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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.

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More general power rules require you to take care about the sign of $f(x)$ and how you define $y^\alpha$. For arbitrary $\alpha$ and $y>0$, this tends to be defined as $y^\alpha = e^{\alpha \log y}$. But that is stepping outside of beginning calculus, so it tends to not be covered when you are first learning limits. –  Thomas Andrews Nov 18 '11 at 20:19

1 Answer 1

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.

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You can only get from $\lim_{x\to a}\left(f(x)^{\alpha}\right)^q = \left(\lim_{x\to a}f(x)^{\alpha}\right)^q$ if you know that $\lim_{x\to a} f(x)^\alpha$ exists, which you haven't yet shown. So you can use the integer theorem to show that, for rational values, when the limit exists, you have equality. –  Thomas Andrews Nov 18 '11 at 20:33
    
@Thomas: True; I didn't check Spivak and assumed this was stated but not proven (in which case the theorem would have said the limits exist). Let me fix it. –  Arturo Magidin Nov 18 '11 at 20:38

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