# A dumb question on continuity and differentiability of function

Consider $$f(x)=\begin{cases} \frac {x}{e^x-1} & \text{if x \ne 0}\\ c & \text{if x = 0} \end{cases}$$
I know that if $$f(x)$$ is continuous, then $$c = \lim_{x\to 0} \frac {x}{e^x-1} = 1$$ because $$\frac {x}{e^x-1}$$ is not continuous at $$x=0$$. Now I want to find the derivative $$f'(0)$$. Do I just differentiate $$c$$ to get $$f'(0)=0$$ or differentiate $$\frac {x}{e^x-1}$$ then substitute $$x=0$$ to get "undefined" as the answer?

The derivative $$f'(0)$$ is defined by

$$\lim_{h\to 0} \frac{f(0+h)- f(0)}{h}$$

so you are evaluating $$f$$ at $$x$$-values other than $$c$$. So the question comes down to whether the values of $$\frac{f(0+h)-f(0)}{h}$$ converge to a value (from both sides) around $$0$$.

Differentiating $$c$$ is therefore a non-starter (as you've realised); differentiating $$\frac{x}{e^x -1}$$ is on the right track but you shouldn't just be thinking in terms of substituting $$0$$ into that derivative.

• Thanks! That means f(x) must be differentiated using limit but not "normal formula" like the quotient rule! Sep 29, 2020 at 4:51
• @Gerald You can use the normal formulae to get an expression for the derivative, but should then think about limits rather than just substituting $x=0$ into the result. Sep 29, 2020 at 5:01

Note that $$f'(0)=\lim_{h\to0}\frac{f(h)-f(0)}{h}=\lim_{h\to0}\frac{\frac{h}{e^h-1}-c}{h}=\lim_{h\to0}\frac{e^h-1-he^h}{(e^h-1)^2}=\lim_{h\to0}\frac{he^h}{2(e^h-1)e^h}=\frac{1}{2}$$

(Using L'Hopital's rule)