一直以来，对于极限的概念都十分模糊，所以在解决此涉及极限的问题时，无从下手。

由于此前CHATGPT的数学能力一直不尽如人意，今天怀着试一试的心态，在对话框中输入了本题，令人惊讶的是，它居然给出了令人十分满意的答案。特此记录，献给CHATGPT！

题目如下：

If $a$ is a positive number, show that $$\lim _{x\rightarrow 0}\dfrac{a^{x}-1}{x}=\ln a$$Hint: The limit is a value of a certain derivative.

CHATGPT给出的解题步骤（详细且逻辑性极强）：

To show that $$\lim _{x\rightarrow 0}\dfrac{a^{x}-1}{x}=\ln a$$ we can use the definition of the derivative of a function at a point. Recall that the derivative of a function $f(x)$ at a point $x=c$ is given by $$f'(c)=\lim _{x\rightarrow c}\dfrac{f(x)-f(c)}{x-c}$$
Now, let $f(x)=a^x$, where $a$ is a positive number. Then, the derivative of $f(x)$ at $x=0$ is $$f'(0)=\lim _{x\rightarrow 0}\dfrac{f(x)-f(0)}{x-0}=\lim _{x\rightarrow 0}\dfrac{a^x-a^0}{x}=\lim _{x\rightarrow 0}\dfrac{a^x-1}{x}$$ Therefore, we have $$\lim _{x\rightarrow 0}\dfrac{a^{x}-1}{x}=f'(0)$$
But what is $f'(0)$? We can use the property of logarithms that says $$a^x=e^{x\ln a}$$
Then, we have $$f(x)=a^x=e^{x\ln a}$$
Using the chain rule, we can find the derivative of $f(x)$ as follows: $$f'(x)=\dfrac{d}{dx}(e^{x\ln a})=\dfrac{d}{dx}(x\ln a)\cdot \dfrac{d}{dx}(e^u)=\ln a\cdot e^{x\ln a}=\ln a\cdot a^x$$
Plugging in $x=0$, we get $$f'(0)=\ln a\cdot a^0=\ln a\cdot 1=\ln a$$
Therefore, we have $$\lim _{x\rightarrow 0}\dfrac{a^{x}-1}{x}=f'(0)=\ln a$$ Hence, we have shown that $$\lim _{x\rightarrow 0}\dfrac{a^{x}-1}{x}=\ln a$$
I hope this helps you understand the solution. 😊

可称之为完美解答