指数未定式

Cases of the form $f(x)^{g(x)}$ are strange because they are often not seen as indeterminate. $0^0$ is generally assigned an integer value, $1$, whereas forms such as $0/0$ do not have such a value associated with them. $0^0$ is given this value mainly for combinatorial reasons; many formulas such as the binomial theorem work nicer if we just make it equal to $1$. Regardless of this standard definition, $\lim_{x\rightarrow a}f(x) = 0$ and $\lim_{x\rightarrow a}g(x) = 0$ do not together imply that $\lim_{x\rightarrow a}f(x)^{g(x)} = 1$. For example, $\lim_{t \to 0^+} {t}^{t} = 1, \\\lim_{t \to 0^+} \left(e^{-\frac{1}{t^2}}\right)^t = 0, \\\lim_{t \to 0^+} \left(e^{-\frac{1}{t^2}}\right)^{-t} = +\infty, \\ \lim_{t \to 0^+} \left(e^{-\frac{1}{t}}\right)^{at} = e^{-a}.$ To deal with such limits, similarly to when taking the derivative of something like $x^x$, we convert the exponent into something we know how to deal with, a product. $f(x)^{g(x)}$ becomes $e^{f(x)ln(g(x))}$. $e^x$ is defined and continuous for all $x \in \mathbb{R}$, so $e^{\lim_{x\rightarrow a}f(x)ln(g(x))}=\lim_{x\rightarrow a} e^{f(x)ln(g(x))}=\lim_{x\rightarrow a} f(x)^{g(x)}$ as desired. The limit of $f(x)ln(g(x))$ is then easily computed with L'Hospital's rule.

In this exercise we prove that for any positive reals $a_1,\ldots,a_n$, we have $\lim_{x\to0} \left(\frac{a_1^x+\cdots+a_n^x}{n}\right)^{1/x} = (a_1\cdots a_n)^{1/n}$, or in other words that "the geometric mean is the $0$th power mean". (See also here.)

Why is this an exponential indeterminate form?

Because $a^x \to 1$ (as $x\to0$) for any positive real $a$, we have $\lim_{x\to0}\left(\frac{a_1^x+\cdots+a_n^x}{n}\right) = 1$.

So for $x\approx 0$, the "base" $\frac{a_1^x+\cdots+a_n^x}{n} \approx 1$, and the exponent $1/x$ is either "very positive" or "very negative" (since $x$ is small). So we have an exponential indeterminate form $1^{\pm\infty}$.

We could take logarithms and use l'Hopital's rule, but below we present a more instructive approach using linear approximation.

So we'll need to be more precise to evaluate the limit: for instance, by linear approximation, $a^x \approx 1 + (\ln{a}) x$ for small $x\approx 0$ (we can worry about the error later, but it's not so important here).

Averaging the $n$ linear approximations for $a_1^x,\ldots,a_n^x$ gives us a more precise estimate on the "base", i.e. $\frac{a_1^x+\cdots+a_n^x}{n} \approx 1 + C x$, where $C = \frac{\ln{a_1}+\cdots+\ln{a_n}}{n}$ is the average of the "slopes" in the approximation.

So we basically want to find $\lim_{x\to0} (1+Cx)^{1/x}$. Recall from another exercise that the difficulty is that $(1+Cx)^{1/x}$ is hard to work with algebraically (it's hard to simultaneously account for the base and exponent); there are a couple standard ways to deal with this.

1. For instance, we could use the classical linear vs. exponential approximation $1+Cx \approx e^{Cx}$ (which is good when $x\approx0$ is small), so that as $x\to0$ we have $(1+Cx)^{1/x} \approx (e^{Cx})^{1/x} = e^C \to e^C$.

2. Alternatively, since fractions are often easier to think about the exponentials, we can take logarithms: $\lim_{x\to0} (1+Cx)^{1/x} = \exp(\lim_{x\to0} \frac{\log(1+Cx)}{x})$. And we know $\log(1+Cx) \approx Cx$ for small $x$, by linear approximation (alternatively, this is equivalent to $1+Cx \approx e^{Cx}$). So $\lim_{x\to0} (1+Cx)^{1/x} = \exp(C)$.

Finally, $e^C = e^{\frac{\ln{a_1}+\cdots+\ln{a_n}}{n}} = (e^{\ln{a_1}+\cdots+\ln{a_n}})^{1/n} = (a_1\cdots a_n)^{1/n}$, establishing the desired limit from the beginning.