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

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    $\begingroup$ How did you came up with last limit? $\endgroup$ – user518372 Jan 14 '18 at 11:25
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    $\begingroup$ $1/t^{s/2-1}=t^{1-s/2}\to 0$ as $t\to 0$. That $f(t)$ is increasing does not necessarily imply $\lim_{t\to\infty}f(t)=+\infty$. $\endgroup$ – mathlove Jan 14 '18 at 11:56
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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, $$ \lim_{u\to0^+}\frac{(1-u)^{s/2}-1}{u^{s/2}}= \lim_{u\to0^+}\frac{-(s/2)u}{u^{s/2}}= \lim_{u\to0^+}-\frac{s}{2}u^{1-s/2}=0 $$

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$\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 $\lim_{t \to \infty} f(t) = 1$.

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  • $\begingroup$ where did the minus in the second line went to ? I mean the fiorst fact in first line should be $-s/2t^{s/2-1}$ $\endgroup$ – Guy Fsone Jan 14 '18 at 22:13
  • $\begingroup$ I'm not quite sure what you mean? You've defined both $f(t)$ and its derivative as $f(t)$, so it's pretty unclear which function you're talking about here. From your question and other people's comments, I assumed that you wanted to disprove the statement that $\lim_{t \to \infty} f(t) = -\infty $, with $f(t)$ as you have initially defined it. $\endgroup$ – user202654 Jan 14 '18 at 23:47
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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:
$\lim_{t\to \infty} f(t)=\lim_{t\to \infty}[(t-1)^{s/2}-t^{s/2}+1]= \lim_{t\to \infty}[t^{s/2}-t^{s/2}+1]=1 >0$

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  • $\begingroup$ Thanks for giving another wrong answer? You too you compute it in a wrong way. with you logic we have $2x+1=(x+1)^2-x^2 \to 0$ which is completely wrong $\endgroup$ – Guy Fsone Jan 14 '18 at 11:38
  • $\begingroup$ Dude!!! First of all, you can also compute it in Matlab or Mathematica. (I did it right now to double check and the answer was again 1). Second, for your example, $(x+1)^2 - x^2$ is $\infty - \infty$ and cannot be considered 0! $\endgroup$ – Mehrdad Zandigohar Jan 14 '18 at 11:46
  • $\begingroup$ Guy's point is that this is the same mistake you have made in equating $(t-1)^c$ with $t^c$. $\endgroup$ – Rahul Jan 14 '18 at 12:53
  • $\begingroup$ @MehrdadZandigohar i am just pointing out where you did wrong with elementary argument. also note all that softwares (Matlab, mathematica .....) are mankind made. I count the number of time i got wrong answer from software. excuse me if i dont trust them at 100% $\endgroup$ – Guy Fsone Jan 14 '18 at 13:34

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