0~~~~t\ge 1. $$
Problem: Patently in the following I found a contradictory argument. Since $$ \lim_{t\to \infty} f(t)=\lim_{t\to \infty}[(t-1)^{s/2}-t^{s/2}+1]= \lim_{t\to 0^+}\frac{(1-t)^{s/2}-1}{t^{s/2}}+1\\\sim \lim_{t\to 0^+}\frac{-{s/2}}{t^{s/2-1}}+1 = -\infty$$
The problem is that, according to the above result, this limit of $f$ at infinity limit should instead $\lim_{t\to \infty} f(t)=\infty$. Rather this give a contradiction.
Which of the following is false ? $ f(t)\ge 0~~~~t\ge 1. $ or $\lim_{t\to \infty} f(t)=-\infty$. and why ? what did I miss.
Patently I am facing the same paradox with the function $g(x) = (x+1)^{s/2}-x^{s/2}-1~~~x>0. $ where I also got $ g(x)\le 0~~~~x\le 0. $ and $\lim_{x\to \infty} g(x)=\infty$.