# With derivatives, why doesn't '$h$' in the denominator not invalidate the whole function as diving by zero?

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.

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