What do I miss on this function? $f(t) = (t-1)^{s/2}-t^{s/2}+1$ 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 $0&lt;s&lt;1.$ Differentiating I have 
$$f(t) = \frac{s}{2}\left[(t-1)^{s/2-1}-t^{s/2-1}\right]\ge 0$$ 
indeed, $0&lt;t-1&lt;t$ and $-1&lt;\frac{s}{2}-1&lt;-\frac{1}{2}&lt;0.$

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&gt;0. $ where I also got $ g(x)\le 0~~~~x\le 0. $ and  $\lim_{x\to \infty} g(x)=\infty$.
$\begin{align}
f(t) &amp; = t^{s/2} \left(1-\frac{1}{t} \right)^{s/2} - t^{s/2}+1 \\
&amp; = t^{s/2} \left[ 1 - \frac{s}{2t} + \frac{s/2(s/2-1)}{2!} \frac{1}{t^2} - ... \right] - t^{s/2}+1 \\
&amp; = \frac{s}{2} t^{s/2 - 1} + \frac{s/2(s/2-1)}{2!} t^{s/2 -2} - ... + 1 \\
\end{align}
$

But since $0 &lt; \frac{s}{2} &lt; \frac{1}{2}$, all powers of $t$ in the above are strictly less than $0$, from this I think it's clear that 
$\lim_{t \to \infty} f(t) = 1$.
You found the result of the limit of $f(x)$ wrong.

In infinity, we have:
$\lim_{t\to \infty} (t-1)^{c}\sim\lim_{t\to \infty} t^{c}$

So for the limit we can simplify like below:
$\lim_{t\to \infty} f(t)=\lim_{t\to \infty}[(t-1)^{s/2}-t^{s/2}+1]= \lim_{t\to \infty}[t^{s/2}-t^{s/2}+1]=1 &gt;0$