2
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.