求助：证明定义域为ℝ的可导函数的特定极限等式

First off, I suspect there's a tiny typo in your problem statement: if the limit is truly $\lim_{x\to 0}$, then as $a>0$, $\ln x - \ln a$ blows up to $-\infty$, and the base $\frac{f(x)}{f(a)}$ approaches $\frac{f(0)}{f(a)}$ (we only know $f$ is continuous at $a$, not necessarily at 0). This scenario doesn't align with the conclusion $\frac{f'(a)}{f(a)} \cdot a$ at all. The far more logical limit here is $\lim_{x\to a}$—that makes this a classic $1^\infty$ indeterminate form, which matches the result you're supposed to prove. I'll proceed with $\lim_{x\to a}$ below; if you really do mean $x\to 0$, feel free to add more context about $f(0)$.

Step 1: Use the Natural Log Trick for Exponential Limits #

For limits of the form $\lim u^v$ (especially indeterminate ones like $1^\infty$), taking the natural log is always a solid first move—it lets us turn the exponent into a manageable fraction:

Let $L = \lim_{x\to a} \left( \frac{f(x)}{f(a)} \right)^{\frac{1}{\ln x - \ln a}}$

Take the natural log of both sides:
$$
\ln L = \lim_{x\to a} \frac{\ln\left( \frac{f(x)}{f(a)} \right)}{\ln x - \ln a} = \lim_{x\to a} \frac{\ln f(x) - \ln f(a)}{\ln x - \ln a}
$$

Step 2: Apply the Definition of the Derivative #

When $x\to a$, both the numerator and denominator approach 0:

• The numerator: $\ln f(x) - \ln f(a) \to 0$ because $f$ is continuous at $a$ (since it's differentiable there) and $\ln t$ is continuous for $t>0$ (and $f(a)>0$).
• The denominator: $\ln x - \ln a \to 0$ because $\ln x$ is continuous at $a>0$.

This is a $\frac{0}{0}$ indeterminate form, so we can rewrite the limit as a ratio of two derivatives (using the definition directly, no need for L'Hospital's Rule if you want to stick to basics):
$$
\ln L = \lim_{x\to a} \frac{\frac{\ln f(x) - \ln f(a)}{x - a}}{\frac{\ln x - \ln a}{x - a}}
$$

Now, recognize each part as a derivative:

• The top fraction's limit is the derivative of $\ln f(x)$ at $x=a$. By the chain rule, that's $\frac{f'(a)}{f(a)}$.
• The bottom fraction's limit is the derivative of $\ln x$ at $x=a$, which is $\frac{1}{a}$.

Plug those in:
$$
\ln L = \frac{\frac{f'(a)}{f(a)}}{\frac{1}{a}} = \frac{a f'(a)}{f(a)}
$$

Step 3: Convert Back to the Original Limit #

To get $L$, we just exponentiate both sides:
$$
L = e^{\frac{a f'(a)}{f(a)}}
$$

Wait a second—this doesn't match the conclusion you wrote ($\frac{f'(a)}{f(a)} \cdot a$). That means either:

1. The problem statement has a typo (maybe the exponent is $\frac{1}{x - a}$ instead of $\frac{1}{\ln x - \ln a}$? That would give $e^{\frac{f'(a)}{f(a)}}$), or
2. The intended conclusion was $e^{\frac{a f'(a)}{f(a)}}$ instead of the product you listed.

Clearing Up Your "Limit = 1" Confusion #

You mentioned thinking the limit is 1 when $a>0$—that's a common mistake with $1^\infty$ forms. Just because the base approaches 1 doesn't mean the whole limit is 1; when the exponent is blowing up to infinity, even tiny deviations from 1 in the base can lead to a non-1 result. We have to use the log trick to properly evaluate the indeterminate form instead of making assumptions.

内容的提问来源于stack exchange，提问作者Mat Research