What do I miss on this function? $f(t) = (t-1)^{s/2}-t^{s/2}+1$

Question

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 $\lim_{x\to \infty} g(x)=\infty$.

Answer 1

Let's do it more slowly; you're doing the substitution $t=1/u$, so the limit becomes $$ \lim_{u\to0^+} \left[ \left(\frac{1}{u}-1\right)^{s/2}-\frac{1}{u^{s/2}}+1 \right]= \lim_{u\to0^+}\left[ 1+\frac{(1-u)^{s/2}-1}{u^{s/2}} \right] $$ Now, using l'Hôpital, $$ \lim_{u\to0^+}\frac{(1-u)^{s/2}-1}{u^{s/2}} = \lim_{u\to0^+}\frac{(s/2)(1-u)^{s/2-1}}{(s/2)u^{s/2-1}}= \lim_{u\to0^+}(1-u)^{s/2-1}u^{1-s/2}=0 $$

Where are you making a mistake? For $0<s<1$, you have $$ \frac{s}{2}-1<-\frac{1}{2} $$ so your last denominator has limit infinity and the fraction has limit $0$.

Using Taylor, $$ \lim_{u\to0^+}\frac{(1-u)^{s/2}-1}{u^{s/2}}= \lim_{u\to0^+}\frac{-(s/2)u}{u^{s/2}}= \lim_{u\to0^+}-\frac{s}{2}u^{1-s/2}=0 $$

Answer 2

$\begin{align} f(t) & = t^{s/2} \left(1-\frac{1}{t} \right)^{s/2} - t^{s/2}+1 \\ & = t^{s/2} \left[ 1 - \frac{s}{2t} + \frac{s/2(s/2-1)}{2!} \frac{1}{t^2} - ... \right] - t^{s/2}+1 \\ & = \frac{s}{2} t^{s/2 - 1} + \frac{s/2(s/2-1)}{2!} t^{s/2 -2} - ... + 1 \\ \end{align} $

But since $0 < \frac{s}{2} < \frac{1}{2}$, all powers of $t$ in the above are strictly less than $0$, from this I think it's clear that $\lim_{t \to \infty} f(t) = 1$.

Answer 3

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:
$\lim_{t\to \infty} f(t)=\lim_{t\to \infty}[(t-1)^{s/2}-t^{s/2}+1]= \lim_{t\to \infty}[t^{s/2}-t^{s/2}+1]=1 >0$