- #1
PcumP_Ravenclaw
- 106
- 4
Hello mates,
is the function ## f(x) = \frac{(2^x - 1)}{x} ## differentiable at x = 0? For it to be differentiable it has to be continuous? From the graph f(0) is undefined although limit exists. I have read that at points like a corner, gap and vertical tangents it is not differentiable. So why is it differentiable at f(0)?
Sehr Danke!
is the function ## f(x) = \frac{(2^x - 1)}{x} ## differentiable at x = 0? For it to be differentiable it has to be continuous? From the graph f(0) is undefined although limit exists. I have read that at points like a corner, gap and vertical tangents it is not differentiable. So why is it differentiable at f(0)?
Sehr Danke!