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!
Best Answer
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.