isnt another way of saying f is differentiable at if this limit exists?

then . The right side is going to 0. Since was an arbitrary point.

It means the function has 0 derivative for all points. Since is path connected, for any , there exists a parametrization g(t), such that and

= 0, which implies thus f is constant

We need to show this path connectivity part because f might be a different constant value at different parts in the domain (f=3 in one neighborhood, f=5 in another so f'(x) = 0 for both but they are not the same right?)