关于在不连续点使用导数极限定义的技术问询

Hey there, great question—let's break this down intuitively since you already know the formal proof that differentiability implies continuity.

First, let's recall the limit definition of the derivative at a point ( x = a ):
[
f'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h}
]

Now, if ( f ) is not continuous at ( a ), that means at least one of two core issues is at play:

• ( f(a) ) doesn't exist (the function has a hole or isn't defined at ( a )), or
• ( \lim_{h \to 0} f(a+h) \neq f(a) ) (the function has a jump or oscillates wildly near ( a ))

Let's walk through both scenarios to see what happens when we apply the derivative limit definition:

Case 1: ( f(a) ) is undefined #

If the function isn't even defined at ( a ), the numerator ( f(a+h) - f(a) ) is immediately meaningless—you can't compute a difference involving a value that doesn't exist. The entire limit expression is undefined from the start, so there's no way to get a determinate or indeterminate form here.

Case 2: ( \lim_{h \to 0} f(a+h) \neq f(a) ) #

Here, as ( h ) approaches 0, ( f(a+h) ) approaches some value ( L ) that's not equal to ( f(a) ). That means the numerator ( f(a+h) - f(a) ) approaches ( L - f(a) ), a non-zero constant. The denominator ( h ) approaches 0, so we're looking at a limit of the form ( \frac{\text{non-zero constant}}{0} ).

This doesn't give an indeterminate form—instead, it leads to one of three outcomes:

• The limit blows up to ( +\infty ) (if the numerator and denominator have matching signs as ( h \to 0 ))
• The limit blows up to ( -\infty ) (if signs don't match)
• The left-hand and right-hand limits don't agree (so the overall limit doesn't exist)

You mentioned thinking the result might be a "determinate but incorrect" value—let's clarify that. That scenario only happens if you mistakenly assume the function is continuous at ( a ), forcing ( \lim_{h \to 0} f(a+h) = f(a) ) (which makes the numerator approach 0, creating the ( \frac{0}{0} ) indeterminate form). But that's a wrong assumption—when you apply the limit definition correctly to a discontinuous function, the numerator doesn't approach 0, so you never hit that indeterminate case.

Let's use a concrete example to make this tangible. Take the step function:
[
f(x) = \begin{cases}
1 & \text{if } x \geq 0 \
0 & \text{if } x < 0
\end{cases}
]
Trying to compute the derivative at ( x = 0 ):

• Right-hand limit: ( \lim_{h \to 0^+} \frac{f(0+h) - f(0)}{h} = \lim_{h \to 0^+} \frac{1 - 1}{h} = 0 )
• Left-hand limit: ( \lim_{h \to 0^-} \frac{f(0+h) - f(0)}{h} = \lim_{h \to 0^-} \frac{0 - 1}{h} = +\infty )

The left and right limits don't match, so the derivative doesn't exist at ( x=0 ). Notice we didn't get an indeterminate form here—one side gives a finite value, the other blows up to infinity. Another example: ( f(x) = \frac{1}{x} ) at ( x=0 ). The function isn't defined at 0, so we can't even form the numerator of the derivative limit, let alone compute it.

Another intuitive angle: the derivative measures the slope of the tangent line at a point. If the function has a jump or hole at ( a ), there's no meaningful tangent line there—either the slope would have to be infinitely steep (for a jump) or the point doesn't exist to anchor a tangent. The limit definition captures this by returning an infinite limit, non-matching left/right limits, or an undefined expression—not an indeterminate form.

To sum up: Applying the derivative limit definition to a discontinuous point won't result in a permanent indeterminate form. Instead, you'll either get an infinite limit, a non-existent limit (due to mismatched left/right sides), or an undefined expression entirely. The "incorrect determinate" value you're imagining comes from incorrectly assuming continuity, but that's not what happens when you apply the definition properly.

备注：内容来源于stack exchange，提问作者Princess Mia