Q:

Proof of power rule for limits?

I'm working through Spivak's Calculus book which proved the following:
$$\lim_{x \to a}\ (f+g)(x) = \lim_{x \to a}\ f(x) + \lim_{x \to a}\ g(x)$$
$$\lim_{x \to a} \ (f \cdot g)(x) = \lim_{x \to a}\ f(x) \cdot  \lim_{x \to a}\ g(x)$$
$$\lim_{x \to a} \ \Bigg( \frac{1}{g} \Bigg) (x) = \frac{1}{\lim \limits_{x \to a} \ g(x)}$$
However, the proof that
$$\lim_{x \to a}[f(x)^\alpha] = \left[\lim_{x \to a}f(x) \right]^\alpha$$
where $\alpha$ is a real number is missing.
It's easy to prove from the above properties when $\alpha$ is an integer, but what about otherwise? I've looked online and only found proofs when it is an integer.

A:

The question is: what is the definition of $a^{\alpha}$ for non-integer exponents? 
For rational exponents, $\alpha = \frac{p}{q}$ with $p$ and $q$ integers, $q\gt 0$, $\gcd(p,q)=1$, let's define $a^{\alpha}=a^{p/q}$ as (i) the unique real numbers $b$ such that $b^{q} = a^p$ if $q$ is odd; and (ii) the unique nonnegative real $b$ such that $b^q=a^p$ if $q$ is even and $a\geq 0$. (The expression is not defined for $q$ even and $a\lt 0$).
For rational exponents the limit rule can be deduced from the integer case. Write $\alpha=\frac{p}{q}$ with $p$ and $q$ integers, $\gcd(p,q)=1$. Assuming $f(x)$ is positive in a neighborhood of $a$ when $q$ is even, and that the limit
$$\lim_{x\to a}f(x)$$
exists, consider $g(x) = (f(x)^{\alpha})^q = f(x)^p$. Then we have (by the integer case),
$$\lim_{x\to a}g(x) = \lim_{x\to a}f(x)^p = \left(\lim_{x\to a}f(x)\right)^p.$$
On the other hand,
$$\lim_{x\to a}g(x) = \lim_{x\to a}\left(f(x)^{\alpha}\right)^q = \left(\lim_{x\to a}f(x)^{\alpha}\right)^q$$
provided that $\lim\limits_{x\to a}(f(x)^{\alpha})$ exists. 
Therefore, if that limit exists, we have:
$$\left(\lim_{x\to a}f(x)\right)^p = \left(\lim_{x\to a}f(x)^{\alpha}\right)^q.$$
Taking $q$th roots on both sides, we get
$$\left(\lim_{x\to a}f(x)\right)^{\alpha} = \lim_{x\to a}\left(f(x)^{\alpha}\right).$$
Added. To establish the existence of $\lim_{x\to a}f(x)^{\alpha}$, however, you would need to do some work with continuous functions. Spivak postpones it until Theorem 3 of Chapter 12, from which you can deduce that $y\mapsto y^{1/q}$ for integer $q$ is continuous where defined (because $x\mapsto x^q$ is continuous); the fact that the limit of $g$ exists then guarantees the limit of $g^{1/q} = f^{\alpha}$ also exists. So the above argument is somewhat "holey".
For irrational exponents, one definition is as a limit of rational exponents. That is, pick a sequence $q_n\to\alpha$ with $q_n$ rationals; then we define $$f(x)^{\alpha} = \lim_{n\to\infty}f(x)^{q_n}.$$
This leads to double limits, which are complicated. (We can only do this for positive bases).
Another definition is via the exponential and logarithm functions. We define $a^b = \exp(b\ln(a))$ (so only for positive bases again), and then use the fact that the exponential and logarithm functions are continuous, so 
$$\begin{align*}
\lim_{x\to a}\left(f(x)^{\alpha}\right) &= \lim_{x\to a}\exp\left(\alpha\ln(f(x))\right)\\
 &= \exp\left(\lim_{x\to a}\alpha\ln(f(x))\right)\\
&= \exp\left(\alpha\ln\left(\lim_{x\to a}f(x)\right)\right) \\
&= \left(\lim_{x\to a}f(x)\right)^{\alpha}\end{align*}$$
which gives the desired result.

