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Let: $$ \begin{align} &\space \color{red}{f(x)=x^{\alpha}} \quad\colon\space\alpha\in\mathbb{R} \\[2mm] &\qquad \lim_{x\rightarrow\infty}f'(x)=\lim_{x\rightarrow\infty}\alpha\,x^{\alpha-1}=0 \qquad\qquad\qquad\qquad\qquad\colon\space\alpha\lt1 \quad{\small\text{ and }}\rightarrow\infty\space\colon\space\alpha\gt1 \\[2mm] &\qquad \lim_{x\rightarrow\infty}\left[f(x+1)-f(x)\right]=\lim_{x\rightarrow\infty}\left[(x+1)^{\alpha}-x^{\alpha}\right]=0 \quad\colon\space\alpha\lt1 \quad{\small\text{ and }}\rightarrow\infty\space\colon\space\alpha\gt1 \end{align} $$

Prove/Explain, why the resulting subtraction limit exists on an extra step? $$ \lim_{x\rightarrow\infty}\left[f(x+1)-f(x)-f’(x)\right]=0\color{red}{\quad\colon\space\alpha\lt2} $$

More generally, $$ \lim_{x\to\infty}f(x+1)-f(x)-f'(x)-\frac12f''(x)-\dots-\frac1{n!}f^{(n)}(x)=0\quad:\ \alpha<n $$

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    $\begingroup$ More generally, can you prove that$$\lim_{x\to\infty}f(x+1)-f(x)-f'(x)-\frac12f''(x)-\dots-\frac1{n!}f^{(n)}(x)=0\quad:\ \alpha<n$$ $\endgroup$ – Simply Beautiful Art Feb 1 '17 at 22:43
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The general statement, originally proposed by @Simply Beautiful Art, follows by Taylor's theorem using Lagrange form of the remainder.

Let me conjecture that the general result is true not only for $\alpha<n$ but also for $\alpha<n+1$, since the remainder involves $(n+1)$th derivative, that is, $(\alpha-n-1)$th power, which needs to be negative for the remainder to converge.

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  • $\begingroup$ @SimplyBeautifulArt You are of course correct, my previous answer was correct but irrelevant to the question. $\endgroup$ – Jan Feb 1 '17 at 23:19
  • $\begingroup$ @SimplyBeautifulArt $2:0$ for you. I was again wrong. The new answer is my third try, but maybe the history should have taught me? $\endgroup$ – Jan Feb 2 '17 at 0:05
  • $\begingroup$ Good job! Personally, this would've been my answer! Note you may now proof the comment I posted below the original question! $\endgroup$ – Simply Beautiful Art Feb 2 '17 at 0:20
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I find that using the substitution $t=1/x$ usually makes things clearer. The limit becomes \begin{align} \lim_{t\to0^+}\left(\frac{(1+t)^\alpha}{t^\alpha}-\frac{1}{t^\alpha}-\frac{\alpha}{t^{\alpha-1}}\right) &= \lim_{t\to0^+}\frac{(1+t)^\alpha-1-\alpha t}{t^\alpha}\\ &= \lim_{t\to0^+}\frac{1+\alpha t+\binom{\alpha}{2}t^2+o(t^2)-1-\alpha t}{t^\alpha}\\ &=\lim_{t\to0^+}\binom{\alpha}{2}t^{2-\alpha} \end{align}

Now you should clearly see where $2$ enters the scene.

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