Proving $\lim_{h \to 0} \frac{|h|^\alpha}{h}=0$ given $\alpha > 1$.

Question

I've stumbled upon that limit in the book that I'm studying (Spivak's Calculus). At the current stage of the book, this is a trivial task if $\alpha \in \mathbf Q$.
But the problem statement doesn't specify anything about $\alpha$, except that $\alpha > 1$, so what if $\alpha \in \mathbf R$?

At the current stage there is no Chain Rule, L'Hopital's Rule, exp/log, integrals, series.

I've attempted to show that $$\lim_{h^+ \to 0} \frac{|h|^\alpha}{h}=\lim_{h^+ \to 0} h^{\alpha - 1}=0=\lim_{h^- \to 0} -(-h)^{\alpha - 1}=\lim_{h^- \to 0} \frac{|h|^\alpha}{h}.$$

I've tried:

1. Something like $0<h<\delta = \epsilon^{\frac{1}{\alpha - 1}} \implies 0<h^{\alpha - 1}<\epsilon$, but I'm not sure how to prove the general statement that $\forall x,y,a \in \mathbf R, a > 0, \ \ 0 \leq x < y \implies x^a < y^a$.
   Also I can only prove the existence of the roots like $\epsilon^{\frac{1}{n}}, n \in \mathbf N$.
2. Prove that $f(x)=x^\alpha, \alpha > 0$, is continuous.
   Using limits multiplication rule for that would not work for irrational $\alpha$.
   Showing continuity through differentiability is problematic, I can prove $(x^n)'=nx^{n-1}$ only for rational n (this also hinders the proof of the statement in 1., i.e. that $f(x)=x^\alpha, \alpha > 0$ is an increasing function on $x \geq 0$).

Nevertheless, the problem is given at this stage, so what am I missing here?

Answer 1

As your 1-st attempt solve problem, let me suggest some steps here:

a)We define $y=x^n$ for $n \in \mathbb{N}$, for $x \geqslant 0$ and proof, that it is non negative, increasing and continuous.

b) Define inverse function $y=x^{\frac{1}{n}}$, again increasing and continuous.

c) Define $y=x^r$ for $r \in \mathbb{Q}$ and prove it is increasing.

d) At last, define $y=x^a$ for $a \in \mathbb{R}$ and show it is increasing.

For proof let's take any $x \in \mathbb{R}$ and consider all $\alpha, \beta \in \mathbb{Q}$, for which holds $\alpha < x < \beta$. We define $a^x$, for $a>1$ as $y$ for which $a^{\alpha} \leqslant y \leqslant a^{\beta}$. It should be proved, that such $y$ exists and is unique. Then that it is increasing and continuous. As for us important is increasing property, then I consider only it in detail, assuming on first 2 you are agree.

Suppose $x_1<x_2$ are any numbers from $\mathbb{R}$. Exists such $\alpha, \beta \in \mathbb{Q}$, for which $x_1< \alpha < \beta < x_2$. From increasing property on $\mathbb{Q}$ (c) and definition of $a^x$ above we have $a^{x_1} \leqslant a^{\alpha} \leqslant a^{\beta} \leqslant a^{x_2} $.

Write, please, if some from these steps are not acceptable for you, or cause some doubts.

Answer 2

The squeeze theorem is the way to go: you know that $$ -1\le\dfrac{|h|}{h}\le 1 $$ so $$ -|h|^{\alpha-1}\le\dfrac{|h|^{\alpha}}{h}\le|h|^{\alpha-1} $$ Now the problem is reduced to show that the limit of $|h|^\beta$ is zero when $\beta>0$ and you should be able to reduce it to the case when $\beta$ is an integer, by choosing $\lfloor\beta\rfloor$ (but that depends on how the material is organized in Spivak’s manual).