Is differentiability defined at an isolated point of the function's domain?

Say we have the function $$f:(0,1)\cup\{2\}\rightarrow \mathbb R$$ defined by $$f(x)=5$$.

Is $$f$$ differentiable or not differentiable (or neither) at $$2$$?

Is $$f$$ a differentiable function?

• Differentiabilty at an isolated point is not defined. – Kavi Rama Murthy Feb 3 at 7:45
• No. One can only take a limit over such $h\to0$ that $f(a+h)$ is defined. Since for an isolated point all such small enough $h$ must be zero the difference quotients $\frac{f(a+h)-f(a)}{h}$ are not defined for small $h$, and neither is the limit. – Conifold Feb 3 at 8:23

Differentiability is defined in terms of sequences of differences $$h_1,h_2,h_3\dots$$ that are all nonzero and yet converge to zero. Such a sequence does not exist in the given domain, as there does not exist a number $$k$$ in the domain where $$|2-k|=0.2$$ or $$0.1$$ or indeed any number smaller than $$1$$. Hence differentiability is not defined at isolated points.
• Is $f$ a differentiable function? – iqntt1s Feb 3 at 9:24
• @iqntt1s Over $(0,1)$, yes. At $\{2\}$, no. – Parcly Taxel Feb 3 at 9:24