The '$h$' in the $\frac{df}{dx}$ formula, i.e., $$f'(x) = \lim_{h\to0}\frac{f(x + h) - f(x)}{h}$$ stands for an infinitesimal change, correct? Why isn't this '$h$' equal $0$? If $h\neq 1$ either, why can it be overlooked as inconsequential in the denominator?

I'm self-learning, stop me if I'm mistaken or if the question has already been asked.

  • 3
    $\begingroup$ The term "limit" really matters. $\endgroup$ – Randall Mar 15 at 16:51
  • $\begingroup$ You can't divide by zero, so we have to think about limits instead. $\endgroup$ – Jair Taylor Mar 15 at 16:53
  • $\begingroup$ In the given context, $h$ represents a non-zero number such that $x+h$ is in the domain of $f$. $\endgroup$ – Michael Hoppe Mar 16 at 11:23
  • $\begingroup$ Oh wow, thank you Michael! That's really helpful! $\endgroup$ – user133876 Mar 16 at 18:26

You’re correct that we can’t simply divide by $h$ if $h$ is zero. But recall that $\frac 0 0$ is an indeterminate form; it does not necessarily diverge to infinity, and in fact when a limit takes on this form it can take on any real value. This is the form of the definition of the derivative: $\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}$ becomes $\frac 0 0$ upon plugging in $0$ for $h$, so we cannot simply plug in and evaluate the limit like this. But as we approach $h=0$ via a limit, both the numerator and denominator will shrink towards zero, while the ratio as a whole will approach some other value. This is how we “get around” the issue of the indeterminate form and division by zero.


Even though dividing by $0$ introduces challenges, it doesn’t necessarily completely “invalidate” what you are doing. Take $1/0$ this can’t be analyzed by itself, but it is still valuable. To “approach” (pun intended) this issue, we use a limit, which upon closer inspection reveals that the limit grows without bound, a much more informative term than DNE, or invalid.

In the case of a derivative, many people describe it as a “paradox” because of exactly your question above, however I think this is wrong, it not a paradox at all, it’s an indeterminate limit as described in the previous answer which demands some sort of “closer look.”

  • $\begingroup$ This answer is enlightening as well, thank you! $\endgroup$ – user133876 Mar 16 at 1:14

Your Answer

