Let $\alpha\in \mathbb{R}$ and $f:(0,\infty)\to\mathbb{R}$, $f(x)=x^\alpha$. Show $\lim_{x\to 1} \frac{f(x)-1}{x-1}=\alpha$
I have a definition for the function $x^\alpha$ for irrationals as $\lim_{n\to\infty} x^{q_n}$, where $(q_n)_{n=1}^\infty$ is any sequence of rational numbers converging to $\alpha$
I proved that $\lim_{x\to 1} \frac{x^q-1}{x-1}=q$ previously and I wanted to then say:
$\lim_{x\to 1} \frac{f(x)-1}{x-1}=\lim_{n\to\infty}\lim_{x\to 1} \frac{x^{q_n}-1}{x-1}=\lim_{n\to\infty}q_n=\alpha$ for some sequence $(q_n)_{n=1}^\infty$ converging to $\alpha$.
I'm not sure if this is sufficient for a proof though, I'm not sure if this step is fully justified just by using the definition of exponentiation I'm given, or if it even makes sense as a limit.
$$\lim_{x\to 1} \frac{f(x)-1}{x-1}=\lim_{n\to\infty}\lim_{x\to 1} \frac{x^{q_n}-1}{x-1}$$
Best Answer
The evaluation of the standard limit $$\lim_{x\to 1}\frac{x^{\alpha}-1}{x-1}=\alpha\tag{1}$$ is a difficult problem and needs a little more effort than one can imagine by looking at the problem statement.
First we begin by proving some inequalities in a systematic fashion.
Let $s, t$ be real numbers such that $0<s<1<t$ and let $\alpha$ be a positive integer. Then we can see that $$1+t+t^2+\dots+t^{\alpha - 1}<\alpha t^{\alpha} $$ Multiplying by the positive number $(t-1)$ we get $$t^{\alpha}-1<\alpha t^{\alpha }(t-1)$$ Adding $\alpha(t^{\alpha} - 1)$ to both sides and dividing by $\alpha(\alpha+1)$ we get $$\frac{t^{\alpha} - 1}{\alpha}<\frac{t^{\alpha+1}-1}{\alpha+1}$$ It follows that if $\alpha, \beta$ are positive integers with $\alpha<\beta$ then $$\frac{t^{\alpha} - 1}{\alpha}<\frac{t^{\beta}-1}{\beta}\tag{2}$$ and similarly one can prove that $$\frac{1-s^{\alpha}}{\alpha}>\frac{1-s^{\beta}}{\beta}\tag{3}$$ The inequalities $(2)$ and $(3)$ also hold for positive rational numbers $\alpha, \beta$ with $\alpha<\beta $. To establish this note that if $\alpha=a/b, \beta=c/d$ where $a, b, c, d$ are positive integers and $ad<bc$ then we can replace $t$ by $t^{1/bd}$ in $(2)$ and get $$\frac {(t^{1/bd})^{ad}-1}{ad}<\frac{(t^{1/bd})^{bc}-1}{bc}$$ or $$\frac {t^{\alpha} - 1}{\alpha}<\frac{t^{\beta} - 1}{\beta}$$ In this manner the inequalities $(2),(3)$ are established for all positive rational $\alpha, \beta$ with $\alpha <\beta$.
Next we show that the same inequalities hold for positive real numbers $\alpha, \beta$ with $\alpha<\beta$. Clearly we can find two rationals $p, q$ such that $\alpha<p<q<\beta$. Let $a_n, b_n$ be sequences of rationals tending to $\alpha, \beta$ respectively. Then from a certain value of $n$ onwards we must have $$0<a_n<p<q<b_n$$ and therefore we have $$\frac{t^{a_n} - 1}{a_n}<\frac{t^p-1}{p}<\frac{t^q-1}{q}<\frac{t^{b_n}-1}{b_n}$$ Letting $n\to\infty $ we get $$\frac{t^{\alpha}-1}{\alpha}\leq \frac{t^{p} - 1}{p}<\frac{t^{q}-1}{q}\leq\frac{t^{\beta}-1}{\beta}$$ so that the inequality $(2)$ is established for positive real numbers $\alpha, \beta$ with $\alpha< \beta $. The same holds for $(3)$ also.
Thus we have proved that if $s, t, \alpha, \beta$ are real numbers with $$0<s<1<t,\,\,0<\alpha<\beta$$ then $$\frac{t^{\alpha} - 1}{\alpha}<\frac{t^{\beta} - 1}{\beta}\tag{4}$$ and $$\frac{1-s^{\alpha}}{\alpha}>\frac{1-s^{\beta}}{\beta}\tag{5}$$ Let us now put $\beta=1$ to get $$t^{\alpha} - 1<\alpha(t-1),\, 1-s^{\alpha}>\alpha (1-s)\tag{6}$$ for $0<\alpha<1$. Similarly putting $\alpha=1$ in $(4),(5)$ we get $$t^{\beta} - 1>\beta(t-1),\,1-s^{\beta}<\beta(1-s)\tag{7}$$ for $\beta>1$. Putting $t=1/s,s=1/t$ in $(6),(7)$ we get $$1-s^{\alpha}<\alpha s^{\alpha-1} (1-s),\,t^{\alpha} - 1>\alpha t^{\alpha-1}(t-1)\tag{8}$$ and $$1-s^{\beta}>\beta s^{\beta-1}(1-s),\,t^{\beta} - 1<\beta t^{\beta-1}(t-1)\tag{9}$$ We can rewrite the inequalities $(6),(7),(8),(9)$ in following manner. If $t, \alpha$ are real numbers with $t>1,\alpha>1$ then $$\alpha(t-1)<t^{\alpha}-1<\alpha t^{\alpha-1}(t-1)\tag{10}$$ and if $0<\alpha <1$ then $$\alpha t^{\alpha - 1}(t-1)<t^{\alpha}-1<\alpha(t-1)\tag{11}$$ If $s, \alpha$ are real numbers with $0<s<1,\alpha>1$ then $$\alpha s^{\alpha - 1}(1-s)<1-s^{\alpha}<\alpha(1-s)\tag{12}$$ and if $0<\alpha <1$ then $$\alpha(1-s)<1-s^{\alpha} <\alpha s^{\alpha - 1}(1-s)\tag{13}$$
Next we prove that $$\lim_{x\to 1}f(x)=1\tag{14}$$ The case when $\alpha$ is an integer follows from limit laws. Let $\alpha $ be a positive real number with $0<\alpha<1$ and let $x>1$. Then we have via the inequality $(11)$ $$0<x^{\alpha}-1<\alpha(x-1)$$ Taking limits as $x\to 1^{+}$ and applying Squeeze Theorem we get $\lim_{x\to 1^{+}}f(x)=1$. The case $x\to 1^{-}$ can be handled by putting $x=1/t$ or using squeeze theorem with inequality $(12)$.
If $\alpha >1$ then let $n$ be an integer greater than $\alpha - 1$ and $x>1$. Using inequality $(10)$ we get $$0<x^{\alpha}-1<\alpha x^n(x-1)$$ By using Squeeze Theorem we get $\lim_{x\to 1^{+}}f(x)=1$. The case $x\to 1^{-}$ can be handled as explained earlier. It follows that result $(14)$ has been established for all non-negative real values of $\alpha$. For negative values of $\alpha $ just note that $f(x) =1/x^{-\alpha}$ and the result easily follows.
We can now prove the target result $(1)$. Let $x>1$ and $\alpha>1$. Then we see from inequality $(10)$ that $$\alpha <\frac {x^{\alpha} - 1}{x-1}<\alpha x^{\alpha - 1}$$ Taking limits as $x\to 1^{+}$ and using $(14)$ we get desired result $(1)$ via Squeeze Theorem. The case $x\to 1^{-}$ can be handled using inequality $(12)$. The case when $0<\alpha<1$ can be handled using inequalities $(11),(13)$. For negative values of $\alpha $ we need to note that $f(x) =1/x^{-\alpha}$.
Note also that the limit in question is essentially about derivative of $f(x) $ and the above constitutes a proof of $f'(1)=\alpha$ which further leads to $f'(x) =\alpha x^{\alpha-1}$ for all real $\alpha$.
The above presentation also shows that the treatment of real exponents via the route of exponential and logarithmic functions is simpler. However the above can not be avoided in case we are dealing with rational $\alpha$ (another route is use of inverse function theorem).