极限与导数（献给CHATGPT） - Koichi's free world for solitary meditation

使用自定义字体使用自定义字体

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