# What happens when one uses the limit definition of derivative on non continuous point?

When one uses the limit definition of a derivative on a non continuous point, will one just permanently get an indeterminate form? Intuitively, it seems that the derivative will just be 'incorrect' but determinate as it will assume the function is continuous and therefore have a wrong limit. I am aware of the proof that a function that is differentiable at a point is continuous at that point, however I want to understand intuitively what goes on when one takes derivative using limit definition at a non-continuous point. thank you

## 1 Answer

Let $$f \colon D \to \mathbb{R}$$ be some real function and $$x \in D$$ such that $$f$$ is differentiable at $$x$$. This means that $$\lim_{h \to 0} \frac{f(x+h)-f(x)}{h}$$ exists. Since the limit $$\lim_{h \to 0} h$$ exists the limit laws tell us that $$\left(\lim_{h \to 0} \frac{f(x+h)-f(x)}{h}\right)\left(\lim_{h \to 0} h \right) = 0$$ is equal to the limit of the products, so $$\lim_{h \to 0} f(x+h)-f(x) = 0.$$ But this is precisely what it means for $$f$$ to be continuous at $$x$$.

You're now interested in the contrapositive: if $$f$$ isn't continuous at $$x$$ then the limit defining the derivative can't exist. So yes, if you try to find that limit you'll see that it diverges in some way. To see both cases consider the functions $$f(x) = \cases{1 & x > 0 \\ -1 & \text{otherwise}}$$ and $$f(x) = \cases{-1 & x > 0 \\ 1 & \text{otherwise}}$$ and you'll find that their left-hand "derivative limits" at $$0$$ diverge to plus and minus infinity.

• To be precise, it's only the limits as $h \to 0^-$ that are plus or minus infinity. The limits as $h \to 0^+$ are zero. Commented Jan 19, 2023 at 21:06
• @HansLundmark Thanks, I edited my answer to include this detail. Commented Jan 20, 2023 at 12:19