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Synopsis

In Tao's Analysis 1, we are asked the following:

Let $\alpha$ be a real number, and let $f:(0,\infty) \to \mathbb{R}$ be the function $f(x) = x^{\alpha}$. Show that $\lim_{x \to 1} \frac{f(x)-f(1)}{x-1} = \alpha$ using Exercise 10.4.2 and the comparison principle; you may need to consider right and left limits separately. The fact that the rationals are dense in the reals might also help.

Now, Exercise 10.4.2 proves the result for $f(x) = x^q$ where $q \in \mathbb{Q}$ and the comparison principle is given as follows:

Lemma 6.4.13 (Comparison principle) - Suppose that $(a_n)_{n=m}^{\infty}$ and $(b_n)_{n=m}^{\infty}$ are two sequences of real numbers such that $a_n \leq b_n$ for all $n \geq m$. Then we have the inequalities $$\text{sup}(a_n)_{n=m}^{\infty} \leq \text{sup}(b_n)_{n=m}^{\infty}\\ \text{inf}(a_n)_{n=m}^{\infty} \leq \text{inf}(b_n)_{n=m}^{\infty} \\ \lim \text{sup}_{n \rightarrow \infty}a_n \leq \lim \text{sup}_{n \rightarrow \infty}b_n\\ \lim \text{inf}_{n \rightarrow \infty}a_n \leq \lim \text{inf}_{n \rightarrow \infty}b_n$$

It should be noted that Terence Tao builds the reals through Cauchy sequences of rational numbers, and defines real exponentiations in the following manner:

Definition 6.7.2 (Exponentiation to a real exponent) - Let $x> 0$ be real, and let $\alpha$ be a real number. We define the quantity $x^\alpha$ by the formula $x^{\alpha} = \lim_{n \to \infty} x^{q_n}$, where $(q_n)_{n=1}^{\infty}$ is any sequence fo rational numbers converging to $\alpha$.

Note that this exercise is given when we have not been taught L'Hopital's rule yet, and when we have yet even to prove that the derivative of $x^{\alpha} = \alpha x^{\alpha -1}$ for real $\alpha$. The previous exercise proved the statement for rationals and the exercise before that proved it for integers. This is not a question of computation but of generalization. Please take this into account.

My Attempt

I immediately thought of using double limits. Let $(q_k)_{k=1}^{\infty}$ be a sequence of rationals converging to $\alpha$. Then $$ \lim_{x \to 1} \frac{x^{\alpha}-1}{x-1} = \lim_{x \to 1} \frac{\lim_{k \to \infty}(x^{q_k})-1}{x-1} = \lim_{k \to \infty} \lim_{x \to 1} \frac{x^{q_k}-1}{x-1} = \lim_{k \to \infty} q_k = \alpha.$$ But this solution doesn't utilize the comparison principle or left and right limits or anything else Tao gives as hints except for the previous exercise. Furthermore, I'm not even sure if the limits are valid. Can any of you show me what's wrong with my solution (if it is wrong), or what Tao may have intended?

Update

Because of a helpful comment, I've realized that I can't abuse limits like I did above. But I'm out of ideas. I can't seem to think of anything for some reason. It's been bothering me for a hour. Any help would be appreciated!

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    $\begingroup$ You can't isolate limits into two different variables. One of these days I'll figure out the perfect sentence for that and recommend people get it tatooed onto their arms. $\endgroup$
    – fleablood
    Jan 26, 2021 at 3:17
  • $\begingroup$ Hmm. Any hints on how to solve this then? I've been stuck on this question for a while trying to figure out how to utilize his hints. Thanks for letting me know that I can't do that though ahaha $\endgroup$
    – mijucik
    Jan 26, 2021 at 3:18

3 Answers 3

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If you break the problem into approaching from the right and the left of 1, then the inequalities will be more clear I think.

Note that for $x>1$ $\frac{x^{\alpha}-1}{x-1}$ is increasing in the $\alpha$ variable so for $q>\alpha>p$ (p and q rational) the inequality:

$$\frac{x^{p}-1}{x-1} < \frac{x^{\alpha}-1}{x-1} < \frac{x^{q}-1}{x-1}$$

Then you may want to take a sequence approaching 1 from the right $x_{n}$ and with the comparison:

$$p = \lim\inf \frac{x_{n}^{p}-1}{x_{n}-1} \leq \liminf \frac{x_{n}^{\alpha}-1}{x_{n}-1} \leq \limsup \frac{x_{n}^{\alpha}-1}{x_{n}-1} \leq \limsup\frac{x_{n}^{q}-1}{x_{n}-1} = q$$

So then liminf and limsup of $\frac{x_{n}^{\alpha}-1}{x_{n}-1}$ are trapped between any such p and q rational. This is enough to show that the distance between the liminf and limsup is zero and the limit and $\alpha$ is zero, you can choose sequences $p_{k}$ and $q_{k}$ converging to alpha if you want to show this.

Approaching from the left the order of the inequalities will be reversed but it is the same idea.

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  • $\begingroup$ This helps a lot! Thank you. It's very late where I am but I'll try and see if I can come up with a proof tomorrow. You've given me a good idea of where to start now though :)) $\endgroup$
    – mijucik
    Jan 26, 2021 at 3:51
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It sounds like they want you to prove it for the integers, then the rationals, then the real numbers.

If $\alpha$ is a positive integer $x^\alpha - 1 = (x-1)(x^{\alpha-1} + x^{\alpha-2} + \cdots + 1)$

$\lim_\limits{x\to 1} \frac {x^{\alpha} - 1}{x-1} = \alpha$

To cover the negative integers, lets assume $\alpha > 0$ and find $\lim_\limits{x\to 1} \frac {x^{-\alpha} - 1}{x-1}$

$x^{-\alpha} - 1 = -x^{\alpha}(x^{\alpha} - 1)$

$\lim_\limits{x\to 1} \frac {-x^{\alpha}(x^{\alpha} - 1)}{x-1} = \lim_\limits{x\to 1} -x^{\alpha}\lim_\limits{x\to 1}\frac {x^{\alpha} - 1}{x-1} = -\alpha$

Next the rationals... Suppose $\alpha = \frac {1}{q}$ with $q\in \mathbb Z^+$

$x^{\frac {1}{q}} - 1 = \frac {(x^{\frac {1}{q}} - 1)(x^{q-1}{q} + x^{q-2}{q} + \cdots + 1)}{(x^{q-1}{q} + x^{q-2}{q} + \cdots + 1)} = \frac {x-1}{(x^{q-1}{q} + x^{q-2}{q} + \cdots + 1)}\\ \lim_\limits{x\to 1}\frac {x^{\frac {1}{q}} - 1}{x-1} = \frac {1}{q}$

Suppose $\alpha = \frac {p}{q}$

$x^{\frac {p}{q}} - 1 = \frac {x^p - 1}{(x^{q-1}{q} + x^{q-2}{q} + \cdots + 1)}\\ \lim_\limits{x\to 1}\frac {x^{\frac {p}{q}} - 1}{x-1} = \lim_\limits{x\to 1}\frac {1}{(x^{q-1}{q} + x^{q-2}{q} + \cdots + 1))}\lim_\limits{x\to 1}\frac {x^p - 1}{x-1} = \frac {p}{q}$

We have proven our proposition is true for all $\mathbb Q \setminus \{0\}$

Now we get into the the hints. Since the rationals are dense in the reals. For all real $r$ there is a rational that is arbitrarily close to it.

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Actually, I think here is the proof that Tao may have had in mind:

Let $(a_n)_{n=0}^{\infty}$ be an arbitrary sequence of positive real numbers converging to $1$ such that $a_n\neq{1}$ for all $n\in{\mathbb{N}}$. To prove that: $$\lim_{x\rightarrow{1};x\in{(0,\infty)\setminus{\{1\}}}}\frac{x^{\alpha}-1}{x-1}=\alpha$$ it will suffice by Proposition $9.3.9$ to show that the sequence: $$\biggr(\frac{a_n^{\alpha}-1}{a_n-1}\biggr)_{n=0}^{\infty}$$ converges to $\alpha$. By the axiom of countable choice, we can find a sequence of rationals $(p_m)_{m=0}^{\infty}$ converging to $\alpha$ such that $p_m<\alpha$ for all $m\geq{0}$ and similarly we can find a sequence of rationals $(q_m)_{m=0}^{\infty}$ converging to $\alpha$ "from above" (i.e. $\alpha<q_m$ for all $m\geq{0}$). Let $n,m\in{\mathbb{N}}$ be arbitrary. There are two cases: $a_n>1$ and $0<a_n<1$. If $a_n>1$, then we have by Proposition $6.7.3$ that $$a_n^{p_m}<a_n^{\alpha}<a_n^{q_m}$$ Thus $$a_n^{p_m}-1<a_n^{\alpha}-1<a_n^{q_m}-1$$ And because $a_n>1$, it follows that $a_n-1>0$, so we may "divide" the entire inequality by a positive number without needing to flip the direction of the inequality: $$\frac{a_n^{p_m}-1}{a_n-1}<\frac{a_n^{\alpha}-1}{a_n-1}<\frac{a_n^{q_m}-1}{a_n-1}$$ When $0<a_n<1$, then the above inequalities are still true (why?). But as $n$ was an arbitrary natural number, we thus conclude that for every $n\geq{0}$, the above inequalities are true. By the comparison principle, we thus have that $$\limsup_{n\rightarrow{\infty}}\frac{a_n^{p_m}-1}{a_n-1}\leq{\limsup_{n\rightarrow{\infty}}\frac{a_n^{\alpha}-1}{a_n-1}}\leq{\limsup_{n\rightarrow{\infty}}\frac{a_n^{q_m}-1}{a_n-1}}$$ But when a sequence is convergent, then its limit superior equals its limit, and furthermore for the leftmost and rightmost sequences above, we know what those limits are by Exercise $10.4.2$ (and Proposition $9.3.9$): $$p_m\leq{\limsup_{n\rightarrow{\infty}}\frac{a_n^{\alpha}-1}{a_n-1}}\leq{q_m}$$ but since $m\in\mathbb{N}$ was arbitrary, we know this inequality must hold for all $m\geq{0}$. Also, since $p_m$ and $q_m$ are clearly finite for all $m\geq{0}$, it follows as a corollary that the limit superior in the middle is also finite (or we would have a contradiction). Thus, we can view it as a constant sequence with respect to $m$. Now take limits on all sides of the inequality as $m\rightarrow{\infty}$ (in particular this is legitimate because all the sequences in question are convergent). By the squeeze test, we thus have that $\limsup_{n\rightarrow{\infty}}\frac{a_n^{\alpha}-1}{a_n-1}=\alpha$ But a similar argument to the above works for the limit inferior too, so that $\liminf_{n\rightarrow{\infty}}\frac{a_n^{\alpha}-1}{a_n-1}=\alpha$. But since the limit superior and limit inferior both agree and are both finite (since $\alpha$ is by hypothesis a finite real number), we conclude that the sequence itself converges to $\alpha$, as desired.

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