# $\lim_{x\to 1; x\in (0,\infty)-\{1\}} \frac{x^{\alpha} -1}{x-1}, \alpha\in\mathbb{R}$

I'm trying to evaluate this limit (taken from T.Tao's Analysis 1 book) without logarithms and without knowing that the function $f:(0,\infty)\to\mathbb{R}, f(x):=x^{\alpha}, \alpha\in\mathbb{R}$ is differentiable on $(0,\infty)$ (infact the book asks to use this limit to show that $f$ is differentiable with limit $f'(x)=\alpha x^{\alpha -1}$.

Now, I've tried to use the fact that $x^{\alpha}, \alpha\in\mathbb{R}$ is defined as $x^{\alpha}:=\lim_{n\to\infty} x^{q_n}$ where $(q_n)_{n=1}^\infty$ is any sequence of rational numbers converging to $\alpha$ (hence a bounded sequence) and the fact (that I've already proved) that $\lim_{x\to 1; x\in (0,\infty)-\{1\}} \frac{x^q-1}{x-1}=q\ \forall q\in\mathbb{Q}$ to use the squeeze theorem somehow but I haven't gotten anywhere so far.

Any hints?

Best regards,

lorenzo.

Using the fact that you have proved, define two convergent sequences $a_n, b_n$ in $\Bbb Q$ such that $a_n$ is strictly increasing to $\alpha$, and $b_n$ is strictly decreasing to $\alpha$, then $a_n<b_n$ for all $n$. And we have to consider the left and right limit:
For $x>1$, we have $$\frac{x^{a_n}-1}{x-1}<\frac{x^\alpha -1}{x-1}<\frac{x^{b_n}-1}{x-1}$$ Taking limit $x\to 1$, we have $$a_n\le \lim_{x\to 1}\frac{x^\alpha -1}{x-1}\le b_n$$ Then take $n$ to infinity and apply Squeeze Theorem;
For $0<x<1$, we have $$\frac{x^{a_n}-1}{x-1}>\frac{x^\alpha -1}{x-1}>\frac{x^{b_n}-1}{x-1}$$ And repeat the same step